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3. 500.0 mL of a gas is collected at 745.0 mmHg. What will the volume, in liters, be at standard pressure?

(Boyle’s)

Givens and Unknowns:

P 1 =

V 1 =

P 2 =

V 2 =

Equation:

P 1 V 1 = P 2 V 2 (Boyle’s Law)

Substitute & Solve:

4. A gas occupies 900.0 mL at a temperature of 27.0 o C. What is the volume in milliliters, at 132.0 o C?

(Charles’)

Givens and Unknowns:

V 1 =

T 1 =

Equation:

V 1 V 2 Charles’ Law

____ = ____

T 1 T 2

Substitute & Solve:

V 2 =

T 2 =

5. Calculate the decrease in temperature, in Kelvin, when 2.00L at 20.0 o C is compressed to 1.00 L. (Charles’)

Givens and Unknowns:

V 1 =

T 1 =

Equation:

V 1 V 2 Charles’ Law

____ = ____

T 1 T 2

Substitute & Solve:

V 2 =

T 2 =

6. You have 10.2 L of methane gas at 300.0 K and 102.5 kPa pressure. What would be the temperature, in

Kelvin, of this same gas in a 2.60 L container at 104.0 kPa pressure? (Combined)

Givens and Unknowns:

P 1 =

V 1 =

n 1 =

T 1 =

Equation:

P 1 V 1 P 2 V 2

__________ = __________

n 1 T 1 n 2 T 2

Substitute & Solve:

P 2 =

V 2 =

n 2 =

T 2 =

1 Answer

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  • Jan
    Lv 7
    4 years ago
    Favorite Answer

    3. V = 490.1 mL

    4. 900.0 mL at a temperature of 27.0 o C. What is the volume in milliliters, at 132.0 o C?

    Answer:

    V/T = c, so

    V / 405.15 = 900 / 300.15

    V = 1214.8 mL

    5. Calculate the decrease in temperature, in Kelvin, when 2.00L at 20.0 o C is compressed to 1.00 L.

    answer:

    T/V = c, so

    T/1 = 293.15/2

    T = 146.6 K

    6. You have 10.2 L of methane gas at 300.0 K and 102.5 kPa pressure. What would be the temperature, in

    Kelvin, of this same gas in a 2.60 L container at 104.0 kPa pressure?

    Answer:

    pV/T = c, so

    T/pV = c'

    T/(1.026*2.6) = 300 /(1.0116*10.2)

    T = 77.6 K

    • Zach4 years agoReport

      you get the best answer!

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