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ololo11 [35]
3 years ago
14

If 3.13 mol of an ideal gas has a pressure of 2.33 atm and a volume of 72.31 L, what is the temperature of the sample in degrees

Celsius?
Chemistry
1 answer:
Ratling [72]3 years ago
8 0

Answer:

382.49 C degree Celsius

Explanation:

Hello,

This problem deals with understanding the ideal gas law which hopes to predict how ideal gases might behave in any given condition. I listed the formula below and we are basically just going to solve for temperature by rearranging the equation as seen on the picture (there's also other rearranged ones in case you need to solve for those).

Universal gas constant R has a value of 0.0821 L * atm/(mole * K) when working with these given units so it will be part of this equation. R value changes based on what units you have.

T = PV/nR

   = (2.33) (72.31) / (3.13)(0.0821)

   = 655.64 K

Question is asking temperature in celsius so we employ the formula attached below:

C = K - 273.15

   = 655.64-273.15

    = 382.49 degree Celsius

382.49 degree Celsius is the answer!

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3 degress?

Explanation:

5 0
3 years ago
The water bottle contains 575 grams of water at 80°C. The water eventually cools to
Jet001 [13]

Answer:

–23000 Calories.

NOTE : The negative sign indicates that heat has been loss to the student back.

Explanation:

The following data were obtained from the question:

Mass (M) of water = 575 g

Initial temperature (T1) = 80 °C

Final temperature (T2) = 40 °C

Heat (Q) transferred =.?

Next, we shall determine the change in temperature (ΔT).

This is illustrated below:

Initial temperature (T1) = 80 °C

Final temperature (T2) = 40 °C

Change in temperature (ΔT) =?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 40 – 80

Change in temperature (ΔT) = –40°C

Finally, we shall determine the heat transferred. This can be obtained as follow:

Mass (M) of water = 575 g

Change in temperature (ΔT) = –40°C

Specific heat capacity (C) of water = 1 Cal/g°C

Heat (Q) transferred =.?

Q = MCΔT

Q = 575 × 1 × –40

Q = –23000 Calories

NOTE : The negative sign indicates that heat has been loss to the student back.

8 0
4 years ago
. An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.
Tju [1.3M]

Answer:

The answer to your question is Volume = 11.4 L

Explanation:

Data

Volume 1 = V1 = 6 L

Pressure 1 = P1 = 1 atm

Temperature 1 = T1 = 22°C

Volume 2 = V2 = ?

Pressure 2 = 0.45 atm

Temperature 2 = -21°C

Process

1.- Convert temperature (°C) to °K

T1 = 273 + 22 = 295°K

T2 = 273 + (-21) = 252°K

2.- Use the combined gas law to solve this problem

                   P1V1 / T1 = P2V2 / T2

-Solve for V2

                   V2 = P1V1T2 / T1P2

-Substitution

                   V2 = (6)(1)(252) / (295)(0.45)

- Simplification

                   V2 = 1512 / 132.75

- Result

                   V2 = 11.38 L

3 0
3 years ago
A piece of potassium metal is added to water in a beaker. The reaction that takes place is 2K(s) 2H20(/) ..... 2KOH(aq) H2(g) Pr
kakasveta [241]

Answer:

Explanation:

The given reaction is exothermic . So ΔH is negative .

Gas is evolving so work done by gas is positive or w is positive.

Change in internal energy that is ΔU is negative.

q = u - w

u is negative , w is positive so q is negative .

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3 years ago
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