PLEASE HELP!!! The image represents the reaction between a certain number of molecules of N2 and H2.

Answer:

9.88

Explanation:

The higher the Ksp, the more soluble a compound is. Hence, Co(OH)₂ ranks as the least soluble hydroxide.

Its maximum concentration must be 1x10⁻⁶ M, and the reaction can be represented as:

Co(OH)₂(s) ⇄ Co⁺²(aq) + 2OH⁻(aq)

Thus, [Co⁺²] = 1x10⁻⁶M

Ksp =  [Co⁺²] *[OH⁻]²

[[OH⁻]² = 5.9x10⁻¹⁵/1x10⁻⁶

[[OH⁻] = √(5.9x10⁻⁹)

[[OH⁻] = 7.6811x10⁻⁵

pOH = -log[OH⁻]

pOH = -log(7.6811x10⁻⁵)

pOH = 4.11

Knowing that pH + pOH = 14

pH = 14 - 4.11

pH = 9.88

Responses: a. 1.28 mol/L; b. 17.0 %; c. 0.0227; d. 1.29 mol/kg Explanation: a. Molar concentration: c = moles/litres. Moles = 167 × 1/159.61. After performing the calculation, Moles = 1.046 mol. Litres = 820 × 1/1000. Hence, Litres = 0.8200 L. Calculating the molar concentration gives c = 1.046/0.8200, resulting in c = 1.28 mol·L⁻¹. b. Percent by mass: Mass % = mass of solute / mass of solution × 100 %. Mass of solution = volume × density, therefore, Mass of solution = 820 × 1.195. By calculating this, Mass of solution = 979.9 g. Thus, Mass % = 167/979.9 × 100, which results in Mass % = 17.0 %. c. Mole fraction: χ = moles of solute / (moles of solvent + moles of solute). Mass of solvent = mass of solution – mass of solute; namely, Mass of solvent = 979.9 – 167. Converting this to moles gives Moles of water = 812.9 × 1/18.02, which results in Moles of water = 45.11 mol. The total moles are 1.046 + 45.11, leading to Total moles = 46.16 mol. Finally, the mole fraction is calculated as χ = 1.046/46.16, equating to χ = 0.0227. d. Molal concentration: b = moles of solute / kilograms of solvent. Mass of solvent = 812.9 g = 0.8129 kg. Therefore, the molal concentration yields: b = 1.046/0.8129 = 1.29 mol/kg.

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

To begin, this isn't really a chemistry forum, but anyway.

This represents a limiting reagent scenario.

Set it up as a Dimensional Analysis issue.

Begin with your desired outcome.

Your goal is to find the mass of acrylonitrile (C3H3N)

so you should initiate with that (I'll abbreviate Acrylonitrile as ACL for convenience)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

If you calculate that, you will find that 15 grams of C3H6 yields 18.9 grams of acrylonitrile produced.

Utilize the same approach for the remaining two reactants.

So, I figured it out, and for

oxygen, I calculated 11.04 grams

and for ammonia, I calculated 15.29 grams

This means that the maximum possible production is 11.04 grams, since to create any additional amount, more O2 would be necessary, but with only 10 grams available, that's the upper limit for this reaction.

The other two reactants are in excess.

Please rate as brainliest!

This appears to be an essay question based on an experiment. Different oxidizing agents possess varying strengths. When compared to nitric acid (HNO3), sulfuric acid (H2SO4) serves as a weaker oxidizing agent. Therefore, using H2SO4 in place of HNO3 would not yield the same experimental results.

1) Calcium carbonate comprises 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (which is 40.0%)!

2) The mass fraction mentioned is superfluous information.

3) The resulting solution is:

m(Ca)=1.2 g

m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)

m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g