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3 years ago

At what temperature will 2.40 moles of chlorine gas exert a pressure of 2.70 atm at a volume of 0.750 L?

Chemistry

2 answers:

Kamila [148]3 years ago

8 0

Answer:

—262.71°C

Explanation:

Step 1:

The following data were obtained from the question:

Number of mole (n) = 2.4 moles

Pressure (P) = 2.70 atm

Volume (V) = 0.750 L

Temperature (T) =?

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the temperature.

Using the ideal gas equation, the temperature can be obtained as follow:

PV = nRT

2.7 x 0.750 = 2.4 x 0.082 x T

Divide both side by 2.4 x 0.082

T = (2.7 x 0.750) /(2.4 x 0.082)

T = 10.29K

Step 3:

Conversion of Kelvin temperature to celsius temperature.

Temperature (celsius) = temperature (Kelvin) - 273

temperature (Kelvin) = 10.29K

Temperature (celsius) = 10.29 - 273

Temperature (celsius) = —262.71°C

DochEvi [55]3 years ago

6 0

Answer:

10.28Kelvin

Explanation:

Using the ideal gas equation;

PV = nRT

P is the pressure

V is the volume of the gas

n is the number of moles

T is the temperature in Kelvin

R is the Gas constant

Given n = 2.4moles

P = 2.70atm

V = 0.750L

R = 0.0821atm.L/mol.K

From the formula above:

T = PV/nR

T = 2.70×0.750/2.4×0.0821

T = 2.025/0.197

T = 10.28K

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Margaret [11]

The identity of the noble gas is the sample is Krypton

<h3>Ideal Gas law</h3>

From the question, we are to determine the identity of the noble gas in the sample

From the ideal gas equation, we have that

PV = nRT

∴ n = PV / RT

Where P is the pressure

V is the volume

n is the number of moles

R is the gas constant

and T is the temperature

From the given information,

P = 678 mmHg = 0.892105 atm

V = 363 mL = 0.363 L

R = 0.08206 L.atm/mol.K

T = 35 °C = 35 + 273.15 K = 308.15 K

Putting the parameters into the equation, we get

n = (0.892105 × 0.363)/ (0.08206 × 308.15)

n = 0.0128 moles

Now, we will determine the Atomic mass of the sample

Using the formula,

Atomic = Mass / Number of moles

Atomic mass of the substance = 1.07 / 0.0128

Atomic mass of the substance = 83.6 amu

The noble gas with the closest atomic mass to this value is Krypton.

Molar mass of Krypton = 83.798 amu

Hence, the identity of the noble gas is the sample is Krypton

Learn more on Ideal Gas law here: brainly.com/question/20212888

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2 years ago

A solution contains 90 milliequivalents of HC1 in 450ml. What is its normality?

Rashid [163]

Answer:

Normality N = 0.2 N

Explanation:

Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.

Normality is represented by N.

Mathematically, we have :

$\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}$

Given that:

number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent

volume of solution (HCl) = 450 mL 450 × 10⁻³ L

$\mathbf{Normality \ N = \dfrac{90 \times 10^{-3}}{450 \times 10^{-3}}}$

Normality N = 0.2 N

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3 years ago

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ivolga24 [154]

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