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2 years ago

What is the temperature of 0.750 mol of a gas stored in a 6,050 mL cylinder al 221 atm?

Chemistry

1 answer:

Rasek [7]2 years ago

4 0

Answer:

T = 246 K

Explanation:

Given that,

Number of moles, n = 0.750 mol

The volume of the cylinder, V = 6850 mL = 6.85 L

Pressure of the gas, P = 2.21 atm

We need to find the temperature of the gas stored in the cylinder. We know that,

PV= nRT

Where

R is gas constant

T is temperature

So,

$T=\dfrac{PV}{nR}\\\\T=\dfrac{2.21\times 6.85}{0.75\times 0.0821}\\T=245.85\ K$

T = 246 K

So, the temperature of the gas is equal to 246 K.

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Answer:

Fractional distillation

Explanation:

Heat the mixture upto 700°C

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2 years ago

What is the frequency of a particular type of electromagnetic

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Answer:

f = 0.6× 10¹⁶ Hz

Explanation:

Given data:

Wavelength of radiation = 5.00 × 10⁻⁸ m

Frequency of radiation = ?

Solution:

Formula:

Speed of light = wavelength × frequency

speed of light = 3 × 10⁸m/s

Now we will put he values in formula.

3 × 10⁸m/s = 5.00 × 10⁻⁸ m × f

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2 years ago

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

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Answer:

208.7°C was the initial temperature of the limestone.

Explanation:

Heat lost by limestone will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of limestone = $m_1=62.6 g$

Specific heat capacity of limestone = $c_1=0.921 J/g^oC$

Initial temperature of the limestone = $T_1=?$

Final temperature = $T_2$ =T = 51.9°C

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.0 g$

Specific heat capacity of water= $c_2=4.186 J/g^oC$

Initial temperature of the water = $T_3=23.1^oC$

Final temperature of water = $T_2$ =T = 51.9°C

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)$

$T_1=208.7^oC$

208.7°C was the initial temperature of the limestone.

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