A8.65-l container holds a mixture of two gases at 11 °c. the partial pressures of gas a and gas b, respectively, are 0.205 atm and 0.658 atm. if 0.200 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

 The  total pressure  = 1.402 atm

calculation

Total  pressure = partial  pressure  of gas A + partial pressure of gas B +  partial pressure  of third gas

partial  pressure  of gas A= 0.205 atm

Partial pressure of gas B =0.658 atm

partial pressure for third gas is calculated using ideal  gas equation

that is PV=nRT   where,

p(pressure)=? atm

V(volume) = 8.65 L

n(moles)= 0.200 moles

R(gas constant)=0.0821 L.atm/mol.k

T(temperature) = 11°c into kelvin =11+273 =284 k

make  p the subject of the formula by  diving both side by V

p =nRT/v

p = [(0.200 moles x 0.0821 L.atm/mol.K x 284 K)/8.65L)] =0.539 atm

Total  pressure  is therefore = 0.205 atm +0.658 atm +0.539 atm

=1.402 atm

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

the answer would be a