Question

how many formula units (ionicmolecules) of mercury(i) perchlorate, hg₂(clo₄)₂, are in 1.24 g of mercury(i) perchlorate? report your answer in proper scientific notation to the correct number of sig figs by filling in the following blanks, but do not include units in your answer.

question 8
1 pts
how many milligrams are present in 6.24×10¹⁹ molecules of fecl₃? report your answer to the correct number of significant figures and rounding with the units reported in the second box.

question 9
1 pts
calculate the mass (in ng) of 6.29×10¹³ molecules of oxygen, o₂. report your answer to 2 decimal places, but do not report the units.

Explanation:

Question 7

Step1: Calculate molar mass of $Hg_2(ClO_4)_2$

The molar mass of Hg is 200.59 g/mol, Cl is 35.45 g/mol, and O is 16.00 g/mol.
$M = 2\times200.59+2\times35.45 + 8\times16.00=401.18 + 70.9+128.0 = 600.08$ g/mol

Step2: Calculate number of moles

$n=\frac{m}{M}$, where $m = 1.24$ g and $M=600.08$ g/mol.
$n=\frac{1.24}{600.08}\approx2.07\times 10^{-3}$ mol

Step3: Calculate formula - units

Use Avogadro's number $N_A = 6.022\times 10^{23}$ formula - units/mol.
$N=n\times N_A=2.07\times 10^{-3}\times6.022\times 10^{23}=1.25\times 10^{21}$

Step1: Calculate molar mass of $FeCl_3$

The molar mass of Fe is 55.85 g/mol and Cl is 35.45 g/mol.
$M = 55.85+3\times35.45=55.85 + 106.35=162.2$ g/mol

Step2: Calculate number of moles

$n=\frac{N}{N_A}$, where $N = 6.24\times 10^{19}$ molecules and $N_A=6.022\times 10^{23}$ molecules/mol.
$n=\frac{6.24\times 10^{19}}{6.022\times 10^{23}}\approx1.04\times 10^{-4}$ mol

Step3: Calculate mass in grams

$m=n\times M$, so $m = 1.04\times 10^{-4}\times162.2 = 0.0168688$ g

Step4: Convert to milligrams

$m = 0.0168688\times1000 = 16.9$ mg

Step1: Calculate molar mass of $O_2$

The molar mass of O is 16.00 g/mol, so for $O_2$, $M = 2\times16.00 = 32.00$ g/mol

Step2: Calculate number of moles

$n=\frac{N}{N_A}$, where $N = 6.29\times 10^{13}$ molecules and $N_A = 6.022\times 10^{23}$ molecules/mol.
$n=\frac{6.29\times 10^{13}}{6.022\times 10^{23}}\approx1.04\times 10^{-10}$ mol

Step3: Calculate mass in grams

$m=n\times M$, so $m=1.04\times 10^{-10}\times32.00 = 3.328\times 10^{-9}$ g

Step4: Convert to nanograms

Since 1 g = $10^{9}$ ng, $m = 3.33$ ng

Answer:

$1.25$ $21$

Question 8