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sergij07 [2.7K]
3 years ago
11

If 74.5 m of oxygen are collected at a pressure of 98.0 kPa, what volume will the gas occupy if the

Chemistry
1 answer:
sergey [27]3 years ago
4 0

Answer:

80.8 mL

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 74.5 mL

Initial pressure (P₁) = 98.0 kPa

Final pressure (P₂) = 90.4 kPa

Final volume (V₂) =?

The final volume of the gas can be by using the Boyle's laws equation as follow:

P₁V₁ = P₂V₂

98 × 74.5 = 90.4 × V₂

7301 = 90.4 × V₂

Divide both side by 90.4

V₂ = 7301 / 90.4

V₂ = 80.8 mL

Thus, the volume of the gas is 80.8 mL

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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
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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}}

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

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<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

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Explanation:

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Answer:

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