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2 years ago

Approximately what volume do 3.0 moles of kr gas occupy at stp? 80.0 40.0 l none of the above 20.0

Chemistry

2 answers:

denpristay [2]2 years ago

6 0

Answer:

The answer is: none of the above

Explanation:

1 mol of gas ideal occupies a volume of 22.4 L, then 3 moles of Kr gas occupies, using the following rule of three:

1 mol ------------ 22.4 L

3 moles--------- X

Clearing X:

$X=\frac{3moles*22.4L}{1mol} =67.2L$

podryga [215]2 years ago

4 0

None of the above.
1 mole filled with gas at STP occupies
=22.4 L

∴ 3mole of kr gas at STP occupies
= 3 × 22.4
= 67.2 L

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2 years ago

Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated

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Complete Question

The complete question is shown on the first uploaded image

Answer

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

$\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0$

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Explanation

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i.e $w = -P_{external} \Delta V$

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i.e $\Delta E = q + w$

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Generally the change in enthalpy is mathematically represented as

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2 years ago

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Ludmilka [50]

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2 years ago

Read 2 more answers

A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atmatm at 20°C.Calculate the van't Hoff factor iii for AlCl

Alja [10]

Answer:

The actual Van't Hoff factor for AlCl3 is 3.20

Explanation:

Step 1: Data given

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Step 2: Calculate the Van't Hoff factor

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The theoretical value is 4 ( because 1 Al^3+ ion + 3 Cl- ions) BUT due to the interionic atractions the actual value will be less

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π = i.M.R.T

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⇒ with i = the van't Hoff factor

⇒ with M = the molar concentration of the solution = 0.050 M

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 20 °C = 293.15 Kelvin

i = π /(M*R*T )

i = (3.85) / (0.050*0.08206*293.15)

i = 3.20

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2 years ago

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Answer:

Less than

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2 years ago