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Author Question: The temperature of a sample of argon gas in a 365 mL container at 740. mmHg and 25C is lowered to ... (Read 76 times)

acwiles

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Question 1

The pressure of a gas sample was measured to be 654 mmHg. What is the pressure in kPa? (1 atm = 1.01325 × 105 Pa)
◦ 87.2 kPa
◦ 118 kPa
◦ 6.63 × 104 kPa
◦ 8.72 × 104 kPa
◦ 8.72 × 107 kPa

Question 2

The temperature of a sample of argon gas in a 365 mL container at 740. mmHg and 25°C is lowered to 12°C. Assuming the volume of the container and the amount of gas is unchanged, calculate the new pressure of the argon.
◦ 0.468  atm
◦ 0.931 atm
◦ 1.02 atm
◦ 1.54 atm
◦ 2.03 atm


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Marked as best answer by acwiles on Nov 5, 2023

viper01

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Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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acwiles

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Reply 2 on: Nov 5, 2023
Great answer, keep it coming :)


T4T

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Reply 3 on: Yesterday
Excellent

 

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