At 25°C and 754 torr, 2.50 moles of Ne occupies a volume of 60.6 L. At constant temperature and pressure, how many moles of Ne must be remov

Question

At 25°C and 754 torr, 2.50 moles of Ne occupies a volume of 60.6 L. At constant temperature and pressure, how many moles of Ne must be removed to reduce the volume to 45.0 L?

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Jezebel 4 years 2021-08-28T10:02:52+00:00 2 Answers 11 views 0

Answers ( )

  1. Verity
    0
    2021-08-28T10:03:52+00:00
    Reply

    Answer:


    number of moles removed = 0.6436 moles

    Explanation:

    given data

    volume v1 = 60.6 L

    volume v2 = 45.0 L

    moles of Ne n1 = 2.50 moles

    temperature =  25°C

    solution

    we will use here gas law  that is express as

    PV = nRT   …………………….1

    so here we get here  number of moles

    so equation will be as

    P1 × V1 = n1 × R × T1

    and

    P2 × V2 = n2 × R × T2

    so from above two equation we can say at pressure and temperature constant so

    \frac{V1}{V2} =\frac{n1}{n2}    ……………2

    put here value we get

    \frac{60.6}{45} =\frac{2.50}{n2}

    n2 = 1.856 moles

    so 1.856 moles occupied when volume is 45 L

    so here number of moles removed will be

    number of moles removed = initial moles – final moles    ……………3

    put here value

    number of moles removed = 2.50 – 1.856


    number of moles removed = 0.6436 moles

  2. danthu
    0
    2021-08-28T10:04:15+00:00
    Reply

    Answer:

    Explanation:

    P1 = 754 torr

    T1 = 25°C

    n1 = 2.5 moles

    V1 = 60.6 L

    V2 = 45 L

    Use ideal gas equation

    P1 x V1 = n1 x R T1 …. (1)

    Here pressure and the temperature remains constant

    P1 x V2 = n2 x R x T1 ….. (2)

    Divide equation (1) by (2)

    V1 / V2 = n1 / n2

    60.6 / 45 = 2.5 / n2

    n2 = 1.856

    So, the number of moles removed is 2.5 – 1.856 = 0.644

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