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viktelen [127]
3 years ago
8

A glass sphere is filled to full volume with a gas. The pressure of the gas inside the sphere is 30.0 atm, and the temperature i

s 30.0°C. The sphere is taken outside on a cold day. The temperature of the gas decreases to 10.0°C. What is the new pressure of the gas? Assume that the volume is constant.
Chemistry
1 answer:
harkovskaia [24]3 years ago
4 0

Answer:

28atm

Explanation:

Using Gay lussac's law equation as follows:

P1/T1 = P2/T2

Where;

P1 = initial pressure (atm)

T1 = initial temperature (K)

P2 = final pressure (atm)

T2 = final temperature (K)

Based on the information provided in this question;

P1 = 30.0 atm, T1 = 30.0°C, P2 = ?, T2 = 10.0°C

NOTE: Absolute temperature i.e. Kelvin is required for this law

T1 = 30°C + 273K = 303K

T2 = 10°C + 273K = 283K

Using P1/T1 = P2/T2

30/303 = P2/283

Cross multiply

P2 × 303 = 30 × 283

303P2 = 8490

P2 = 8490/303

P2 = 28.02

New pressure of the gas = 28atm

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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
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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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A 48.0g sample of quartz, which has a specific heat capacity of 0.730·J·g−1°C−1, is dropped into an insulated container containi
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Answer:

The equilibrium temperature of the water is 26.7 °C

Explanation:

<u>Step 1:</u> Data given

Mass of the sample quartz = 48.0 grams

Specific heat capacity of the sample = 0.730 J/g°C

Initial temperature of the sample = 88.6°C

Mass of the water = 300.0 grams

Initial temperature = 25.0°C

Specific heat capacity of water = 4.184 J/g°C

<u>Step 2:</u> Calculate final temperature

Qlost = -Qgained

Qquartz = - Qwater

Q =m*c*ΔT

Q = m(quartz)*c(quartz)*ΔT(quartz) = -m(water) * c(water) * ΔT(water)

⇒ mass of the quartz = 48.0 grams

⇒ c(quartz) = the specific heat capacity of quartz = 0.730 J/g°C

⇒ ΔT(quartz) = The change of temperature of the sample = T2 -88.6 °C

⇒ mass of water = 300.0 grams

⇒c(water) = the specific heat capacity of water = 4.184 J/g°C

⇒ ΔT= (water) = the change in temperature of water = T2 - 25.0°C

48.0 * 0.730 * (T2-88.6) -300.0 * 4.184 *(T2 - 25.0)

35.04(T2-88.6) = -1255.2 (T2-25)

35.04T2 -3104.544 = -1255.2T2 + 31380

1290.24T2 = 34484.544

T2 = 26.7 °C

The equilibrium temperature of the water is 26.7 °C

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