Answer:
Percent Yield = 94.237%
Explanation:
CO = Carbon Dioxide = Molar Mass 28g/mol
C = Carbon = 12g/mol
O = Oxygen = 16g/mol
Theoretical yield = 93.7 grams
Actual yield = 88.3 grams
Percent yield =
(actual yield
/theoretical yield
)x100
Percent Yield = (88.3/93.7)x100
Percent Yield = 94.237%
Answer:
B. There is a very large percentage of C-12.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.
In such a way, the answer will be B. There is a very large percentage of C-12.
Regards!
Answer:p1 v1 / T1 = p2v2/T2
Explanation:
CH4 + 2O2 = CO2 + 2H2O
According to molar weights :
16 gm CH4 + 64 gm O2 = 44 gm CO2 + 36 gm H2O
Since 16 gm CH4 produce 36 gm H2O
Hence 2.5 gmCH4 produce 36×2.5/16 gm H2O
= 5.265 gm of H2O
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