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3 years ago

A sample of carbon monoxide gas has a volume of

Chemistry

1 answer:

jek_recluse [69]3 years ago

8 0

Answer: 2.3L

Explanation:

Given that:

Volume of carbon monoxide gas V1 = 3.45L

Temperature T1 = 28°C

Convert Celsius to Kelvin

(28°C + 273 = 301K)

Pressure P1 = 2.82 atm

Final Volume V2 = ?

Final temperature T2 = 28°C = 301K

Final pressure = 4.2atm

Then, apply the general gas equation

(P1V1)/T1 = (P2V2)/T2

(2.82 x 3.45)/301 = (4.2 x V2)/301

9.73/301 = 4.2V2/301

0.032 = 0.014V2

To get the value of V2, divide both sides by 0.014

0.032/0.014 = 0.014V2/0.014

2.286 = V2

Rounded to nearest tenth, V2 becomes 2.3L

Thus, the new volume of the carbon monoxide gas is 2.3 liters

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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:

Step 1: 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

Step 2: Calculate final temperature

Qlost = -Qgained

Qquartz = - Qwater

Q =m*c*ΔT

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

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⇒ 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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Answer:
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