# 22. What is the molar mass of oxygen (O2)?A- 15.9994 g/molB- 1.204 x 1024 g/molC- 6.02 x 1023 g/molD- 31.9988 g/mol

D- 31.9988 g/mol

Explanation:

The molar mass of a substance is the mass in grams of one mole of the substance. For a compound, the molar mass is the gram - formula mass or gram - molecular weight. It is determined by the addition of its component atomic masses;

To find molar mass of O₂;

Atomic mass of O = 15.9994g/mol

Molar mass of O₂ = 2(15.9994) = 31.9988g/mol

## Related Questions

The following figure represents the formation of an ionic compound. Substances A and B are initially uncharged, but when mixed electrons are transferred. Using the figure, identify which substance will form the cation and which will form the anion. Provide a brief (one or two sentences) explanation for your response. (Hint: How does losing electrons affect atomic radii?)

This question is incomplete

Explanation:

This question is incomplete but some general explanation provides a clear answer to what is been asked in the question.

An ionic/electrovalent compound is a compound whose constituent atoms are joined together by ionic bond. Ionic bond is a bond involving the transfer of valence electron(s) from an atom (to form a positively charged cation) to another  atom (to form a  negatively charged anion). The atom transferring is usually a metal while the atom receiving is usually a non-metal.

For example (as shown in the attachment), in the formation of NaCl salt, the sodium (Na) transfers the single electron (valence) on it's outermost shell to chlorine (Cl) which ordinarily has 7 electrons on it's outermost shell but becomes 8 after receiving the valence electron from sodium. It should also be noted that Na is a metal while Cl is a non-metal.

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Explanation:

Hello,

In this case, we write the reaction again:

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

Best regards.

When liquid phosphorus trichloride is added to water, it reacts to form aqueous phosphorous acid, h3po3(aq), and aqueous hydrochloric acid. express your answer as a balanced chemical equation. identify all of the phases in your answer?

Answer: The balanced chemical equation for the reaction of phosphorus trichloride with water is written below.

Explanation:

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

The balanced chemical equation for the reaction of phosphorus trichloride with water follows:

By Stoichiometry of the reaction:

1 mole of liquid phosphorus trichloride reacts with 3 moles of water to produce 1 mole of aqueous phosphorus acid and 3 moles of aqueous hydrochloric acid.

Hence, the balanced chemical equation for the reaction of phosphorus trichloride with water is written above.

Hey there!:

I'll balance it:

PCl3 + H2O ---> H3PO3 + HCl

* PCl3 (l )+ 3 H2O (l)--->H3PO3(aq) +3 HCl(aq)

Hope that helps!

The destruction of the ozone layer by chlorofluorocarbons (CFC’s) can be described by the following reactions:ClO(g) + O3(g) ? Cl(g) + 2 O2(g) ?H°rxn = –29.90 kJ2 O3(g) ? 3 O2(g) ?H°rxn = 24.18 kJDetermine the value of heat of reaction for the following:Cl(g) + O3(g) ? ClO(g) + O2(g) ?H°=_____________?

ΔH°rxn = 54.08 kJ

Explanation:

Let's consider the following equations.

a) ClO(g) + O₃(g) ⇄ Cl(g) + 2 O₂(g)                     ΔH°rxn = –29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g)                                            ΔH°rxn = 24.18 kJ

We have to determine the value of heat of reaction for the following reaction: Cl(g) + O₃(g) ⇄ ClO(g) + O₂(g)

According to Hess's law, the enthalpy change in a chemical reaction is the same whether the reaction takes place in one or in several steps. That means that we can find the enthalpy of a reaction by adding the corresponding steps and adding their enthalpies. According to Lavoisier-Laplace's law, if we reverse a reaction, we also have to reverse the sign of its enthalpy.

Let's reverse equation a) and add it to equation b).

-a) Cl(g) + 2 O₂(g) ⇄ ClO(g) + O₃(g)                    ΔH°rxn = 29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g)                                            ΔH°rxn = 24.18 kJ

-------------------------------------------------------------------------------------------------

Cl(g) + 2 O₂(g) + 2 O₃(g) ⇄ ClO(g) + O₃(g) + 3 O₂(g)

Cl(g) + O₃(g) ⇄ ClO(g) +O₂(g)

ΔH°rxn = 29.90 kJ + 24.18 kJ = 54.08 kJ

The heat of the reaction (ΔH°rxn) for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is calculated using Hess's Law. The sum of the heat of reversed first reaction and the second reaction provided is 54.08 kJ.

### Explanation:

The chemistry question asks to determine the heat of the reaction for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). In Hess's Law, the heat of the reaction or ΔH for a reaction can be calculated from the sum of the heats of other reactions that sum to the desired reaction. In this case, we want to reverse the first reaction provided (which changes the sign of ΔH) and add it to the second reaction provided.

So, reversing the first reaction we get: Cl(g) + 2 O2(g) ? ClO(g) + O3(g) ?H°rxn = 29.90 kJ

Adding this to the second reaction: 2 O3(g) ? 3 O2(g), ?H°rxn = 24.18 kJ, gives the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). Adding the ΔH values gives the ΔH for this reaction: 29.90 kJ + 24.18 kJ = 54.08 kJ. So, ?H°rxn for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is 54.08 kJ.

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Which two structures will provide a positive identification of a plant cell under a microscope? A.) Lysosomes, cell wall. B.) large central vacuole, cell wall. C.) large central vacuole ribosomes. D.)nucleoid, chloroplasts.