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eimsori [14]
2 years ago
11

A gas occupies 12.3 L at a temperature of 40.0 K. What is the volume when the temperature is increased to 60.0 K?

Chemistry
1 answer:
dimaraw [331]2 years ago
7 0

Answer:

the volume is 18.45 L

Explanation:

The computation of the volume when the temperature is increased to 60.0 K is shown below:

Since the gas occupied 12.3 L at a temperature of 40.0 k

And, the volume when the new temperature is 60.0 k

So, the volume is

= 60.0 k × 12.3 L ÷ 40.0 k

= 18.45 L

hence,  the volume is 18.45 L

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Answer:

9.39 × 10²² molecules

Explanation:

We can find the moles of gases (n) using the ideal gas equation.

P . V = n . R . T

where,

P is the pressure (standard pressure = 1 atm)

V is the volume

R is the ideal gas constant

T is the absolute temperature (standard temperature = 273.15 K)

n=\frac{P.V}{R.T} =\frac{1atm.3.50L}{(0.08206atm.L/mol.K).273.15K} =0.156mol

There are 6.02 × 10²³ molecules in 1 mol (Avogadro's number). Then,

0.156mol.\frac{6.02\times10^{23}molecules}{mol} =9.39\times10^{22}molecules

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disa [49]

Answer:9%

Explanation:

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2 years ago
What is the molar mass of cholesterol if 0.00105 mol weighs 0.406 g?
goblinko [34]
<h3>Answer: 386.67 g/mol </h3>

Explanation:

Molar Mass = Mass ÷ Mole

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0.32gcm3 .           I miss the old Kanye, straight from the 'Go Kanye

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