How many moles of gas are present in 57.20 L of argon at a pressure of 1 atm and a temperature of 0°C?

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20 POINTS

n= 2.55 moles

Explanation:

Using the formula of ideal gas law:

PV = nRT

  nRT=PV

n= PV/RT

  n= number of moles

  R= Avogadro constant = 0.0821

  T= Temperature in K => ºC + 273.15 K

  V= volume in L

  P= pressure in atm

n= (1 atm)(57.20 L) / (0.0821)(237.15 K)

n= 2.50 moles

b. examine and describe cells

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml