Because Kevin is so strong, he is able to move the crate in 10 seconds. How much power, in watts, does he use?

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.

6 0

3 years ago

What is the total number of atoms in the molecule

serg [7]

Answer:

Sulfuric acid contains 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms.

Explanation:

8 0

2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$