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!
I don't understand what you mean??
Answer:
C14H30 -----> C7H14 + C2H4 +C5H10
Explanation:
According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than 112°C is caused by reusing samples leads to errors.
A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).
However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above 112°C.
Learn more: brainly.com/question/5325004