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Vlad [161]
4 years ago
9

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]4 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]4 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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Why can’t there be a temperature lower than absolute zero
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Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
Eddi Din [679]

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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