The correct answer is B. The atomic radius increases from left to right across the periodic table.
There are a few factors that contribute to this trend. As you move from left to right across a period (horizontal row), the atomic number increases, meaning there are more protons in the nucleus. The increase in positive charge in the nucleus exerts a stronger pull on the electrons present in the same energy level, causing the electron cloud to be pulled closer to the nucleus.
Additionally, as you move from left to right across a period, the number of energy levels or shells remains mostly constant, while the number of protons and electrons increases. The increase in electron-electron repulsion forces the electrons to spread out over the same energy levels, thus increasing the size of the electron cloud and atomic radius.
However, it is important to note that when moving down a group (vertical column), the atomic radius generally increases. This is because each subsequent row adds a new energy level or shell, increasing the average distance between the nucleus and the outermost electrons. The additional energy levels contribute to shielding the outer electrons from the pull of the nucleus, leading to a larger atomic radius.
So, the atomic radius trend in the periodic table is that it generally increases from left to right across a period and increases from top to bottom within a group.
what was the general trend from 1985 to 2014 in the size of the area with ph levels less than or equal to 4.5 (below which the environment is considered to be acidic)?
From 1985 to 2014, the general trend in the size of the area with pH levels less than or equal to 4.5 (indicating acidic environment) showed an increase. This trend suggests that over this time period, the acidic areas expanded or became more prevalent.
Several factors contribute to this trend. One major factor is human activities, particularly the burning of fossil fuels and industrial emissions. These activities release pollutants such as sulfur dioxide and nitrogen oxides into the atmosphere, which can then react with water and form acids, leading to acid rain. Acid rain can contribute to the acidification of soil and water bodies, leading to a decrease in pH levels.
Another contributing factor is deforestation. When forests are cleared, the soil is exposed to rainfall, which can lead to increased leaching of acidic compounds into the soil and water systems. Additionally, without the buffering capacity of trees, the pH of the soil can become more susceptible to fluctuations.
Climate change can also impact pH levels. Rising temperatures can affect the carbon dioxide levels in the atmosphere, leading to increased absorption of CO2 by the oceans. This can result in ocean acidification, where the pH of seawater decreases, negatively impacting marine life.
It is important to note that the trend may vary in different regions due to specific local factors and mitigation efforts. However, overall, the general trend from 1985 to 2014 indicates an increase in the size of areas with pH levels less than or equal to 4.5, signaling a growing concern for acidic environments.
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Which conversion factor should be used to convert an area from in^2 to cm^2 if 1 in = 2.54 cm?
The conversion factor that should be used to convert an area from square inches ([tex]\text in^2[/tex]) to square centimeters ([tex]\text cm^2[/tex]) is [tex]\rm (2.54 cm)^2[/tex] / [tex]\rm (1 in)^2[/tex] .
A conversion factor is a numerical ratio that expresses the relationship between two different units of measurement. It is used to convert a quantity from one unit to another unit.
In this case, conversion factor between inches (in) and centimeters (cm) is 1 in = 2.54 cm, so the conversion factor between square inches and square centimeters is:
[tex]\rm (1 in)^2 = (2.54 cm)^2[/tex]
Squaring both sides of the equation gives:
[tex]\rm 1 in^2 = (2.54 cm)^2[/tex]
Therefore, to convert an area from square inches to square centimeters, we can multiply the area in square inches by the conversion factor:
area in [tex]\rm cm^2[/tex] = area in [tex]\rm in^2 \times (2.54 cm)^2 / (1 in)^2[/tex]
Simplifying the expression gives:
area in [tex]\rm cm^2[/tex] = area in[tex]\rm in^2 \times 6.4516[/tex]
So, to convert an area from square inches to square centimeters, we should multiply the area in square inches by 6.4516.
Therefore, The conversion factor between square inches and square centimeters is[tex]\rm (2.54 cm)^2[/tex] / [tex]\rm(1 in)^2[/tex], and to convert an area from square inches to square centimeters, we should multiply the area in square inches by 6.4516.
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Calculate the ph of a solution prepared by dissolving of acetic acid and of sodium acetate in water sufficient to yield of solution. the ka of acetic acid is
The PH of the given prepared solution is: pH = 5.056
How to find the PH of the solution?Acetic acid (CH₃COOH), is in equilibrium with water and as a result, the chemical equation is expressed as:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺
From here, we can get the H-H equation to find pH of a solution:
pH = pka + log₁₀ [CH₃COO⁻] / [CH₃COOH]
Where:
Molarity of acetic acid, [CH₃COOH] = 0.150mol / 1.00L = 0.150M
Molarity of acetate ion, [CH₃COO⁻] = 0.300mol / 1.00L = 0.300M.
pKa of the buffer = -log Ka: 4.754
Plugging in the relevant values gives us:
pH = 4.754 + log₁₀ [0.300M] / [0.150M]
pH = 5.056
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Complete question is:
Calculate the pH of a solution prepared by dissolving 0.150 mol of acetic acid and 0.300 mol of sodium acetate in water sufficient to yield 1.00 L of solution. The Ka of acetic acid is 1.76 ⋅ 10-5. Calculate the pH of a solution prepared by dissolving 0.150 mol of acetic acid and 0.300 mol of sodium acetate in water sufficient to yield 1.00 L of solution. The Ka of acetic acid is 1.76 10-5. 3.892 10.158 5.056 2.516 4.502
assuming that the cu(nh3)4 2 solution has one strong absorption in the visible region, explain why looking for the maximum absorption (analytical wavelength) in the range of 550-650 nm was more reasonable than looking for the maximum absorption in the range of 350-450 nm?
The choice of looking for the maximum absorption in the range of 550-650 nm, rather than the range of 350-450 nm, is based on the understanding of the electronic transitions and the absorption properties of the Cu(NH₃)₄²⁺ complex.
In the visible region, electronic transitions involving the d-orbitals of transition metals commonly occur. Cu²⁺ ions have partially filled d-orbitals, which can undergo electronic transitions when exposed to light. These transitions result in the absorption of specific wavelengths of light and give rise to visible color.
Typically, transition metal complexes exhibit intense absorption bands in the visible region due to d-d electronic transitions. The exact wavelengths of maximum absorption are influenced by the ligand environment around the central metal ion and the nature of the metal ion itself.
In the case of the Cu(NH₃)₄²⁺ complex, the presence of four ammonia (NH₃) ligands significantly affects the electronic structure and energy levels. This ligand field splitting leads to transitions with characteristic energies in the visible region. The strong absorption observed in the visible range implies that the complex absorbs light of longer wavelengths (lower energies).
Therefore, searching for the maximum absorption (analytical wavelength) in the range of 550-650 nm is more reasonable as it corresponds to the visible region where the complex is expected to exhibit strong absorption. On the other hand, searching in the range of 350-450 nm would likely be less productive for this particular complex since it falls outside the expected absorption range based on the ligand field and the nature of Cu²⁺ ions.
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complete the mechanism for the reaction between acetophenone in acid and chlorine by adding the missing bonds, atoms, charges, nonbonding electrons, and curved arrows (forward reaction only). the mechanism for steps 1 to 3 are pre‑drawn for you.
The mechanisms include:
Chlorine adds to the carbonyl carbon of acetophenone, forming a chlorinated intermediate.The chlorinated intermediate loses a proton, forming a carbocation.A chloride ion attacks the carbocation, forming chloroacetophenone.The water molecule abstracts a proton from chloroacetophenone, forming the final product, 2-chloro-1-phenylethanol.How to create the mechanism?Step 1: The chlorine atom in Cl2 is electrophilic, meaning that it has a partial positive charge. The carbonyl carbon of acetophenone is nucleophilic, meaning that it has a partial negative charge. The chlorine atom and the carbonyl carbon form a new covalent bond, and a chloride ion is formed.
Step 2: The chlorinated intermediate is unstable because it has a positive charge on a carbon atom. The carbon atom loses a proton, forming a carbocation.
Step 3: A chloride ion attacks the carbocation, forming chloroacetophenone. The chloride ion is a good nucleophile because it has a lone pair of electrons. The lone pair of electrons on the chloride ion attacks the carbocation, forming a new covalent bond.
Step 4: The water molecule abstracts a proton from chloroacetophenone, forming the final product, 2-chloro-1-phenylethanol. The water molecule is a good base because it has a lone pair of electrons. The lone pair of electrons on the water molecule attacks the proton on chloroacetophenone, forming a new covalent bond and a hydroxide ion. The hydroxide ion then leaves, taking the proton with it.
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what did bohr’s model of the atom include that rutherford’s model did not have? a nucleus energy levels electron clouds smaller particles
Bohr's model of the atom included the energy levels that Rutherford's model of the atom did not have.
The Bohr's model is a model of atomic structure proposed by Niels Bohr in 1913.
This model has since been replaced by the quantum mechanical model, but it still helps to clarify certain concepts and predict certain outcomes.
It depicted the electrons in the atom as being situated in specific layers, or energy levels, surrounding the nucleus.
The Rutherford's model of the atom did not include the concept of energy levels, but Bohr's model did.
According to this model, electrons traveled around the nucleus in specific, discrete energy levels.
Electrons could also move from one energy level to another by either absorbing or releasing energy.
As a result, Bohr's model is frequently referred to as the planetary model of the atom.
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Version #3 heward university orgnic chemistry ii final exam 2017 summer session ii name q1. Give iupac names for the following compounds q2. Draw structures corresponding to cach of the given names a. 3-methyl-2-buten-1-ol c 2,3 c. 3,5 acid.
The IUPAC names for the following compounds:a. 3-methyl-2-buten-1-ol: is 3-methyl-2-buten-1-ol. b. 2,3-dichlorobutane: is 2,3-dichlorobutane. c. 3,5-dimethylbenzoic acid: is 3,5-dimethylbenzoic acid.
1)The IUPAC names for the following compounds:
a. 3-methyl-2-buten-1-ol: The IUPAC name of this compound is 3-methyl-2-buten-1-ol.
b. 2,3-dichlorobutane: The IUPAC name of this compound is 2,3-dichlorobutane.
c. 3,5-dimethylbenzoic acid: The IUPAC name of this compound is 3,5-dimethylbenzoic acid.
2)Structures corresponding to the given names:
a. 3-methyl-2-buten-1-ol:
CH3
|
CH3-C=C-CH2-CH2-OH
|
H
b. 2,3-dichlorobutane:
Cl Cl
| |
CH3-CH-CH2-CH2-CH3
|
H
c. 3,5-dimethylbenzoic acid:
CH3
|
CH3-C-COOH
|
H
|
CH3
The above structures represent the compounds described by the given IUPAC names.
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Finish those equations (the most struggling part of my HW) some cannot react
45. Mg(OH)2 + NaCl → …………… + ……………
46. H2SO4 + ……………. → BaSO4↓ + HCl
47. AgNO3 + Fe(NO3)2 → …………… + ……………
48. MgCl2 + NaOH → …………… + ……………
49. NaOH + Fe2(SO4)3 → …………… + ……………
50. Pb(NO3)2 + …………… → PbCl2↓ + 2KNO3
51. MgSO4 + BaCl2 → …………… + ……………
52. Cu(NO3)2 + KOH → …………… + ……………
53. Al(OH)3 + NaOH → …………… + ……………
54. BaCO3 + H2SO4 → …………… + ……………
55. ZnCl2 + …………… → AgCl + ZnSO4
56. BaSO4 + KOH → …………… + ……………
57. Fe(NO3)3 + KOH → …………… + ……………
58. NaHCO3 + KHCO3 → ……………
59. ¬Ba(OH)2 + Na2CO3 → …………… + ……………
60. Al + H2SO4 → …………… + H2
The following chemical equations have been balanced using the appropriate coefficients.
Mg(OH)2 + NaCl → MgCl2 + 2NaOH
H2SO4 + BaCl2 → BaSO4↓ + 2HCl
AgNO3 + Fe(NO3)2 → AgFeO2↓ + 2NO3
How to balance chemical equationsBalancing chemical equations in chemistry involves adjusting the coefficients of the reactants and products in a chemical reaction so that the number of atoms of each element is equal on both sides of the equation.
Note: Never change the subscripts in a chemical formula to balance an equation. Only coefficients can be changed.
Mg(OH)2 + NaCl → MgCl2 + 2NaOH
H2SO4 + BaCl2 → BaSO4↓ + 2HCl
AgNO3 + Fe(NO3)2 → AgFeO2↓ + 2NO3
MgCl2 + 2NaOH → Mg(OH)2↓ + 2NaCl
3NaOH + Fe2(SO4)3 → Fe(OH)3↓ + 3Na2SO4
Pb(NO3)2 + 2KCl → PbCl2↓ + 2KNO3
MgSO4 + BaCl2 → MgCl2 + BaSO4↓
Cu(NO3)2 + 2KOH → Cu(OH)2↓ + 2KNO3
Al(OH)3 + NaOH → NaAlO2 + 2H2O
BaCO3 + H2SO4 → BaSO4↓ + CO2↑ + H2O
ZnCl2 + 2AgNO3 → 2AgCl↓ + Zn(NO3)2
BaSO4 + 2KOH → Ba(OH)2↓ + K2SO4
Fe(NO3)3 + 3KOH → Fe(OH)3↓ + 3KNO3
NaHCO3 + KHCO3 → NaKHCO3 + H2O + CO2↑
Ba(OH)2 + Na2CO3 → BaCO3↓ + 2NaOH
Al + H2SO4 → Al2(SO4)3 + H2↑
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For Mg, because of the number of valence electrons you answered earlier, which scenario is most likely to occur in Nature? Mg will gain 6 electrons and become negatively charged Mg will gain 6 electrons and become positively charged Mg will lose 2 electrons and become negatively charged Mg will lose 2 electrons and become positively charged
Answer:
Mg will lose 2 electrons and become positively charged.
Explanation:
Using its electron configuration, it will want to use the closest Noble Gas to "replace" part of its electron config. Since Neon is #10 and Mg is #12, Mg will be losing 2 valence electrons, becoming positively charged as it is going down.
Uranium-235 decays by alpha emission. what isotope is also produced by this transformation?
When uranium-235 (U-235) undergoes alpha decay, it emits an alpha particle, which consists of two protons and two neutrons. As a result, the uranium-235 nucleus loses two protons and two neutrons. The correct answer is thorium-231.
When uranium-235 undergoes alpha decay, it releases an alpha particle, which is essentially a helium-4 nucleus consisting of two protons and two neutrons. This emission causes the uranium-235 nucleus to lose two protons and two neutrons.
The resulting isotope produced by this transformation is thorium-231 (Th-231). Thorium-231 has an atomic number of 90 and an atomic mass of 231, as it contains 90 protons and 141 neutrons.
So, when uranium-235 undergoes alpha emission, it transforms into thorium-231 as a byproduct of the decay process.
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chegg a solution is 0.0111 m in both br– and so42–. a 0.217 m solution of lead(ii) nitrate is slowly added to it with a buret.
The concentration of the first anion (Br⁻) when the second one (SO₄²⁻) starts to precipitate at 25°C is approximately 0.059 M.
How to determine concentration?To determine the concentration of the first anion when the second one starts to precipitate, compare the solubility product constants (Ksp) of the two compounds involved.
Given the Ksp values for
PbBr₂ (6.60 × 10⁻⁶) and
PbSO₄ (2.53 × 10⁻⁸),
For PbBr₂:
IP = [Pb²⁺][Br⁻]²
Ksp = [Pb²⁺][Br⁻]²
For PbSO₄:
IP = [Pb²⁺][SO₄²⁻]
Ksp = [Pb²⁺][SO₄²⁻]
Assume that the concentration of the first anion (Br⁻) at this point is x M.
For PbBr₂:
(x)(2x²) = 6.60 × 10⁻⁶
2x³ = 6.60 × 10⁻⁶
x³ = 3.30 × 10⁻⁶
x ≈ 0.059 M
Therefore, the concentration of the first anion (Br⁻) when the second one (SO₄²⁻) starts to precipitate at 25°C is approximately 0.059 M.
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Complete question:
A solution is 0.0151 Min both Br and SO42-. A 0.191 Msolution of lead(II) nitrate is slowly added to it with a buret. The anion will precipitate from solution first. (Kip for PbBr2-6.60 x 10; Kp for PbSO4 -2.53 × 10-8) Part 2 (1 point) What is the concentration in the solution of the first anion when the second one starts to precipitate at 25°C? M
which reagent (reactant) is limiting when 1.85 mol of sodium hydroxide and 1.00 mol carbon dioxide are allowed to react? how many moles of sodium carbonate can be produced? how many moles of the excess reactant remain after the completion of the reaction?
In the given chemical reaction [tex]\rm NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O[/tex], the limiting reagent is carbon dioxide and [tex]\rm Na_2CO_3[/tex] that can be produced is 1.00 mol, and 0.85 mol of NaOH remains unreacted after the completion of the reaction.
The limiting reagent is the reactant that gets completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed.
The balanced chemical equation for the reaction between sodium hydroxide and carbon dioxide is:
[tex]\rm NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O[/tex]
To determine which reactant is limiting, we need to calculate the number of moles of sodium hydroxide and carbon dioxide present and compare the mole ratios of the reactants in the balanced equation.
Moles of NaOH = 1.85 mol
Moles of [tex]\rm CO_2[/tex] = 1.00 mol
The balanced equation shows that the mole ratio of NaOH to [tex]\rm CO_2[/tex] is 2:1, which means that 2 moles of NaOH react with 1 mole of [tex]\rm CO_2[/tex] .
Using the mole ratio, we can calculate the maximum number of moles of Na2CO3 that can be produced by each reactant:
Moles of NaOH / 2 = 0.925 mol [tex]\rm Na_2CO_3[/tex]
Moles of[tex]\rm CO_2 \times 1[/tex] /1 = 1.00 mol [tex]\rm Na_2CO_3[/tex]
Since the calculated number of moles of [tex]\rm Na_2CO_3[/tex] from [tex]\rm CO_2[/tex] is less than the calculated number of moles of [tex]\rm Na_2CO_3[/tex] from NaOH, [tex]\rm CO_2[/tex] is the limiting reactant.
According to the balanced equation, 1 mole of [tex]\rm CO_2[/tex] reacts to form 1 mole of [tex]\rm Na_2CO_3[/tex] . Therefore, the number of moles of [tex]\rm Na_2CO_3[/tex] produced is 1.00 mol.
To determine the amount of excess reactant remaining after the reaction, we need to calculate the amount of NaOH that did not react.
Moles of NaOH remaining = Moles of NaOH - (Moles of [tex]\rm Na_2CO_3[/tex] produced x 2)
= 0.85 mol NaOH
Therefore, 0.85 mol of NaOH remains unreacted after the completion of the reaction.
In conclusion, carbon dioxide is the limiting reactant, and 1.00 mol of [tex]\rm Na_2CO_3[/tex] can be produced. 0.85 mol of NaOH remains unreacted after the completion of the reaction.
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The question is incomplete. The complete question is:
Sodium hydroxide reacts with carbon dioxide as follow:
[tex]\rm NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O[/tex]
Which reagent (reactant) is limiting when 1.85 mol of sodium hydroxide and 1.00 mol carbon dioxide are allowed to react? How many moles of sodium carbonate can be produced? How many moles of the excess reactant remain after the completion of the reaction?
Did you use the same apparatus for all your measurements? if not, how could it affect your data analysis outcome?
No, using different apparatus for measurements can affect the data analysis outcome.
Using different apparatus for measurements can introduce variability in the data. Each apparatus may have slight variations in calibration, sensitivity, or precision, leading to inconsistent measurements. This can result in inaccurate or unreliable data. In data analysis, consistency and precision are crucial for making reliable conclusions. If different apparatus were used, it is important to account for these variations during data analysis.
This can be done by conducting appropriate calibration or standardization procedures to ensure the measurements are comparable. Additionally, statistical methods like averaging or calculating standard deviations can help mitigate the impact of variability introduced by different apparatus. Overall, using the same apparatus for all measurements ensures consistency and reduces the potential for systematic errors in data analysis.
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A certain liquid has a vapor pressure of 92.0 Torr at 23.0 ∘
C and 299.0 Torr at 45.0 ∘
C. Calculate the value of ΔH vap
for this liquid. ΔH vap
= Calculate the normal boiling point of this liquid. boiling point:
The boiling point of the given liquid is 89.3°C.
Given data: Vapor pressure at 23°C = 92.0 Torr
Vapor pressure at 45°C = 299.0 Torr
We have to find the value of ΔHvap for this liquid.
ΔHvap= -R*(T2-T1) /ln(P2/P1)
Where, R is the gas constant
T1 = 23°C = 23+273 = 296K
T2 = 45°C = 45+273 = 318K
P1 = 92.0 Torr
P2 = 299.0 Torr
Plugging these values into the above formula,
ΔHvap= -8.314*(318-296) /ln(299/92)
= 29.44 kJ/mol (approx)
The normal boiling point of the liquid can be calculated as follows:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Rearranging the above equation we get,
T2 = ΔHvap / R * (ln(P2/P1)) + 1/T1
T2 = 29.44 kJ/mol / (8.314 J/mol K) * ln(299/92) + 1/296
T2 = 362.3 K or 89.3°C
Therefore, the boiling point of the given liquid is 89.3°C.
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1- List the desired properties of precipitate formed in gravimetric methods of analysis. 2- The addition of dimethylglyoxime, H 2
C 4
H 6
O 2
N 2
to a solution of nickel (II) ion give rise to a precipitate. Ni 2+
+2H 2
C 4
H 6
O 2
N 2
>Ni(H 2
C 4
H 6
O 2
N 2
) 2
+2H +
If 0.15 g nickel alloy is treated with dimethylglyoxime and 175mg nickel dimethylglyoxime is collected. Determine the mass and the percent of nickel in the alloy.
The mass of Ni in the alloy is 0.0323 g, and the percent of nickel in the alloy is 21.53%.
1. The desired properties of precipitates that are formed in gravimetric methods of analysis include:
i. Be completely pure and easily filterable
ii. The precipitate should be of known composition
iii. Large and easily recognizable crystalline structure
iv. The precipitate should be free from impurities
v. The precipitate should be completely insoluble in the mother liquor.
2. Given: Amount of Ni alloy = 0.15 gAmount of Ni-dimethylglyoxime obtained = 175 mgTo determine: The mass and percent of nickel in the alloySolution: The mass of Ni present in Ni-dimethylglyoxime can be calculated by using the molecular weight of Ni-dimethylglyoxime.To calculate the molecular weight of Ni-dimethylglyoxime, multiply the atomic weight of each element by its subscript and then sum the values.
Molecular weight of Ni(H2C4H6O2N2)2 = (58.69 + 2(1.01 × 6) + 2(12.01 × 4) + 2(16 × 2) + 2(14.01 × 2)) g/mol = 318.12 g/mol
Mass of Ni in 175 mg of Ni-dimethylglyoxime = (58.69 g Ni / 318.12 g Ni(H2C4H6O2N2)2) × 0.175 g = 0.0323 g
Percent of Ni in the alloy = (Mass of Ni / Mass of alloy) × 100% = (0.0323 g / 0.15 g) × 100% = 21.53% (rounded off to two decimal places)Therefore, the mass of Ni in the alloy is 0.0323 g, and the percent of nickel in the alloy is 21.53%.
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magnesium metal is produced by the electrolysis of molten magnesium chloride using inert electrodes a. sketch the cell. labe the anode and cathode, indicate the sign of the electrodes, and show the direction of electron and ion flow.
The electrolysis of molten magnesium chloride produces magnesium metal at the cathode and chlorine gas at the anode.
The electrolysis of molten magnesium chloride is a process that uses electricity to split the molten salt into its component elements. The molten salt is placed in an electrolysis cell, which has two electrodes: a cathode and an anode. The cathode is connected to the negative terminal of the power supply, and the anode is connected to the positive terminal of the power supply.
When the power supply is turned on, electrons flow from the negative terminal of the power supply to the cathode, and protons flow from the positive terminal of the power supply to the anode. The electrons at the cathode react with the chloride ions in the molten salt to form chlorine gas. The protons at the anode react with the magnesium ions in the molten salt to form magnesium metal.
The overall reaction is:
MgCl₂ → Mg + Cl₂
The magnesium metal is produced at the cathode and rises to the top of the cell, where it can be collected. The chlorine gas is produced at the anode and bubbles out of the cell.
The electrolysis of molten magnesium chloride is a very efficient way to produce magnesium metal. The process is also relatively inexpensive, and it does not produce any harmful pollutants.
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2. (2 pts) Make a series of five 1/17 serial dilutions each between 100 and 1000 mL with one (not the final product) being exactly 444.44 mL.
Therefore, the series of five 1/17 serial dilutions each between 100 and 1000 mL with one (not the final product) being exactly 444.44 mL is as follows: 1000 mL, 58.82 mL, 3.46 mL, 0.20 mL, and 0.01 mL.
A series of five 1/17 serial dilutions each between 100 and 1000 mL with one (not the final product) being exactly 444.44 mL can be made as follows:
To make a series of five 1/17 serial dilutions each between 100 and 1000 mL with one (not the final product) being exactly 444.44 mL, follow the steps below:
Step 1:
Start with 1000 mL of the solution
Step 2: Take 1/17 of the solution and mix it with 16/17 of water, giving you a 58.82 mL solution.
This is a 1:17 dilution.
Step 3: Repeat Step 2 using 58.82 mL of the previous dilution and 941.18 mL of water to get a 1/17 dilution of 1000/17 or 58.82 mL.
Step 4: Repeat the previous step, using 58.82 mL of the previous dilution and 941.18 mL of water to get a 1/17 dilution of 1000/17² or 3.46 mL.
Step 5: Repeat Step 4, using 3.46 mL of the previous dilution and 96.54 mL of water to get a 1/17 dilution of 1000/17³ or 0.20 mL.
Step 6: Finally, repeat Step 5, using 0.20 mL of the previous dilution and 99.80 mL of water to get a 1/17 dilution of 1000/17⁴ or 0.01 mL.
Note: The volumes can be scaled to any range as long as the same 1/17 ratio is used between them.
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Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of ·0.0095Ms−1
2NH3 (g) --> N2 (g) + 3H2 (g)
Suppose a 250.mL flask is charged under these conditions with 200. mmol of ammonia. After how much time is there only 100. mmol left? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
2. The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half-life of 51 minutes. Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 0.75/μgmL. What will the concentration be 204 minutes later? Round your answer to 2 significant digits.
The concentration of the drug in the bloodstream 204 minutes after injection will be 0.227 μgmL.
Reaction order: Zero-order
The zero-order reaction can be defined as a reaction in which the rate of the reaction is independent of the concentration of the reactant(s).Rate of the reaction= k [NH3]0= k
If the initial concentration of NH3 is [NH3]0 and the concentration at time t is [NH3], then the rate of the reaction is given by:rate= -Δ[NH3]/Δt= k [NH3]0
Since the rate is zero order in NH3, we can write:Δ[NH3]= -k Δt
Integrated rate law for zero-order reactions can be given as
[NH3]= -kt + [NH3]0
Here, [NH3]0 is the initial concentration of ammonia (0.200 M), and [NH3] is the concentration at any time t, t is the time for which we want to calculate the concentration of ammonia, and k = 0.0095 M/s.
As we have to find out the time after which the concentration of ammonia reduces to 100 mmol, we will plug this value in the above equation.
[NH3] = -kt + [NH3]0[NH3]
= -0.0095 * t + 0.2 (where [NH3] = 0.1 M and t = time taken to reduce ammonia concentration to 100 mmol)
0.1 = -0.0095t + 0.20.0095t
= 0.1 - 0.02t
= (0.1 - 0.02)/0.0095t
= 7.37 min (approximately)
Therefore, after 7.37 min, the concentration of ammonia will be 0.1 M.2. Half-life of the reaction: 51 minutes
We have to find the concentration of the drug in the bloodstream 204 minutes after injection. The half-life of the reaction is 51 minutes. We can use the integrated rate law for first-order reactions:
ln [A] = -kt + ln[A]0
Here, [A]0 is the initial concentration of the drug, k is the rate constant, and t is time. We can use the formula of half-life to calculate the rate constant:
k = 0.693/τ (where τ is half-life)
k = 0.693/51
k = 0.0136 min-1
Substitute the value of k and t into the integrated rate law for first-order reactions:
ln [A] = -kt + ln[A]0
ln [A] = -0.0136 * 204 + ln(0.75)
ln [A] = -2.7712 + ln(0.75)
[A] = e^-2.7712+ ln(0.75)
[A] = 0.227 μgmL (approximately)
Therefore, the concentration of the drug in the bloodstream 204 minutes after injection will be 0.227 μgmL.
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There are two isotopes of an unknown elesnent, x−19 and x−21. The abundsnce of x−19 is 13.898, A weghted average uses the percentages of each isotope to scale their contribution to the total mass. Each isotope's contribution is the percentage (in decimal formi rultipied by the mass of the isotope. What is the comribution ( n arnu) to the weighted average from the X-19 iscoope, which has a mass of 19.00 amu? of Pont Earned- 313 utarets hemaning What is the cercentage of the X-2t woloce? 3f3 Muenest Almaking 917histeriad 3/3 kikntintakeseded
The contribution to the weighted average for the x-21 isotope = (0.86102) × (21.00) = 18.08642amu
Therefore, the contribution of x-19 isotope to the weighted average is 2.62482amu and the percentage of the x-21 isotope is 86.102%.
Let's solve the given problem. The unknown element has two isotopes; the abundances are given below:x-19; abundance is 13.898%x-21; abundance is 100 - 13.898
= 86.102%
The contribution to the weighted average from the x-19 isotope with a mass of 19.00 amu can be calculated as follows:
Mass of x-19 isotope
= 19.00 amu Abundance of x-19 isotope
= 13.898% or 0.13898Contributions to the weighted average for x-19 isotope
= (0.13898) × (19.00)
= 2.62482amu
The percentage of the x-21 isotope can be calculated as follows:
Mass of x-21 isotope
= 21.00 amu Abundance of x-21 isotope
= 100 - 13.898
= 86.102% or 0.86102.
The contribution to the weighted average for the x-21 isotope
= (0.86102) × (21.00)
= 18.08642amu
Therefore, the contribution of x-19 isotope to the weighted average is 2.62482amu and the percentage of the x-21 isotope is 86.102%.
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9. (2 pts) Make 3 serial dilutions, each generating 400 mL of a 1 -in- 4 dilution of a 3M stock solution
Serial dilution refers to the process of diluting a sample in stages, such that the dilution factor is constant for each stage.
Here is how to make three serial dilutions, each generating 400 mL of a 1 -in- 4 dilution of a 3M stock solution.
1. Start by adding 100 ml of the 3M stock solution to a 300 ml volume of distilled water.
2. Mix well to obtain a 1:4 dilution.
3. Now, remove 100 ml of this solution and place it in a new 300 ml volume of distilled water.
4. Mix well to obtain a 1:4 dilution again.
5. Repeat the above process to make a third serial dilution. That is, remove 100 ml of the second dilution and add it to a new 300 ml volume of distilled water.
6. Mix well to obtain a 1:4 dilution again.
7. Repeat this until you have made the desired dilution.
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When performing absorption spectrometry, Logger Pro should display a plot of: Molar absorptivity vs. Wavelength. Absorbance vs. Wavelength. Absorbance vs. Concentration. Molar absorptivity vs. Concentration.
The correct plot that Logger Pro should display while performing absorption spectrometry is Absorbance vs. Wavelength. This is because Absorbance vs. Wavelength is a graph of how much light of each wavelength is absorbed by the sample.
Absorbance is the logarithm of the ratio of the incident light intensity to the transmitted light intensity, and it is proportional to the concentration of the analyte in the solution.The relationship between absorbance and concentration is given by Beer-Lambert Law. The Beer-Lambert Law, also known as the Beer-Lambert-Bouguer Law, describes the absorption of light by a material as a function of its concentration. The law states that absorbance is proportional to the path length, the concentration of the absorbing species, and the molar absorptivity, which is a property of the absorbing species and the wavelength of light used.
Logger Pro is software that is used to collect and analyze data from various experiments, including absorption spectrometry. It is useful for plotting graphs of data collected from an experiment. In summary, the plot that Logger Pro should display during absorption spectrometry is Absorbance vs. Wavelength, which helps determine the concentration of an analyte in a solution using the Beer-Lambert Law.
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So 0.5mL of the supplement was used and diluted to 50mL in a volumetric flask. How would I set that up for question 4?
Also in question 3, when you take the new concentration found in mg/L and are trying to get the molarity in mol/L. When you take the concentration and divide it by the molar mass, why are you then dividing it by 1000?
Please help solve and answer #5: The molar mass of methylcobalamin is 1344.41 g/mol. Assuming one mole of cobalt per mole of methylcobalamin, find the number of mg of vitamin B12 in a 1 mL dose of the supplement. "Note that 1 g = 1000 mg
The molarity of the diluted solution = 0.005M. The number of mg of vitamin B12 in a 1 mL dose of the supplement is 6.72 mg.
To calculate the molarity of the diluted solution, use the formula:
Initial volume × Initial molarity = Final volume × Final molarity
Where,Initial volume = 0.5mL = 0.0005L
Initial molarity = molarity of the solution before dilution = ?
Final volume = 50mL = 0.050L
Final molarity = molarity of the solution after dilution = 0.005M
Substitute the values in the formula and calculate the initial molarity.
Initial molarity = (Final volume × Final molarity) / Initial volume
= (0.050 × 0.005) / 0.0005 = 0.5 M
The molarity of the diluted solution = 0.005 M.
When you take the new concentration found in mg/L and are trying to get the molarity in mol/L.
When you take the concentration and divide it by the molar mass, you are then dividing it by 1000 because there are 1000 milligrams in a gram.
Therefore, to convert milligrams per liter (mg/L) to moles per liter (mol/L), you need to divide by the molar mass and then by 1000.
This will give you the concentration in mol/L.
Molar mass of methylcobalamin = 1344.41 g/mol
The number of mg of vitamin B12 in a 1 mL dose of the supplement
= (Concentration of vitamin B12 in the supplement × Molar mass of methylcobalamin × Volume of 1 dose in liters) / 1 g
= (5000 × 1344.41 × 0.001) / 1 g
= 6.72 mg
Therefore, the number of mg of vitamin B12 in a 1 mL dose of the supplement is 6.72 mg.
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Which is true for a calibration curve? It is directly related to Beer's Law. It is used to determine the lambda max of a chemical substance. It is synonymous with the term "absorption spectrum." It is a plot of absorbance vs. wavelength.
The calibration curve is a plot of absorbance versus concentration. It is used to determine the relationship between concentration and absorbance for a given substance. The curve is constructed by measuring the absorbance of a series of solutions with known concentrations and plotting the data.
The curve is typically linear, and the equation for the line is used to calculate the concentration of an unknown solution based on its absorbance value.Calibration curves are used in various analytical techniques, including spectrophotometry and chromatography. The curve is not synonymous with the term "absorption spectrum," which refers to the range of wavelengths absorbed by a substance. It is also not used to determine the lambda max of a chemical substance. The lambda max is the wavelength at which a substance absorbs the most light and is typically determined by measuring the absorbance of a solution at different wavelengths.
Therefore, the correct statement for the calibration curve is that it is directly related to Beer's Law, and it is a plot of absorbance vs. concentration.
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For the air as a closed system, the _____ terms can be neglected in the energy balance.
The negligible terms in the energy balance refer to the specific energy change associated with potential energy and kinetic energy.
Which terms can be neglected?For the air as a closed system, the negligible terms in the energy balance typically refer to the specific energy change associated with potential energy and kinetic energy. In many cases, these terms can be neglected or assumed to be constant if the changes in elevation and velocity of the air are small and do not significantly affect the overall energy balance.
The energy balance equation for the closed system of air usually focuses on the conservation of internal energy, which includes terms related to heat transfer and work. Neglecting the potential energy and kinetic energy terms simplifies the analysis by assuming that any changes in elevation and velocity are insignificant compared to the other energy transfers involved.
It is important to note that neglecting these terms is an assumption made for simplification purposes and may not be applicable in certain scenarios where elevation changes or air velocity have a significant impact on the energy balance.
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Calculate the atomic radius in cm for the following: a. BCC metal with a 0
=0.3236 nm. (Enter your answer to three significant figures.) r=∣cm b. FCC metal with a 0
=3.8727 A
˚
. (Enter your answer to three significant figures.)
The atomic radius for BCC is [tex]r=1.4011\times10^-^8\ cm[/tex] and the atomic radius for FCC is [tex]r=1.352\times 10^-^1^0\ cm[/tex] .
The atomic radius measures the separation between an atom's nucleus and its outermost electron shell. Typically, it is expressed in picometers (pm).
The radius of an atom depends on the element. While certain elements, like fluorine, have a small atomic radius, others, like cesium, have huge atomic radii.
For BCC metal:
[tex]r=\frac{\sqrt[a]{3} }{4} \\\\= \frac{0.3236\times 20^-7\times\sqrt{3}) }4\\r=1.4011\times10^-^8\ cm[/tex]
For FCC metal:
[tex]r=\frac{\sqrt{2a} }{4} \\= {\sqrt{2} \times3.827\times10^-^1^0}/{4}[/tex]
[tex]r=1.352\times 10^-^1^0\ cm[/tex]
Thus, the atomic radius for the metal BCC is [tex]r=1.4011\times10^-^8\ cm[/tex] and the atomic radius for FCC is [tex]r=1.352\times 10^-^1^0\ cm[/tex] .
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a chemist is identifying the elements present in a sample of seawater. what characteristic of an element’s atoms always determines the element’s identity?
The characteristic of an element's atoms that always determines the element's identity is the number of protons in the nucleus. This is also known as the atomic number of the element.
An element is a substance made up of a single type of atom.
Elements can be identified by their atomic number, which is the number of protons in the nucleus of each of their atoms.
Each element has a unique atomic number. This means that no two elements can have the same atomic number.
For example, the element oxygen has an atomic number of 8, which means that all oxygen atoms have 8 protons in their nuclei.
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The complete question is-
A chemist is identifying the elements present in a sample of seawater. What characteristic of an element's atoms always determines the element's identity? In other words, what specific property of an element's atoms uniquely distinguishes it from atoms of other elements, allowing chemists to identify and differentiate between different elements in a sample?
Gold has a density of 19.3g/cm^3. what is the mass, in kg, of a cube of gold measuring 2.0 inches on each side? use unit analysis, not the density formula.
The mass of the cube of gold measuring 2.0 inches on each side is 2.56 kg.
Given that gold has a density of 19.3 g/cm³ and the cube of gold has sides of 2.0 inches, we need to calculate the mass of the cube in kg without using the density formula. We can convert the length of each side of the cube from inches to cm by using the conversion factor 1 inch = 2.54 cm.
Therefore, each side of the cube measures:2.0 inches × 2.54 cm/inch = 5.08 cm. The volume of the cube can be calculated by raising the length of one side to the third power, i.e.,
V = (side)³ = (5.08 cm)³ = 132.5 cm³
The mass of the cube can be calculated by using the density of gold and the volume of the cube. However, we are not supposed to use the density formula, so we can use unit analysis to determine the mass in kg. We have:
19.3 g/cm³ = 19.3 g/cm³ × 1 kg/1000 g = 0.0193 kg/cm³
Multiplying this by the volume of the cube in cm³, we get:
0.0193 kg/cm³ × 132.5 cm³ = 2.56 kg
Therefore, the mass of the cube of gold measuring 2.0 inches on each side is 2.56 kg.
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The total concentration of a phosphate buffer containing NaH 2
PO 4
and Na 2
HPO 4
is 15mM and pH of the solution is 7.3. What would be the concentration of each phosphate species in the buffer. Given that pK a
=7.2
The concentration of each phosphate species in the buffer are H2PO4- = 5.34 mM and HPO42- = 9.66 mM.
The total concentration of a phosphate buffer containing NaH2PO4 and Na2HPO4 is 15 mM and pH of the solution is 7.3. We need to calculate the concentration of each phosphate species in the buffer.
Given that pKa= 7.2.The reaction equation for the phosphate buffer is as follows;
H2PO4- ⇔ HPO42- + H+
The pKa of the acid is 7.2.
Hence, pH = pKa + log([HPO42-]/[H2PO4-])
or, log([HPO42-]/[H2PO4-]) = pH - pKa
or, log([HPO42-]/[H2PO4-]) = 7.3 - 7.2
or, log([HPO42-]/[H2PO4-]) = 0.1
or, [HPO42-]/[H2PO4-] = 10^(0.1)
or, [HPO42-]/[H2PO4-] = 1.258
or, [H2PO4-] = 15/(1 + 1.258)
= 5.34 m
M [HPO42-] = 15 - 5.34
= 9.66 mM
Therefore, the concentration of each phosphate species in the buffer are as follows:
H2PO4- = 5.34 mM HPO42- = 9.66 mM
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the atomic models after dalton’s time included ideas about the atomic structure. which atomic model that shows the atomic structure is missing from this set? bohr’s model schrödinger’s model rutherford’s model thomson’s model
The atomic models after Dalton's time that included ideas about the atomic structure are Rutherford's model, Thomson's model, Bohr's model, and Schrödinger's model.
Thomson's atomic model is missing from this set.
Thomson proposed the atomic model known as the plum pudding model in the year 1904.
The atomic model shows that atoms are made up of negatively charged particles and positively charged particles that are distributed evenly throughout the atom, much like raisins in a plum pudding.
In 1911, Ernest Rutherford discovered the atomic nucleus, which is made up of positively charged particles, through his gold foil experiment.
Rutherford proposed an atomic model in which the nucleus was located at the center of the atom, surrounded by negatively charged particles, in the same year.
This model is also known as the planetary model of the atom.
In 1913, Niels Bohr proposed a new atomic model based on Rutherford's planetary model that included electron orbits.
Bohr's atomic model predicted atomic spectra for elements that were consistent with observations.
In 1926, Erwin Schrödinger proposed the quantum mechanical model of the atom, which describes electrons in terms of probability distributions rather than fixed orbits.
The electrons in this model are described as wavefunctions.
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An expanding gas does 147 j of work on its surroundings at a constant pressure of 1.04 atm. if the gas initially occupied 68.0 ml, what is the final volume of the gas? l
The final volume of the gas is 209.35 mL.
To find the final volume, we can use the formula W = P * ΔV, where W is the work done, P is the pressure, and ΔV is the change in volume.
We can rearrange the formula W = P * ΔV to solve for ΔV:
ΔV = W / P.
Substituting the given values,
ΔV = 147 J / (1.04 atm) = 141.35 mL.
To find the final volume, we add the change in volume to the initial volume:
Final volume = Initial volume + ΔV = 68.0 mL + 141.35 mL = 209.35 mL.
So, the final volume of the gas is 209.35 mL.
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