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olasank [31]
3 years ago
15

A 2.50L sample of nitrogen gas at a temperature of 308K has a pressure of 1.15atm. What is the new volume of the gas if the pres

sure is increased to 1.80atm and the temperature is decreased to 286K?
Chemistry
1 answer:
iris [78.8K]3 years ago
7 0

Answer:

1.48L

Explanation:

From the question given, we obtained the following data:

V1 (initial volume) = 2.50L

T1 (initial temperature) = 308K

P1 (initial pressure) = 1.15atm

P2 (final pressure) = 1.80atm

T2 (final temperature) = 286K

V2 (final volume) =..?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume of the gas can obtain as follow:

P1V1/T1 = P2V2/T2

1.15 x 2.5/308 = 1.8 x V2/286

Cross multiply to express in linear form:

308 x 1.8 x V2 = 1.15 x 2.5 x 286

Divide both side by 308 x 1.8

V2 = (1.15 x 2.5 x 286)/(308 x 1.8)

V2 = 1.48L

Therefore, the new volume of the gas is 1.48L

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Calculate the weight of 3.491 into 10 to the power 19 molecules of cl2​
Rom4ik [11]

Answer:

The mass of 3.491 × 10¹⁹ molecules of Cl₂  of Cl₂ is 4.11 × 10⁻³ grams

Explanation:

The number of particles in one mole of a substance id=s given by the Avogadro's number which is approximately 6.023 × 10²³ particles

Therefore, we have;

One mole of Cl₂ gas, which is a compound, contains 6.023 × 10²³ individual molecules of Cl₂

3.491 × 10¹⁹ molecules of Cl₂ is equivalent to (3.491 × 10¹⁹)/(6.023 × 10²³) = 5.796 × 10⁻⁵ moles of Cl₂

The mass of one mole of Cl₂ = 70.906 g/mol

The mass of 5.796 × 10⁻⁵ moles of Cl₂ = 70.906 × 5.796 × 10^(-5) = 4.11 × 10⁻³ grams

Therefore;

The mass of 3.491 × 10¹⁹ molecules of Cl₂  of Cl₂ = 4.11 × 10⁻³ grams.

4 0
3 years ago
How many moles of H2 are in a flask with a volume of 2500 mL at a pressure of 30.0 kPa and a temperature of 27oC?
quester [9]

Answer:

0.0300 moles of H₂

Explanation:

The original equation is PV = nRT. We need to change this to show moles (n).

n = \frac{PV}{RT}

It's important to convert your values to match the constant (r) in terms of units.

30.0 kPa = 0.296 atm

2500 mL = 2.50 L

27 °C = 300 K

Now, plug those values in to solve:

n = \frac{(0.296)(2.50)}{(0.0821)(300)}    -  for the sake of keeping the problem clean, I didn't include the units but you should just to make sure everything cancels out :)

Finally, you are left with n = 0.0300 moles of H₂

4 0
3 years ago
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