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QveST [7]
3 years ago
7

A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula

of the compound?
(A) SF2 (B) SF3 (C) SF4 (D) SF5 (E) SF6
Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

Answer:

The empirical formula is SF6 (option E)

Explanation:

Step 1: Data given

Mass of sulfur = 3.21 grams

Mass of fluorine = 11.4 grams

Molar mass sulfur = 32.065 g/mol

Molar mass fluorine = 19.00 g/mol

Step 2: Calculate moles

Moles = mass /molar mass

Moles sulfur = 3.21 grams / 32.065 g/mol

Moles sulfur = 0.100 moles

Moles fluorine = 11.4 grams / 19.00 g/mol

Moles fluorine = 0.600 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.100 / 0.100 = 1

F : 0.600 / 0.100 = 6

The empirical formula is SF6 (option E)

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What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm
Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

Explanation:

<u>Part I :</u>

n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}

n = 4.9983

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(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = 5\times 196.96 g

4.99 moles of gold contain = 984.8 g

Mass of {3.01\times 10^{24}} atoms of gold = 984.5 g

<u>Part II :</u>

Density of Gold = 19.32 g/cm^{3}

Volume of the cuboid = length\times breadth\times height

Volume of the gold bar =6.00\times 4.25\times 2.00

Volume of the gold bar = 51cm^{3}

Using formula,

Density = \frac{mass}{Volume}

Mass = Density\times Volume

Mass = 19.32 \times 51

Mass = 985.32 g

So, A  gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

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What is the pH of a solution that has a pOH of 5.9
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otez555 [7]

Answer:

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