Hey there!:
Given the mass of PbCl(OH) :
0.135 Kg = 0.135 Kg*(1000g / 1Kg) = 135 g
Molecular mass of PbCl(OH) = 207+35.5+16+1 = 259.5 g / mol
Atomic mass of Pb = 207 g/mol
Hence mass of Pb in 135 g PbCl(OH) :
(207 g Pb / 259.5 g PbClOH) * 135g PbClOH =
0.79768 * 135 => 107.68 g of Pb
For Pb2Cl2CO3 :
Given the mass of Pb2Cl2CO3 :
0.135 Kg = 0.135 Kgx(1000g / 1Kg) = 135 g
Molecular mass of Pb2Cl2CO3 = 2*207+2*35.5+12+3*16 = 545 g / mol
Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3 = 2*207 = 414 g
Hence mass of Pb in 135 g Pb2Cl2CO3:
(414 g Pb / 545 g PbClOH) * 135g PbClOH =
0.75963 * 135 => 102.55 g of Pb2Cl2CO3
Hope that helps!
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
The last answer
the place where the organisms live
Hope this helps!!! :))))
