First solve the moles of oxgen present in the compound
mol O = 6.93 g O ( 1 mol O / 16 g O )
mol O = 0.43 mol H
then solve the moles of hydrogen present
mol H = ( 7.36 - 6.93) g H ( 1 mol H / 1 g H)
mol H = 0.43 mol H
so the O and H are in the same mole content so the molecular formula would be OH, but the molar mass will not satisfy. so the answer would be
H2O2
It becomes a liquid with the water
Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %
The ribosomes are the ones delivering the products of the endoplasmic reticulum
1-pentyne consists of a carbon chain of 5 carbons one with a triple bond. 1-octyne is a carbon chain of 8 carbons with a triple bond at some point. It is known that the longer the carbon chain the higher the boiling point since more energy will be required to break the bonds between carbons. Based on this it is predicted that 1-octyne will have a higher boiling point than 1-pentyne.
