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2 years ago

How many moles are present in 63.80L of oxygen gas, O2?

Chemistry

1 answer:

matrenka [14]2 years ago

4 0

Answer:

2.85moles of oxygen gas

Explanation:

Given parameters:

Volume of oxygen gas = 63.8L

Unknown:

Number of moles = ?

Solution:

We assume that the gas is under standard temperature and pressure. To find the number of moles, use the expression below:

1 mole of a gas at STP occupies a volume of 22.4L

So;

63.8L of oxygen gas will take up a volume of

$\frac{63.8}{22.4}$ = 2.85moles of oxygen gas

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What does this equation, 2 C(s) + O 2(g) ® 2 CO(g), tell us?

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Answer:

D. All of these.

Explanation:

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A glucose solution has a density of 1.02 g/ml. what is its specific gravity

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In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal. She heats 19.6 grams of z

storchak [24]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc T1= 98.37 °C

Initial temperature of water T1= 24.16 °C

Final temperature of water (and zinc) T2 = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

Qzinc = -Qwater

m(zinc)*c(zinc)* ΔT(zinc) = - m(water)*c(water)* ΔT(water)

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

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2 years ago