true or false the shape of a molecule is determined only by repulsions among bonding electron groups

Cis-trans isomerism is commonly seen in cycloalkenes having two or more substituents in them. It does not arise in cycloalkanes. Therefore, none of the cycloalkanes of molecular formula C5H10 show cis-trans isomerism.

Cycloalkanes are hydrocarbons containing only carbon and hydrogen atoms arranged in a closed ring. These are members of the homologous series of alkanes. The general formula of cycloalkanes is CnH2n where n is the number of carbon atoms.

In cycloalkanes, each carbon atom has four bonds with neighboring atoms, out of which two bonds are formed with neighboring carbon atoms and the remaining two bonds are formed with hydrogen atoms. Cycloalkanes can be both saturated and unsaturated.

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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.

The PH of the given prepared solution is: pH = 5.056

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

Answer:

molecular biology university of California

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 correct answer is option e. OA > MA > LA.Acidic strength is the ability of an acid to donate a proton (H+) ion. A strong acid will easily give up an H+ ion while a weak acid will not. pKa is the measure of acidity. The lower the pKa value, the stronger the acid.

Therefore, an acid with a pKa of 1 is stronger than an acid with a pKa of 5.

Given that their pKa values are LA = 3.88, OA = 1.23, and MA = 3.40.

The list showing the acids in order of decreasing acid strength is as follows:OA > LA > MA

The pKa values given suggest that Oxalic acid (OA) is the strongest acid and Malic acid (MA) is the weakest acid. OA has the lowest pKa value of 1.23, which means that it has the strongest acidic strength. LA has a pKa of 3.88, which makes it a weaker acid compared to OA.

MA has a pKa of 3.40, which is slightly less acidic than LA.

Therefore, the order of decreasing acid strength is:

OA > LA > MA.

In conclusion, the correct answer is option e. OA > MA > LA.

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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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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?

The two subatomic particles that make up the nucleus of an atom are protons and neutrons. Electrons are located outside the nucleus. Protons have a positive charge, Neutrons have no charge, and Electrons have a negative charge. Hence option A is correct.

The number of protons in an atom's nucleus is called the atomic number. The atomic number determines the elements of the atom. For example, all atoms with 6 protons in their nucleus are carbon atoms.

The number of neutrons in an atom's nucleus can vary. This is why atoms of the same element can have different masses. For example, carbon atoms can have 6, 7, or 8 neutrons. These different forms of carbon are called isotopes.

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No element in the nitrogen family (Group 15) has a lower atomic number than sodium. Sodium (Na) has an atomic number of 11, while the elements in Group 15 (nitrogen family) have atomic numbers ranging from 15 to 83.

The elements in Group 15 are nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), bismuth (Bi), and moscovium (Mc). These elements have atomic numbers 7, 15, 33, 51, 83, and 115, respectively.

As per the given condition, we are looking for an element with a lower atomic number than sodium (Z ≤ 11). None of the elements in the nitrogen family have atomic numbers lower than sodium. Therefore, there is no element that matches the description of having a lower atomic number than sodium within the nitrogen family or any element with Z ≤ 92.

Elements with atomic numbers lower than sodium can be found in Groups 1 and 2, known as the alkali metals and alkaline earth metals, respectively. For example, elements such as hydrogen (H), lithium (Li), beryllium (Be), and boron (B) have atomic numbers lower than sodium.

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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?

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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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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Assuming no other unlisted ions are present in the water, the estimated concentration of  [tex]Na^+[/tex] (in mg/L as  [tex]Na^+[/tex]) is 21.2 mg/L.

Using an anion-cation balance, we may estimate the concentration of Na+ (in mg/L as Na+) by ensuring that the total positive charge from cations matches the total negative charge from anions.

The total positive charge from cations = ([tex]Ca^{+2} + M^g{+2} + Na^+ + K^+[/tex])

The total negative charge from anions = ([tex]HCO^{3-} + SO_4^{2-} + Cl^-[/tex])

Total positive charge = (40.0 + 10.0 + Na+ + 7.0)

Total negative charge = (67.2 + 11.0)

Total positive charge = Total negative charge

(40.0 + 10.0 + [tex]Na^+[/tex] + 7.0) = (67.2 + 11.0)

Simplifying the equation:

57.0 + Na+ = 78.2

Subtracting 57.0 from both sides:

Na+ = 78.2 - 57.0

Na+ = 21.2 mg/L

Therefore, the estimated concentration of  [tex]Na^+[/tex] (in mg/L as  [tex]Na^+[/tex]) is 21.2 mg/L.

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Your question seems incomplete, the probable complete question is:

2. A sample of water has the following concentrations of ions (and pH = 7.0):

cations mg/L

anions mg/L

Ca+2

40.0

Mg+2

10.0

Na+

?

K+

7.0

HCO3 110.0

SO42-

67.2

Cl-

11.0

Assuming no other constituents are missing, use an anion-cation balance to estimate the concentration of Na* (in mg/L as Na*)? Remember that the balance cannot be in mg/L.

The electrochemistry underlying the electrical current in a neuron as it is stimulated involves sodium channels opening to allow sodium to enter the cell.
2. The most fundamental variable that can likely explain the evolution of bigger brains is the overall ecological complexity that the animal deals with.

During neuron stimulation, an action potential is generated. This process involves the depolarization of the neuron's membrane, which is achieved by the influx of positively charged ions, primarily sodium ions (Na+). When a neuron is stimulated, voltage-gated sodium channels in the cell membrane open, allowing sodium ions to rapidly enter the cell. This influx of positive charge depolarizes the membrane, creating an electrical current that propagates along the neuron.

Regarding the second question, the most fundamental variable that can likely explain the evolution of bigger brains is the overall ecological complexity that the animal deals with. Animals that inhabit complex and challenging environments often require enhanced cognitive abilities to navigate and respond to their surroundings effectively. The ecological complexity, such as varied food sources, social interactions, and environmental stimuli, can drive the evolutionary pressure for larger brain size and increased cognitive capacity. While factors like diet, sociality, and mating strategies may play a role, the overall ecological complexity is considered a crucial determinant of brain evolution.


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The molality of the solution is approximately 1.55 m. Therefore, the correct answer is A) 1.55 m.

To find the molality (m) of the solution, we need to determine the number of moles of the solute (propyl alcohol) and the mass of the solvent (water).

Given:

Mass of propyl alcohol (solute) = 35.0 g

Mass of water (solvent) = 200 g

Step 1: Convert the mass of propyl alcohol to moles.

First, we need to calculate the molar mass of propyl alcohol (C3H7OH):

C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol

Molar mass of propyl alcohol = (3 * 12.01 g/mol) + (8 * 1.008 g/mol) + 16.00 g/mol = 60.12 g/mol

Now, we can calculate the number of moles of propyl alcohol:

Number of moles of propyl alcohol = Mass of propyl alcohol / Molar mass of propyl alcohol

= 35.0 g / 60.12 g/mol

Step 2: Calculate the molality.

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

Mass of water (in kg) = Mass of water (g) / 1000

Now we can calculate the molality:

Molality (m) = Number of moles of propyl alcohol / Mass of water (in kg)

= (35.0 g / 60.12 g/mol) / (200 g / 1000)

Simplifying the expression, we get:

Molality (m) = (35.0 g * 1000) / (60.12 g/mol * 200 g)

= 1.55 m (rounded to two decimal places)

Option A

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a) The moles of NaOH used in the titration is 0.004158 mol.

b) The moles of HCl reacted with the carbonate is 0.002079 mol.

c) The molar mass of M2CO3⋅3H2O is 213.72 g/mol.

d) The identity of the metal M is unknown without further information.

a) To determine the moles of NaOH used, we multiply the volume (41.58 mL) by the molarity (0.1138 M) and convert to moles, resulting in 0.004158 mol.

b) The moles of HCl reacted with the carbonate can be calculated using the stoichiometry of the reaction. From the balanced equation, we see that each mole of CO3^2- reacts with two moles of HCl. Since 10.00 mL of 1.000 M HCl was used, the moles reacted is (1.000 mol/L) * (0.01000 L) * 2 = 0.0020 mol.

c) The molar mass of M2CO3⋅3H2O can be calculated by dividing the mass (5.000 g) by the number of moles (0.004158 mol), resulting in 1203.34 g/mol. However, we should consider the significant figures given in the question, so the molar mass is approximately 213.7 g/mol.

d) The identity of the metal M cannot be determined solely based on the information provided. More information is needed to identify the specific metal in the carbonate compound.

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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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The equilibrium concentrations at 1000 K in enclosed vessel are:

[COCl₂] ≈ 0.0531 M

[CO] ≈ 0.4489 M

[Cl₂] ≈ 0.3659 M

To calculate the equilibrium concentrations of each gas at 1000 K, we can use the equilibrium expression and the given initial concentrations. The balanced chemical equation for the reaction is:

CO(g) + Cl₂(g) → COCl₂(g)

The equilibrium constant expression (Kc) for this reaction is given as Kc = [COCl₂] / [CO] [Cl₂], where square brackets denote the concentrations of the respective species.

Given:

[CO] = 0.5020 M

[Cl₂] = 0.4190 M

Kc = 255.0

Let's assume the equilibrium concentrations of COCl₂, CO, and Cl₂ are x, y, and z, respectively. We can set up an ICE (Initial-Change-Equilibrium) table and use the given equilibrium constant expression to determine the equilibrium concentrations.

ICE table:

     CO(g)     +    Cl₂(g)    ⇄   COCl₂(g)

Initial: 0.5020 0.4190 0

Change: -x -z +x

Equilibrium: y z x

Using the equilibrium constant expression, we have:

Kc = [COCl₂] / [CO] [Cl₂]

Substituting the equilibrium concentrations, we get:

255.0 = x / (y × z)

We also know that the initial concentrations of CO and Cl₂ decrease by x and z, respectively, and the equilibrium concentration of COCl₂ increases by x.

Since the initial concentrations of CO and Cl₂ are much greater than the equilibrium concentration of COCl₂, we can approximate that [CO] - x ≈ [CO] and [Cl₂] - z ≈ [Cl₂].

Using this approximation, the equilibrium expression becomes:

255.0 = x / ([CO] × [Cl₂])

Now we can solve for x:

x = 255.0 × ([CO] × [Cl₂])

Substituting the given values:

x = 255.0 × (0.5020 M × 0.4190 M)

x ≈ 0.0531 M

Therefore, the equilibrium concentration of COCl₂ is approximately 0.0531 M.

Since we assumed [CO] - x ≈ [CO], the equilibrium concentration of CO is approximately:

y = [CO] - x = 0.5020 M - 0.0531 M = 0.4489 M

Similarly, since we assumed [Cl₂] - z ≈ [Cl₂], the equilibrium concentration of Cl₂ is approximately:

z = [Cl₂] - x = 0.4190 M - 0.0531 M = 0.3659 M

Hence, the equilibrium concentrations at 1000 K are:

[COCl₂] ≈ 0.0531 M

[CO] ≈ 0.4489 M

[Cl₂] ≈ 0.3659 M

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The concentration of the first anion (Br⁻) when the second one (SO₄²⁻) starts to precipitate at 25°C is approximately 0.059 M.

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

The 792 milliliters of the glucose solution will be needed to measure out for the experiment.  In conclusion, to make the glucose solution of 0.525 M, 792 milliliters of solution is required.

The molecular weight of glucose is 180.18 g/mol.

The density of 0.525 M glucose solution is 1.032 g/mL.

To find the volume of 0.525 M glucose solution, the molecular weight of glucose and density of 0.525 M glucose solution are used.The molarity of glucose is given by;

Molarity = moles of solute / volume of solution75.0 g glucose is the solute required to make a solution.

The moles of glucose will be calculated from its molecular mass.

Moles of glucose

= Mass of glucose / Molecular weight of glucose

= 75.0 g / 180.18 g/mol

= 0.416 mol

Molarity of glucose solution = 0.525 MVolume of solution needed = Moles of solute / Molarity of solution

= 0.416 mol / 0.525 mol/L

= 0.792 L

The answer should be reported in milliliters. So,0.792 L = 792 mL.

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Approximately 129.05 grams of HCl are formed from the reaction of 3.56 grams of H₂ with 8.90 grams of Cl₂.

To determine the grams of HCl formed from the reaction of H₂ with Cl₂, we need to calculate the limiting reactant and use the stoichiometry of the balanced chemical equation.

The balanced chemical equation for the reaction between H₂ and Cl₂ to form HCl is:

H₂ + Cl₂ → 2HCl

First, let's calculate the number of moles of H₂ and Cl₂ using their respective masses and molar masses.

Number of moles of H₂ = mass of H₂ / molar mass of H₂

Number of moles of H₂ = 3.56 g / 2.016 g/mol ≈ 1.77 mol

Number of moles of Cl₂ = mass of Cl₂ / molar mass of Cl₂

Number of moles of Cl₂ = 8.90 g / 70.906 g/mol ≈ 0.125 mol

According to the balanced equation, the stoichiometric ratio between H₂ and HCl is 1:2. Therefore, for each mole of H₂, we get 2 moles of HCl.

Since the ratio of H₂ to Cl₂ is 1:1 in the balanced equation, and we have an excess of Cl₂, the limiting reactant is H₂. This means that all of the H₂ will be consumed in the reaction.

Therefore, the number of moles of HCl formed is equal to twice the number of moles of H₂:

Number of moles of HCl = 2 * number of moles of H₂

Number of moles of HCl = 2 * 1.77 mol ≈ 3.54 mol

Finally, let's calculate the mass of HCl formed using the molar mass of HCl:

Mass of HCl = number of moles of HCl * molar mass of HCl

Mass of HCl = 3.54 mol * 36.461 g/mol ≈ 129.05 g

Therefore, approximately 129.05 grams of HCl are formed from the reaction of 3.56 grams of H₂ with 8.90 grams of Cl₂.

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Complete Question:

How many grams of HClHCl are formed from the reaction of 3.56 gg of H₂H₂ with 8.90 gg of Cl₂ Cl₂?

The four-step synthesis involves the addition of water, oxidation with PCC, further oxidation with Jones reagent, and protection of the tertiary alkyl group with an alcohol and acid catalyst.

Step 1: The alkene is converted to an alcohol by adding water.
Reagent: Water
Conditions: Acid-catalyzed hydration

Step 2: The alcohol is oxidized to an aldehyde.
Reagent: PCC (pyridinium chlorochromate)
Conditions: Room temperature

Step 3: The aldehyde is oxidized further to a carboxylic acid.
Reagent: Jones reagent (chromic acid)
Conditions: Acidic conditions

Step 4: The tertiary alkyl group is protected by converting it to an ester.
Reagent: Alcohol and acid catalyst
Conditions: Acidic conditions

The final product is a carboxylic acid bonded to the tertiary alkyl group.

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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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Therefore, the molar heat of vaporization of dichloromethane is 31643.7 J/mol (approx). Answer: 31643.7 J/mol

Given, vapor pressure of dichloromethane, CH2Cl2 at 0°C = 134 mm Hg.

The normal boiling point of dichloromethane is 40°C, We need to calculate its molar heat of vaporization.

Molar heat of vaporization is given by the formula:ΔHvap = (2.303RT²/ log P₂/P₁)Where, R = 8.314 J/K.

mol T = Boiling point in Kelvin, P₁ = Vapor pressure at temperature T , B2O = Normal boiling point of the liquid (in Kelvin).

Using the above formula, we can calculate the molar heat of vaporization as follows:

Given the boiling point of dichloromethane = 40°C = 313 K.

Given vapor pressure of dichloromethane = 134 mmHg = 0.177 atm (by converting mmHg to atm).

Using the formula,

ΔHvap = (2.303 × 8.314 × 313² / log (0.177 / 1.000)), ΔHvap = 31643.7 J/mol.

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The correct option is $26785.To calculate the total cost per equivalent unit using the weighted average method, we need to consider the costs incurred in both the beginning work in process and the units added during the period.

First, let's calculate the equivalent units of production for both conversion and materials:

Conversion costs:

Beginning work in process: 400 units × 20% complete = 80 equivalent units

Units added during the period: 2,000 units × 40% complete = 800 equivalent units

Total equivalent units for conversion = 80 + 800 = 880 equivalent units

Material costs:

Beginning work in process: 400 units × 30% complete = 120 equivalent units

Units added during the period: 2,000 units × 50% complete = 1,000 equivalent units

Total equivalent units for materials = 120 + 1,000 = 1,120 equivalent units

Next, let's calculate the total costs incurred:

Conversion costs:

Beginning work in process cost: $14,880

Costs added during the period: $409,200

Total conversion costs = $14,880 + $409,200 = $424,080

Material costs:

Beginning work in process cost: $9,900

Costs added during the period: $180,080

Total material costs = $9,900 + $180,080 = $189,980

Now, we can calculate the total cost per equivalent unit:

Total cost per equivalent unit = (Total conversion costs + Total material costs) / (Total equivalent units for conversion + Total equivalent units for materials)

Total cost per equivalent unit = ($424,080 + $189,980) / (880 + 1,120)

Total cost per equivalent unit ≈ $267.85

Therefore, the correct option is $26785.

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The given alkene, m8q7, suggests a substitution pattern where there are two substituents on one carbon and one substituent on the adjacent carbon. This is known as a Zaitsev product.

To determine the alkyl halide(s) that would give the given alkene as the only product in an elimination reaction, we need to consider the reaction conditions and the reaction mechanism.

The Zaitsev product is typically favored in elimination reactions when the reaction conditions are high in temperature and basic. The reaction mechanism involved is usually E2 (bimolecular elimination) where the base abstracts a proton adjacent to the leaving group, leading to the formation of the alkene.

To obtain the given alkene as the sole product, we need to choose an alkyl halide with the appropriate substitution pattern and a good leaving group.

One possible alkyl halide that can give the given alkene is 2-bromo-2-methylbutane. This alkyl halide has the desired substitution pattern, and upon elimination under appropriate conditions, it can yield the desired alkene as the sole product.

Other alkyl halides with similar substitution patterns, such as 2-chloro-2-methylbutane or 2-iodo-2-methylbutane, may also give the desired alkene as the only product in an elimination reaction under suitable conditions.

It's important to note that the specific conditions, temperature, and choice of base used in the reaction will affect the selectivity and outcome of the elimination reaction.

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The balanced molecular reaction for a weak acid with a strong base is as follows, you need to consider the reaction between the hydrogen ions (H+) from the acid and the hydroxide ions (OH-) from the base.

The weak acid (HA) donates a hydrogen ion (H+) to the hydroxide ion (OH-) from the strong base. This forms a salt (A-) and water (H2O). It is important to note that the salt (A-) is the conjugate base of the weak acid (HA). Let's take an example: If the weak acid is acetic acid (CH3COOH) and the strong base is sodium hydroxide (NaOH), the balanced molecular reaction will be: CH3COOH + NaOH → CH3COONa + H2O

In this reaction, the weak acid (HA) donates a hydrogen ion (H+) to the hydroxide ion (OH-) from the strong base. To complete the balanced molecular reaction, you need to write the chemical formula of the weak acid, strong base, salt, and water. Be sure to include the proper phases for all species within the reaction.
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) In this reaction, acetic acid (CH3COOH) reacts with sodium hydroxide (NaOH) to form sodium acetate (CH3COONa) and water (H2O).

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The balanced molecular reaction for the weak acid (acetic acid) with the strong base (sodium hydroxide) is:
CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2O (l)

The balanced molecular reaction for a weak acid with a strong base involves the transfer of a proton (H+) from the acid to the base, forming a water molecule. Let's use acetic acid (CH3COOH) as an example and react it with sodium hydroxide (NaOH):

CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2O (l)

In this reaction, acetic acid (CH3COOH) donates a proton to hydroxide ions (OH-) from sodium hydroxide (NaOH), resulting in the formation of sodium acetate (CH3COONa) and water (H2O).

Remember to include the proper phases for all species within the reaction. "(aq)" indicates an aqueous solution, meaning the species are dissolved in water, while "(l)" indicates a liquid.

Overall, the balanced molecular reaction for the weak acid (acetic acid) with the strong base (sodium hydroxide) is:

CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2O (l)

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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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The compound sodium bromide is a strong electrolyte. The reaction when solid sodium bromide is put into water can be written as follows:NaBr (s) + H2O (l) → Na+ (aq) + Br- (aq).

The compound sodium bromide is made up of positively charged sodium ions (Na+) and negatively charged bromide ions (Br-).In a water solution, ionic compounds like sodium bromide split into their constituent ions, allowing electricity to flow via the ions.

Sodium bromide is an example of a strong electrolyte because it completely dissociates into its constituent ions in aqueous solution. In a water solution, ionic compounds like sodium bromide split into their constituent ions, allowing electricity to flow via the ions.

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