Question

question 1 (1 point) saved calculate the number of atoms in 0.2505 mol of rhodium 1.51×10²² atoms 1.51×10²³ atoms 1.61×10²³ atoms 1.51×10²⁴ atoms question 2 (1 point) saved what is the mass of 0.5623 mol of copper (ii) chloride? 135 g 67.0 g 112 g 75.6 g question 3 (1 point) saved how many formula units are in 34.15 g of potassium nitrate?

Explanation:

Question 1

Step1: Recall Avogadro's number

The number of atoms in 1 - mole of any substance is $6.022\times10^{23}$ atoms/mol.

Step2: Calculate number of atoms

We have $n = 0.2505$ mol of rhodium. The number of atoms $N=n\times N_A$, where $N_A = 6.022\times10^{23}$ atoms/mol. So $N=0.2505\times6.022\times 10^{23}=1.508511\times10^{23}\approx1.51\times 10^{23}$ atoms.

Step1: Find molar - mass of $CuCl_2$

The molar mass of $Cu$ is $M_{Cu}=63.55$ g/mol, and the molar mass of $Cl$ is $M_{Cl}=35.45$ g/mol. For $CuCl_2$, the molar mass $M = 63.55+2\times35.45=63.55 + 70.9=134.45$ g/mol.

Step2: Calculate mass

We know that $n = 0.5623$ mol and $m=n\times M$. So $m=0.5623\times134.45 = 75.599235\approx75.6$ g.

Step1: Find molar - mass of $KNO_3$

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

Step2: Calculate number of moles

$n=\frac{m}{M}$, where $m = 34.15$ g and $M = 101.11$ g/mol. So $n=\frac{34.15}{101.11}\approx0.3378$ mol.

Step3: Calculate formula units

The number of formula units $N=n\times N_A$, where $N_A = 6.022\times10^{23}$ formula units/mol. So $N=0.3378\times6.022\times10^{23}=2.0342216\times10^{23}\approx2.03\times 10^{23}$ formula units.

Answer:

$1.51\times 10^{23}$ atoms

Question 2