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gayaneshka [121]
3 years ago
6

3.50 liters of a gas at 727.0 K will occupy how many liters at 153.0 K?

Chemistry
1 answer:
miv72 [106K]3 years ago
8 0

Answer:

0.737 L

Explanation:

Charles law states for a fixed amount of gas, volume of the gas is directly proportional to the absolute temperature of the gas at constant pressure

we can use the following equation

V1/T1 = V2/T2

where V1 is volume and T1 is temperature at first instance

V2 is volume and T2 is temperature at the second instance

substituting the values

3.50 L / 727.0 K = V2 / 153.0 K

V2 = 0.737 L

new volume at 153.0 K is 0.737 L

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How do I find the moles of OH- which reacted (mol) in the titration. Table Attached
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Critical Thinking Challenge: A student has a part time
mina [271]

Answer:

6⅔ shifts

Explanation:

From the question given:

A shift = 4 hours

Pay = $8.25 per hour

Next, we shall determine the number of hours that will result in a pay of $220. This can be obtained as follow:

$8.25 = 1 hour

Therefore,

$220 = $220 × 1 hour / $8.25

$220 = 220/8.25 hours.

$220 = 80/3 hours

$220 = 26⅔ hours

Therefore, it will take 26⅔ hours to receive a pay of $220.

Finally, we shall determine the number of shifts in 26⅔ hours. This can be obtained as follow:

4 hours = 1 shift

Therefore,

26⅔ hours = 26⅔ ÷ 4

26⅔ hours = 80/3 × 1/4

26⅔ hours = 80/12

26⅔ hours = 20/3

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6 0
3 years ago
A 5.76 liter sample of a gas at 22.00C mL 748 torr pressure was heated to a final volume of 17.28 liters, with the pressure rema
timama [110]

Answer:

The final temperature was 612 °C

Explanation:

Charles's law relates the volume and temperature of a certain amount of ideal gas, maintained at a constant pressure, using a constant of direct proportionality. In this law, Charles says that at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T}=k

When you want to study two different states, an initial and a final one of a gas and evaluate the change in volume as a function of temperature or vice versa, you can use the expression:

\frac{V1}{T1} =\frac{V2}{T2}

In this case:

  • V1= 5.76 L
  • T1= 22 °C= 295 °K (Being 0°C=273°K)
  • V2=17.28 L
  • T2=?

Replacing:

\frac{5.76 L}{295 K} =\frac{17.28 L}{T2}

Solving:

T2 =\frac{17.28 L}{\frac{5.76 L}{295 K}}=\frac{17.28 L*295 K}{5.76 L}

T2= 885 °K = 612 °C

<u><em>The final temperature was 612 °C</em></u>

5 0
3 years ago
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