Question

how many molecules of tartrazine, na3c16h9n4o9s2, are in 0.631 mol of tartrazine? 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 5
1 pts
if there are 5.19 x 10^25 iron atoms, how many moles of iron do you have? report your answer to 2 decimal places.

question 6
1 pts
how many formula units (ionic molecules) of mercury(i) perchlorate, hg2(clo4)2, are in 0.145 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.

Explanation:

Question 1

Step1: Recall Avogadro's number

Avogadro's number $N_A = 6.022\times 10^{23}$ molecules/mol. To find the number of molecules of tartrazine, we multiply the number of moles by Avogadro's number.
$n = 0.631$ mol, and the number of molecules $N=n\times N_A$.
$N = 0.631\times6.022\times 10^{23}$

Step2: Calculate the result

$0.631\times6.022 = 3.790882$. In scientific - notation and correct significant figures (3 sig - figs), $N = 3.79\times 10^{23}$

Step1: Use the relationship between moles and number of atoms

The number of moles $n$ of a substance is related to the number of atoms $N$ by the formula $n=\frac{N}{N_A}$, where $N_A = 6.022\times 10^{23}$ atoms/mol. Given $N = 5.19\times 10^{25}$ atoms.
$n=\frac{5.19\times 10^{25}}{6.022\times 10^{23}}$

Step2: Perform the division

$\frac{5.19\times 10^{25}}{6.022\times 10^{23}}=\frac{5.19}{6.022}\times10^{25 - 23}\approx0.862\times 10^{2}=86.19$

Step1: Calculate the 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 the number of moles

$n=\frac{m}{M}$, where $m = 0.145$ g. So $n=\frac{0.145}{600.08}\approx2.42\times 10^{-4}$ mol

Step3: Calculate the number of formula units

The number of formula units $N=n\times N_A$, where $N_A = 6.022\times 10^{23}$ formula units/mol.
$N=2.42\times 10^{-4}\times6.022\times 10^{23}=2.42\times6.022\times 10^{-4 + 23}=14.57324\times 10^{19}$
In proper scientific - notation and 3 sig - figs, $N = 1.46\times 10^{20}$

Answer:

$3.79$; $23$

Question 5