Question

practice questions:

1. how many moles are present in a sample of co₂, made up of 5.83×10²⁴ molecules?(9.68mol)
2. a sample of cyanic acid, hcn, contains 3.33×10²² atoms. how many moles of cyanic acid are in the sample?(ans.1.84×10⁻²)
3. molar mass of compounds

Explanation:

Question 4:

Step 1: Recall the formula

The formula to find the number of moles (\(n\)) from the number of molecules (\(N\)) is \(n=\frac{N}{N_A}\), where \(N_A = 6.022\times10^{23}\) molecules/mol.

Step 2: Substitute the values

Given \(N = 5.83\times10^{24}\) molecules and \(N_A=6.022\times10^{23}\) molecules/mol.

$$n=\frac{5.83\times10^{24}}{6.022\times10^{23}}$$

$$n=\frac{5.83\times10}{6.022}$$

(since \(10^{24}/10^{23}=10\))

$$n=\frac{58.3}{6.022}\approx9.68$$

mol

Step 1: Determine atoms per molecule of HCN

In HCN (Hydrogen Cyanide), the formula is \(HCN\), so each molecule has \(1\) H atom, \(1\) C atom, and \(1\) N atom. Total atoms per molecule \(= 1 + 1+ 1=3\) atoms/molecule.

Step 2: Find number of molecules

Let the number of molecules be \(N\). We know total atoms \(= 3.33\times10^{22}\) and atoms per molecule \(= 3\). So \(N=\frac{\text{Total Atoms}}{\text{Atoms per Molecule}}=\frac{3.33\times10^{22}}{3}=1.11\times10^{22}\) molecules.

Step 3: Find moles of HCN

Using \(n = \frac{N}{N_A}\), where \(N_A = 6.022\times10^{23}\) molecules/mol.

$$n=\frac{1.11\times10^{22}}{6.022\times10^{23}}$$

$$n=\frac{1.11}{6.022\times10}$$

(since \(10^{22}/10^{23}=\frac{1}{10}\))

$$n=\frac{1.11}{60.22}\approx1.84\times10^{-2}$$

mol

Answer:

The number of moles is approximately \(9.68\) mol.

Question 5: