Question

quiz 6
question 1 calculate the number of atoms in 0.2505 mol of rhodium
question 2 what is the mass of 0.5623 mol of copper (ii) chloride?
question 3 how many formula units are in 34.15 g of potassium nitrate?
question 4 how many molecules of trinitrogen dichlorine occupy 804.2 l at stp?
question 5 if you have 453.2 g of a substance q, what is the gas density at stp if there are 6.751 mol of this substance?
question 6 the gas density of an unknown formula unit is 5.891 g/l at stp. how many atoms are in 51.56 grams of it?
question 7 find the percent composition of tin (iv) arsenide
question 8 a compound contains 16.7 g of iridium and 10.3 g of selenium. what is its empirical formula?

Explanation:

Question 1

Step1: Use Avogadro's number

The number of atoms $N$ in a substance is given by $N = n\times N_A$, where $n$ is the number of moles and $N_A=6.022\times 10^{23}\text{ atoms/mol}$. Given $n = 0.2505\text{ mol}$.
$N=0.2505\text{ mol}\times6.022\times 10^{23}\text{ atoms/mol}$

Step2: Calculate the result

$N = 0.2505\times6.022\times 10^{23}=1.508511\times 10^{23}\text{ atoms}$

Step1: Determine the molar mass of copper (II) chloride ($CuCl_2$)

The molar mass of $Cu$ is $M_{Cu}=63.55\text{ g/mol}$, and the molar mass of $Cl$ is $M_{Cl}=35.45\text{ g/mol}$. For $CuCl_2$, $M = 63.55\text{ g/mol}+2\times35.45\text{ g/mol}=63.55 + 70.90=134.45\text{ g/mol}$

Step2: Calculate the mass

The mass $m$ of a substance is given by $m=n\times M$. Given $n = 0.5623\text{ mol}$ and $M = 134.45\text{ g/mol}$
$m=0.5623\text{ mol}\times134.45\text{ g/mol}=75.50\text{ g}$ (rounded to two - decimal places)

Step1: Determine the molar mass of potassium nitrate ($KNO_3$)

The molar mass of $K$ is $M_{K}=39.10\text{ g/mol}$, $N$ is $M_{N}=14.01\text{ g/mol}$, and $O$ is $M_{O}=16.00\text{ g/mol}$. For $KNO_3$, $M=39.10\text{ g/mol}+14.01\text{ g/mol}+3\times16.00\text{ g/mol}=39.10 + 14.01+48.00 = 101.11\text{ g/mol}$

Step2: Calculate the number of moles

The number of moles $n=\frac{m}{M}$, where $m = 34.15\text{ g}$ and $M = 101.11\text{ g/mol}$. So $n=\frac{34.15\text{ g}}{101.11\text{ g/mol}}=0.3378\text{ mol}$

Step3: Calculate the number of formula units

The number of formula units $N=n\times N_A$, where $N_A = 6.022\times 10^{23}\text{ formula units/mol}$
$N=0.3378\text{ mol}\times6.022\times 10^{23}\text{ formula units/mol}=2.0342\times 10^{23}\text{ formula units}$

Answer:

$1.509\times 10^{23}\text{ atoms}$ (rounded to three - significant figures)

Question 2