the remaining mass of 28 grams of a radioactive element whose half life is 20 years is given by the following equation, after t years. how much of the initial mass remains after 100 years? round your answers to two decimal places

This statement is false. Fats are not more oxidized than polysaccharides. In fact, the opposite is true: polysaccharides are more oxidized than fats.

What are the Fats?

Fats are composed of long chains of hydrocarbons, which contain a high proportion of carbon-hydrogen bonds. These bonds are considered to be reduced, meaning they contain more potential energy compared to bonds between carbon and oxygen or carbon and nitrogen. When fats are oxidized, they undergo a reaction with oxygen that breaks these carbon-hydrogen bonds and produces carbon dioxide and water. This process releases energy, which can be used by the body for various functions.

In contrast, polysaccharides are composed of chains of sugar molecules, which contain a higher proportion of carbon-oxygen and carbon-carbon bonds. These bonds are considered to be more oxidized than carbon-hydrogen bonds and contain less potential energy. When polysaccharides are oxidized, they undergo a reaction with oxygen that breaks these carbon-oxygen and carbon-carbon bonds and produces carbon dioxide and water, along with releasing energy.

Therefore, it can be concluded that polysaccharides are more oxidized than fats, not the other way around.

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Complete question is: The advantage of fats over polysaccharides is that fats are more oxydized is FALSE.

Explanation:

As seen by the previous picture, we see that in the molecule SF2, there are 4 bonding electrons total (2 in the sulfur atom, and 1 each for the fluorine atoms), and 16 nonbonding electrons (2 pairs in the sulfur atom, and 3 pairs each in the fluorine atoms)

The quantitative data in the statement above is the mass of sucrose used, which is 0.501 g. Quantitative data is any data that can be expressed as a numerical value or measurement.

In this case, the mass of sucrose used is a specific value that can be quantified and measured. The statement also includes qualitative data, which is descriptive and cannot be expressed as a numerical value. For example, the statement mentions that the resulting solution is clear and colorless, which is qualitative data.

Quantitative data is important in many fields, including chemistry and biology, as it allows for precise measurements and comparisons. In this case, knowing the exact mass of sucrose used can be important in determining the concentration of the resulting solution.

In summary, the mass of sucrose used in the statement above is an example of quantitative data. It is a specific numerical value that can be measured and quantified.

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A chemical formula that uses the smallest ratio feasible to display the proportional number of each type of atom in a molecule. Therefore, choice A is correct.

Chemical formulas can be divided into three categories: analytical, molecular, and structural.

The empirical formula, which is defined as the ratio of subscripts of the least number of full components in the formula, is the simplest formula for a compound.

Molecular formulas show the quantity of each type of atom in a molecule, while structural formulae indicate the bonds that hold the atoms in a molecule. Empirical equations give the simplest whole-number ratio of the atoms in acompound.

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(a) Initial pH of 0.10 M CH3COOH(aq) solution is 2.87.

(b) pH after the addition of 10.0 mL of 0.10 M NaOH(aq) is 4.56.

(c) Volume of 0.10 M NaOH(aq) required to reach halfway to the stoichiometric point is 12.5 mL.

(d) pH at the halfway point is equal to pKa, which is 4.75.

(e) Volume of 0.10 M NaOH(aq) required to reach the stoichiometric point is 25.0 mL.

(f) pH at the stoichiometric point is 8.72.

(a) Before the addition of NaOH, the initial [H+] in the 0.10 M CH3COOH(aq) solution can be calculated using the Ka expression for CH3COOH:

Ka = [H+][CH3COO-] / [CH3COOH]

Assuming that [H+] = [CH3COO-] at equilibrium, the expression can be simplified to:

Ka = [H+]^2 / [CH3COOH]

Solving for [H+], we get:

[H+] = sqrt(Ka x [CH3COOH])

[H+] = sqrt(1.8 x 10^-5 x 0.10) = 1.34 x 10^-3 M

pH = -log[H+] = -log(1.34 x 10^-3) = 2.87

(b) When 10.0 mL of 0.10 M NaOH(aq) is added, it reacts with CH3COOH according to the following balanced equation:

CH3COOH(aq) + NaOH(aq) → CH3COO- (aq) + Na+ (aq) + H2O(l)

The moles of CH3COOH initially present can be calculated as:

moles of CH3COOH = concentration x volume = 0.10 x 25.0 x 10^-3 = 2.50 x 10^-3 mol

The moles of NaOH added can be calculated as:

moles of NaOH = concentration x volume = 0.10 x 10.0 x 10^-3 = 1.00 x 10^-3 mol

Since CH3COOH and NaOH react in a 1:1 ratio, the remaining moles of CH3COOH can be calculated as:

moles of CH3COOH remaining = moles of CH3COOH initially present - moles of NaOH added

moles of CH3COOH remaining = 2.50 x 10^-3 - 1.00 x 10^-3 = 1.50 x 10^-3 mol

The concentration of CH3COOH after the addition of NaOH can be calculated as:

[CH3COOH] = moles of CH3COOH remaining / volume of solution = 1.50 x 10^-3 / 25.0 x 10^-3 = 0.060 M

The concentration of CH3COO- can be calculated from the amount produced by the reaction:

[CH3COO-] = moles of NaOH added / volume of solution = 1.00 x 10^-3 / 35.0 x 10^-3 = 0.029 M

Using the Ka expression for CH3COOH and the concentrations of CH3COOH and CH.

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The appropriate method for removing a solvent depends on several factors. The following answers are correct; volume of solvent being evaporated, volatility of the solvent, the hazards associated with the solvent, and the thermal stability of the product. Option A, C, E, and F are correct.

If a large volume of solvent needs to be removed, then a rotary evaporator or distillation may be necessary. For smaller volumes, a simple evaporation may be sufficient.

Solvents with high volatility can be removed by simple evaporation or with the help of a vacuum. Solvents with low volatility may require techniques such as rotary evaporation or distillation.

If the solvent is hazardous or toxic, special precautions such as fume hoods or closed systems may be necessary.

If the product is sensitive to heat, then techniques such as vacuum or freeze-drying may be necessary to remove the solvent.

Hence, A. C. E. F. is the correct option.

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When three atomic orbitals mix, three [tex]sp^{2}[/tex] hybrid orbitals are formed; when two atomic orbitals mix, two sp hybrid orbitals are formed; and when one atomic orbital mixes, there is no hybridization.

To determine the number and type of hybrid orbitals formed when three atomic orbitals, two atomic orbitals, and one atomic orbital mix, we will consider each case separately.

1. When three atomic orbitals mix:
In this case, three atomic orbitals combine to form three hybrid orbitals. Since there are three orbitals involved, the hybridization will be [tex]sp^{2}[/tex]. The three [tex]sp^{2}[/tex] hybrid orbitals are formed by the combination of one s orbital and two p orbitals.

2. When two atomic orbitals mix:
When two atomic orbitals mix, they form two hybrid orbitals. This hybridization is known as sp hybridization, as it involves the combination of one s orbital and one p orbital.

3. When one atomic orbital mixes:
In this case, as only one atomic orbital is involved, there is no hybridization or mixing of orbitals. The single atomic orbital remains unhybridized.

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The correct formula for the combination of zinc and hydroxide is b. zn(oh)2(s). This formula represents the solid compound formed when zinc cations (Zn2+) combine with hydroxide anions (OH-) in a 1:2 ratio. The brackets around the (OH) indicate that there are two hydroxide ions for every one zinc ion.

It's important to note that the state of the compound can vary depending on the conditions it is in. In this case, the (s) indicates that the compound is a solid. This means that under normal conditions, zinc hydroxide will exist as a solid rather than a liquid or gas.
The correct state for zinc hydroxide in an aqueous solution would be c. zn oh 2(aq). This formula indicates that the compound is dissolved in water (aq), which means it exists as individual ions rather than a solid.
Finally, the formula d. zn2oh(s) is incorrect because it suggests that there is a covalent bond between the zinc and hydroxide ions, which is not the case. Zinc hydroxide is an ionic compound, meaning that the zinc and hydroxide ions are held together by electrostatic attraction rather than a shared pair of electrons.

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The heat released (q) for the combustion of 15.0 g of 2-propanol is approximately -5.01 x 10^2 kJ.

To calculate q for the combustion of 15.0 g of 2-propanol, we first need to determine the number of moles of 2-propanol. We can do this using the molar mass (M) of 2-propanol, which is 60.0 g/mol.

moles of 2-propanol = (15.0 g) / (60.0 g/mol) = 0.25 mol

From the balanced chemical equation, we know that 2 moles of 2-propanol will produce 6 moles of CO₂. The heat of combustion of 2-propanol is -2020 kJ/mol. To find the heat released (q) for 0.25 mol of 2-propanol, we can use the following equation:

q = (moles of 2-propanol) * (heat of combustion per mole)
q = (0.25 mol) * (-2020 kJ/mol) = -505 kJ

So, the heat released (q) for the combustion of 15.0 g of 2-propanol is approximately -5.01 x 10^2 kJ.

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Molecular bonding is best explained as the sharing of electrons between atoms, which is also known as covalent bonding.

Molecular bonding, specifically covalent bonding, occurs when atoms share electrons in order to achieve a stable electron configuration. This sharing allows both atoms involved to complete their outer electron shells, resulting in a strong bond between the atoms. In this type of bonding, atoms share electrons in order to achieve a stable electron configuration. This type of bonding occurs mainly between nonmetals. The other options listed, such as donating electrons and charged ions attracting, are examples of ionic bonding, which occurs mainly between metals and nonmetals.

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Electron configurations of transition metal complexes that are susceptible to Jahn-Teller (JT) distortions are those with degenerate energy levels and an uneven number of electrons in the d-orbitals.

Jahn-Teller distortions occur in transition metal complexes when the system is in an electronically degenerate state, meaning the energy levels of the orbitals are equal. In these cases, the system tends to lower its overall energy through geometric distortions, which breaks the degeneracy. The most susceptible electron configurations are those with uneven electron occupancy in their degenerate d-orbitals, such as d9, d7, and high-spin d4 complexes.

Transition metal complexes with degenerate energy levels and an uneven number of electrons in the d-orbitals, such as d9, d7, and high-spin d4 configurations, are more likely to exhibit Jahn-Teller distortions. These distortions occur to lower the overall energy of the system by breaking the degeneracy through geometric changes.

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The standard entropy change for the reaction is -364 J/k mol.

To calculate the standard entropy change for the reaction, we can use the formula:

ΔS° = ΣnS°(products) - ΣmS°(reactants)

where ΔS° is the standard entropy change, n and m are the of the products and reactants, and S° is the standard molar entropy.

For the given reaction:

2[tex]H_{2} S[/tex](g) + [tex]SO_{2}[/tex](g) → 3S(s) + 2[tex]H_{2} O[/tex](g)

n = 3 (for S) and 2 (for [tex]H_{2} O[/tex])
m = 2 (for [tex]H_{2} S[/tex]) and 1 (for [tex]SO_{2}[/tex])

Substituting the values:

ΔS° = [3(0 J/k mol) + 2(189 J/k mol)] - [2(206 J/k mol) + 1(248 J/k mol)]
ΔS° = -364 J/k mol

Therefore, the standard entropy change for the reaction is -364 J/k mol.

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Around 5,318,704.94 grams or 5,318.70 kg of carbon dioxide were produced.

A naturally occurring element of the Earth's atmosphere, carbon dioxide (CO2) is a colorless, odorless gas. It has the chemical formula CO2, which stands for carbon one and oxygen two.

We must apply the balanced chemical equation for wood combustion to this issue in order to find a solution:

6O2 + C6H10O5 = 6CO2 + 5H2O

According to this equation, each molecule of wood (C6H10O5) that is burnt requires 6 oxygen molecules to make 6 carbon dioxide molecules and 5 water molecules.

We must first estimate the volume of wood that really burnt in order to compute the amount of carbon dioxide generated. The amount of wood that burnt if only 72% of the home was destroyed by fire is:

0.72 x 4535.924 Kg = 3266.136 Kg

The number of moles of wood burnt can then be determined:

3266.136 Kg x 162.14 g/mol = 20134.73 mol of wood

Finally, we may calculate the quantity of carbon dioxide created using the balanced chemical equation:

6 moles of CO2 are created from 1 mole of wood (20134.73).Wood at mol produces 6 x 20134 (mol).73 mol CO2 = 120808.38 mol CO2

We may use the molar mass of CO2 to express the amount of CO2 in moles to mass:

Molar mass of CO2 = 44.01 g/mol

The mass of CO2 produced = 120808.38 mol x 44.01 g/mol = 5,318,704.94 g.

Consequently, around 5,318,704.94 grams or 5,318.70 kg of carbon dioxide were created.

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The (s) in the chemical equation CdSO4(aq) + K2S(aq) → CdS(s) + K2SO4(aq) indicates that CdS is a solid precipitate that forms as a result of the reaction between the two aqueous solutions.

In other words, the solid CdS is not dissolved in water and can be seen as a separate phase in the mixture.

To write a chemical equation showing the dissolution of cadmium sulfate and potassium sulfide in water, we can use the subscript (aq) to indicate that they are aqueous solutions. The chemical equation for the dissolution of cadmium sulfate in water is:

CdSO4(s) → Cd2+(aq) + SO42-(aq)

This equation shows that when cadmium sulfate is added to water, it dissociates into cadmium ions (Cd2+) and sulfate ions (SO42-), which are both soluble in water.

Similarly, the chemical equation for the dissolution of potassium sulfide in water is:

K2S(s) → 2K+(aq) + S2-(aq)

This equation shows that when potassium sulfide is added to water, it dissociates into two potassium ions (2K+) and one sulfide ion (S2-), which are also both soluble in water.

Overall, the dissolution of these compounds in water allows them to participate in chemical reactions and form new compounds like CdS and K2SO4.

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The correct net ionic equation to describe this precipitation reaction is; Co²+ (aq) + 2 OH¯(aq) → Co(OH)₂ (s). Option C is correct.

This is because the net ionic equation only shows the species that are involved in the chemical reaction and are actually changed. The spectator ions (Na⁺ and NO₃⁻) are omitted from the equation because they do not participate in the reaction and remain in their original form.

A precipitation reaction is a type of chemical reaction in which two aqueous solutions of ionic compounds are mixed together, resulting in the formation of an insoluble solid called a precipitate.

This occurs because one of the products formed in the reaction is insoluble in water and separates from the solution as a solid. Precipitation reactions are often used in chemistry to separate and purify different compounds based on their solubility properties.

Hence, C. is the correct option.

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To create an 8 um (mdil) standard solution, you will need to dilute the stock solution (mconc) provided.
Using the Beer's Law equation, we know that A = εbc, where A is absorbance, ε is the molar absorptivity, b is the path length (in this case, 1 cm), and c is concentration.

We can rearrange this equation to solve for concentration: c = A/εb
Since we are diluting the stock solution, we can set up the equation:
mdil = mconc(Vconc/Vdil)
where mdil is the concentration of the diluted solution (8 um), mconc is the concentration of the stock solution (10 um), Vconc is the volume of stock solution used, and Vdil is the total volume of the diluted solution.
We can solve for Vconc:
Vconc = (mdil x Vdil)/mconc
Plugging in the given values, we get:
Vconc = (8 um x 10 mL)/(10 um x 1/10 mL) = 80/1 = 80 mL
Therefore, you will need 80 mL of the stock solution (mconc) to create an 8 um (mdil) standard solution using a 10 mL volumetric flask and 1 mL and 5 mL volumetric pipets.

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At 27°C and constant pressure, the volume of the gas in the balloon is approximately 240.0 mL.To solve this problem, we need to use the combined gas law, which relates the pressure, volume, and temperature of a gas:[tex](P_1V_1)/T_1 = (P_2V_2)/T_2[/tex]

Where [tex]P_1[/tex]and [tex]T_1[/tex] are the initial pressure and temperature, [tex]V_1[/tex]is the initial volume, [tex]P_2[/tex] and [tex]T_2[/tex] are the final pressure and temperature, and  [tex]V_2[/tex] is the final volume we are trying to find.We are given the initial volume [tex]V_1[/tex] = 120.0 mL, the initial temperature [tex]T_1[/tex]= -123C, and the pressure is constant, so we can assume [tex]P_1 = P_2[/tex]. We need to convert the temperatures to Kelvin, so [tex]T_1[/tex] = 150 K and [tex]T_2[/tex]= 300 K.

Using the combined gas law, we can solve for [tex]V_2[/tex]:
[tex](P_1V_1)/T_1 = (P_2V_2)/T_2[/tex]
[tex](P_1)(120.0 mL)/(150 K) = (P_2)(V_2)/(300 K)[/tex]
Simplifying, we can cancel out the pressures and cross-multiply:
[tex]V_2[/tex] = (120.0 mL)(300 K)/(150 K)
[tex]V_2[/tex] = 240.0 mL
Therefore, the volume of the gas in the balloon at 27C is 240.0 mL.To answer your question, we'll use Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin, as long as the pressure remains constant.Charles's Law formula: [tex]V_1/T_1 = V_2/T_2[/tex]

Where:
[tex]V_1[/tex] = initial volume = 120.0 mL
[tex]T_1[/tex] = initial temperature = -123°C
[tex]V_2[/tex]= final volume (what we want to find)
[tex]T_2[/tex] = final temperature = 27°C

First, we need to convert the temperatures from Celsius to Kelvin:
[tex]T_1[/tex](K) = -123°C + 273.15 = 150.15 K
[tex]T_2[/tex](K) = 27°C + 273.15 = 300.15 K

Now, we can plug in the values into Charles's Law formula:
(120.0 mL / 150.15 K) = ([tex]V_2[/tex] / 300.15 K)
To find [tex]V_2[/tex], we'll rearrange the equation and multiply both sides by 300.15 K:
[tex]V_2[/tex] = (120.0 mL / 150.15 K) × 300.15 K
[tex]V_2[/tex]≈ 240.0 mL

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Since it is not possible to have a negative volume of pure acid, there may be an error in the given information or the desired concentration is not achievable with the provided data. Please re-check your question and try again.

To create a 25% acid solution using pure acid and a 10% acid solution, you can use the following equation:

(amount of pure acid in mm) * 100% + (100 mm * 10%) = (total volume of new solution) * 25%

Let "x" represent the amount of pure acid in millimeters:

x * 100% + (100 mm * 10%) = (100 mm + x) * 25%

Now, solve for x:

x + 10 = 25x/4 + 25

Multiply both sides by 4 to eliminate the fraction:

4x + 40 = 100x/4 + 100

Simplify:

4x + 40 = 25x + 100

Subtract 4x from both sides:

40 = 21x + 100

Subtract 100 from both sides:

-60 = 21x

Divide both sides by 21:

x = -60/21 = -20/7

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The pressure of the compressed gas is 1114 mmHg.If a gas, in a sealed container, has a pressure of 815 mmhg and volume of 40.3 l. the container is compressed to a volume of 29.5 l, at constant temperature.

To find the pressure of the compressed gas, we can use Boyle's Law, which states that at constant temperature, the product of the initial pressure and volume (P1V1) equals the product of the final pressure and volume (P2V2).
Given:
Initial pressure (P1) = 815 mmHg
Initial volume (V1) = 40.3 L
Final volume (V2) = 29.5 L
We need to find the final pressure (P2).
Using Boyle's Law formula: P1V1 = P2V2
(815 mmHg)(40.3 L) = (P2)(29.5 L)
Solve for P2:
P2 = (815 mmHg)(40.3 L) / (29.5 L) ≈ 1113.6 mmHg

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The given concentration-time data and the balanced chemical equation show the stoichiometry and kinetics of the reaction between A and B to form D.

To determine the initial rate of the reaction, we need to find two data points where the concentrations of both reactants are the same. From the given table, we can see that experiment 4 and experiment 5 have the same initial concentrations of both reactants, which are [A] = 0.13 M and [B] = 0.13 M.

In experiment 4, the concentration of D increased from 0 to 0.40 M in 60 seconds. Therefore, the initial rate of the reaction is:

Rate = (0.40 M - 0 M) / 60 s = 0.0067 M/s

So the initial rate of the reaction when the initial concentrations of both reactants are 0.13 M is 0.0067 M/s.

The balanced chemical equation for the reaction is:

2A + 3B → 4D

This equation indicates that two molecules of A and three molecules of B react to form four molecules of D. The stoichiometric coefficients are used to balance the number of atoms on both sides of the equation. This reaction is a type of chemical reaction known as a "multiple-reaction", where multiple reactants combine to form a single product.

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The pH of the solution obtained by dissolving two extra-strength aspirin tablets in 258 mL of water is approximately 2.65.

First, we need to calculate the number of moles of acetylsalicylic acid present in the solution:

Convert the mass of acetylsalicylic acid in each tablet from milligrams (mg) to grams (g):

573 mg = 0.573 g

Calculate the total mass of acetylsalicylic acid in two tablets:

0.573 g/tablet x 2 tablets = 1.146 g

Calculate the number of moles of acetylsalicylic acid in 1.146 g:

moles = mass/molar mass = 1.146 g / 180.16 g/mol = 0.00637 mol

Calculate the concentration of acetylsalicylic acid in the solution:

concentration = moles/volume = 0.00637 mol / 0.258 L = 0.0247 M

Now, we can use the Ka value and the equilibrium expression for the dissociation of acetylsalicylic acid to calculate the pH of the solution:

Ka = [H+][C9H7O4-]/[HC9H7O4]

Let x be the concentration of H+ ions that are produced by the dissociation of acetylsalicylic acid. Then:

Ka = x^2 / (0.0247 - x)

Since the value of Ka is very small, we can assume that x is much smaller than 0.0247. This allows us to simplify the equation:

Ka ≈ x^2 / 0.0247

x^2 = Ka x 0.0247

x = sqrt(Ka x 0.0247) = sqrt(3.3x10^-4 x 0.0247) = 0.00226 M

Therefore, the pH of the solution is:

pH = -log[H+] = -log(0.00226) ≈ 2.65

So the pH of the solution obtained by dissolving two extra-strength aspirin tablets in 258 mL of water is approximately 2.65.""

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Answer:see below

Explanation:

   :Cl:      :Br:

   ||       ||

H3C-Xe-CH3

   ||      

   :Br:      

The compound is benzene, and its structure consists of a hexagonal ring with alternating single and double bonds between carbon atoms, with each carbon atom bonded to a hydrogen atom.



The molecular formula is C6H6, indicating six carbon and six hydrogen atoms.

Based on this information, the chemical structure of the compound C6H6 is benzene.

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Answer:

The total mass of the recovered chemicals should be equal to the original mass of the sample if the experiment is performed correctly and all the chemicals are accounted for. This is due to the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction. However, in practice, there may be small differences due to experimental errors or incomplete recovery of all the chemicals.

Explanation:

The elements with the higher first ionization energy in each pair are Kr, Ca, As, and S.

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The correct answer is Zn(NO3)2(aq) and HNO3(aq) will not be involved in the net ionic equation as they are spectator ions and do not participate in the reaction. The net ionic equation for the given reaction is: Zn2+(aq) + H2S(aq) → ZnS(s) + 2H+(aq)

 First, let's write the balanced chemical equation and then the total ionic equation. The balanced chemical equation is:

Zn(NO3)2(aq) + H2S(aq) → ZnS(s) + 2HNO3(aq)

Now let's write the total ionic equation:

Zn²⁺(aq) + 2NO₃⁻(aq) + 2H⁺(aq) + S²⁻(aq) → ZnS(s) + 2H⁺(aq) + 2NO₃⁻(aq)

The net ionic equation is obtained by removing spectator ions, which are ions that remain unchanged during the reaction. In this case, the spectator ions are 2H⁺(aq) and 2NO₃⁻(aq). So the net ionic equation is:

Zn²⁺(aq) + S²⁻(aq) → ZnS(s)

From the given options, the species that will NOT be involved in the net ionic equation is NO₃⁻(aq). The correct answer is Zn(NO3)2(aq) and HNO3(aq) will not be involved in the net ionic equation as they are spectator ions and do not participate in the reaction. The net ionic equation for the given reaction is: Zn2+(aq) + H2S(aq) → ZnS(s) + 2H+(aq)

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The AH for condensing 20.0 grams of vaporization for water is: H + 2O(l) — H₂O(g) AHvap = +44.0 kJ/mole is -48.8 kJ.

To calculate the enthalpy change (AH) for condensing 20.0 grams of water, we need to first convert the mass to moles.

20.0 g H-48.8 kJO x (1 mol H-48.8 kJO/18.01 g H-48.8 kJO)

= 1.11 mol H-48.8 kJO

Next, we use the enthalpy of vaporization (AHvap) of water to calculate the enthalpy change for condensation:

AHcondensation = -AHvap

AHcondensation = -44.0 kJ/mol

Finally, we can calculate the enthalpy change for condensing 20.0 grams of water:

AH = AHcondensation x n

where n is the number of moles of water condensed.

AH = (-44.0 kJ/mol) x (1.11 mol)

AH = -48.8 kJ

Therefore, the enthalpy change for condensing 20.0 grams of water is -48.8 kJ (the negative sign indicates that this is an exothermic process, meaning heat is released).

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The ratio of the osmotic pressures of 0.30 m K₂SO₄ and 0.15 m NaBr is 3.

To determine the ratio of osmotic pressures, we need to consider the number of ions each solute contributes.

K₂SO₄ dissociates into 2 K⁺ ions and 1 SO₄²⁻ ion, contributing a total of 3 ions.
NaBr dissociates into 1 Na⁺ ion and 1 Br⁻ ion, contributing a total of 2 ions.

Now, we can calculate the osmotic pressure for each solution using the formula:

osmotic pressure = molality × number of ions

For K₂SO₄:
osmotic pressure = 0.30 mol/kg × 3 ions = 0.90 osmol

For NaBr:
osmotic pressure = 0.15 mol/kg × 2 ions = 0.30 osmol

The ratio of osmotic pressures is:
0.90 osmol (K₂SO₄) / 0.30 osmol (NaBr) = 3

So, the ratio of osmotic pressures for 0.30 m K₂SO₄ and 0.15 m NaBr is 3.

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Based on the information provided, we can determine the shape of the product molecule Y2X. Since Y2X has three atoms, we need to consider the possible molecular geometries for a molecule with three atoms.

The possible molecular geometries for a three-atom molecule are:

Linear - all three atoms lie in a straight line.

Bent - the three atoms are not in a straight line and the two outer atoms are bent away from the middle atom.

Trigonal planar - the three atoms lie in a flat plane, forming an equilateral triangle.

Since the reactants X and Y are atoms, it is likely that they will form a diatomic molecule, XY. When another Y atom is added to form the product Y2X, the resulting molecule will have three atoms. Therefore, the possible molecular geometries for Y2X are the same as for a three-atom molecule.

The molecular geometry of Y2X will depend on the electronic configuration of the atoms involved, but we can make some general predictions.

If X is a small atom like hydrogen (H) or helium (He), and Y is a larger atom like chlorine (Cl) or bromine (Br), then the resulting molecule Y2X is likely to be bent. This is because the larger Y atoms will repel each other, causing the molecule to bend.

On the other hand, if X is a larger atom than Y, then the resulting molecule Y2X is likely to be linear. This is because the smaller Y atoms will be attracted to the larger X atom, causing the molecule to form a straight line.

Therefore, based on the limited information provided, we can predict that the shape of the product molecule Y2X is either bent or linear, depending on the relative sizes of X and Y atoms.

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Based on the information provided, we can determine the shape of the product molecule Y2X. Since Y2X has three atoms, we need to consider the possible molecular geometries for a molecule with three atoms.

The possible molecular geometries for a three atom molecule are:

Linear - all three atoms lie in a straight line.

Bent - the three atoms are not in a straight line and the two outer atoms are bent away from the middle atom.

Trigonal planar - the three atoms lie in a flat plane, forming an equilateral triangle.

Since the reactants X and Y are atoms, it is likely that they will form a diatomic molecule, XY. When another Y atom is added to form the product Y2X, the resulting molecule will have three atoms. Therefore, the possible molecular geometries for Y2X are the same as for a three-atom molecule.

The molecular geometry of Y2X will depend on the electronic configuration of the atoms involved, but we can make some general predictions.

If X is a small atom like hydrogen (H) or helium (He), and Y is a larger atom like chlorine (Cl) or bromine (Br), then the resulting molecule Y2X is likely to be bent. This is because the larger Y atoms will repel each other, causing the molecule to bend.

On the other hand, if X is a larger atom than Y, then the resulting molecule Y2X is likely to be linear. This is because the smaller Y atoms will be attracted to the larger X atom, causing the molecule to form a straight line.

Therefore, based on the limited information provided, we can predict that the shape of the product molecule Y2X is either bent or linear, depending on the relative sizes of X and Y atoms.

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All of the following metals will corrode if immersed in freshwater except chromium.  Therefore, the correct answer is option (C)

The metal chromium is very resistant to corrosion and does not corrode in freshwater. Chromium is a potent oxidising agent, which means it can readily provide electrons to other substances, creating a shield of protective oxide that prevents corrosion of the metal.

The metal is protected from the atmosphere by this extraordinarily tough oxide layer, which also serves to shield the metal from further oxidation. Chromium can also create a solid connection with oxygen, which increases its resistance to corrosion.

Chromium does not interact with other chemicals in freshwater, hence it does not corrode. To maintain corrosion resistance, chromium is frequently utilised in products like plumbing fixtures, tools, and automobile components.

Complete Question:

All of the following metals will corrode if immersed in freshwater except

(a) copper

(b) nickel

(C) chromium

(D) aluminum

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