Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
Answer:
Explanation has been given below.
Explanation:
- Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
- Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
- First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
- Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
- Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
Answer:
See Below
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