Lewis diagrams of molecules of three different hydrocarbons are shown above. Which of the following claims about the molecules i

Answer:

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

(ii) Volume of each component

(iii) Volume of solution

Assuming mixing doesn't alter the total volume.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

The isotopic mass of 47Z is calculated to be 46.96 amu. Isotopes of a single element differ in neutron count, and to ascertain the relative atomic mass, we consider each isotope's mass weighted by their natural abundance. This provided a computation to derive the mass of 47Z.

Cu(NO3)2 --> MM187.5558 NiNO3 *COEF2* --> 120.6983

Convert HCl and H2O to moles.

36.0 g of HCl = 0.987 moles HCl

98.0 g of H2O = 5.44 moles H2O

Based on the stoichiometric ratio for HCl,
there are 0.987 moles of H and 0.987 moles of Cl.

For H₂O, according to the stoichiometric ratio, you have 10.88 moles of H and 5.44 moles of O.

Combining them:
11.867 moles H
0.987 moles Cl
5.44 moles O

Revert the moles back to grams, then divide by the total mass and multiply by 100 for the percentage by mass.

11.867 moles H = 11.96 g H
0.987 moles Cl = 34.99 g Cl
5.44 moles O = 87.03 g O

11.96/(36.0+98.0)(100) = 8.93% for H
34.99/(36.0+98.0)(100) = 26.11% for Cl
87.03/(36.0+98.0)(100) = 64.96% for O.

First convert grams of C4H10 to moles using its molar mass of 58.1 g/mol: 3.50 g C4H10 × (1 mol C4H10 / 58.1 g C4H10) = 0.06024 mol C4H10 Next convert moles to molecules using Avogadro’s number: 0.06024 mol C4H10 × (6.022×10^23 molecules C4H10 / 1 mol C4H10) = 3.627×10^22 molecules C4H10 Each butane molecule contains 4 carbon atoms, so: 3.627×10^22 molecules C4H10 × (4 atoms C / 1 molecule C4H10) = 1.45×10^23 carbon atoms present.