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olga55 [171]
3 years ago
13

2. A sample of gas has a pressure of 195 kPa and a volume of 3.50 L. If the pressure

Chemistry
1 answer:
zhannawk [14.2K]3 years ago
7 0

Answer:

<h2>4.55 L</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we're finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

We have

V_2 =  \frac{195000 \times 3.5}{150000}  =  \frac{682500}{150000}  \\  = 4.55

We have the final answer as

<h3>4.55 L</h3>

Hope this helps you

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CaCo3+2HCL Cacl2+h2o+co2
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Here

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3 years ago
650. J is the same amount of energy as A. 155 cal. B. 2720 calC. 650 cal.D. 1550 cal.
sladkih [1.3K]

Answer:

A\text{ : 155 cal}

Explanation:

Here, we want to convert J to cal

Mathematically:

1\text{ cal = 4.186 J}

Thus, to get our answer in cal, we divide the value given by 4.186

We have that as:

\frac{650}{4.186}\text{ = 155 cal}

5 0
1 year ago
2N2H4+ N2O4———3N2+4H2O SalmaKhan99 avatar How many grams of N2 gas will be formed by reacting 100g of N2H4 and 200g of N2 Kindly
Alona [7]

Answer:

131.26 g

Explanation:

From the balanced equation,

2 moles of N₂H₄reacts with 1 mole of N₂O₄ to give 3 moles of N₂

Now number of moles of N₂H₄ present in 100 g N₂H₄ is n = 100 g/molar mass N₂H₄.

Molar mass N₂H₄ = 2 × 14.01 g/mol + 1 × 4 g/mol = 28.02 g/mol + 4 g/mol = 32.02 g/mol

n₁ = 100/32.02 = 3.123 mol

Also

Now number of moles of N₂O₄ present in 200 g N₂O₄ is n = 200 g/molar mass N₂O₄.

Molar mass N₂O₄ = 2 × 14.01 g/mol + 16 × 4 g/mol = 28.02 g/mol + 64 g/mol = 92.02 g/mol

n₂ = 200/92.02 = 2.173 mol

Since the mole ratio of N₂H₄  to N₂O₄ is 2 : 1, We require 2 × 2.173 mol N₂H₄  to react with 2.173 mole N₂O₄  

Number of moles of N₂H₄ required is 4.346. But the number of moles of N₂H₄  present is 3.123 so N₂H₄  is the limiting reagent.

So, from the equation, 2 moles of N₂H₄ produces 3 moles of N₂

Therefore number of mole N₂ = 3/2 moles of N₂H₄ = 3/2 × 3.123 mol = 4.6845 mol

From n = m/M where n = number of moles of nitrogen gas = 4.6845 mol and M = molar mass of nitrogen gas = 28.02 g/mol and m = mass of nitogen gas.

m = nM = 4.6845 mol × 28.02 g/mol = 131.26 g

So the mas of nitrogen gas produced is 131.26 g

4 0
3 years ago
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