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3 years ago

Complete the sentences to best explain the ranking.

Chemistry

1 answer:

Jobisdone [24]3 years ago

4 0

Answer:

dipole-dipole forces, ion-dipole forces, higher molar mass, hydrogen bonding, stronger intermolecular forces

Explanation:

1. H₂S and H₂Se exhibit the following intermolecular forces: dipole-dipole forces and ion-dipole forces. These molecules have a bent geometry, thus, a dipolar moment which makes them dipoles. When they are in the aqueous form they are weak electrolytes whose ions interact with the water dipoles

2. Therefore, when comparing H₂S and H₂Se the one with a higher molar mass has a higher boiling point. In this case, H₂Se has a higher boiling point than H₂S due to its higher molar mass.

3. The strongest intermolecular force exhibited by H₂O is hydrogen bonding. This is a specially strong dipole-dipole interaction in which the positive density charge on the hydrogens is attracted to the negative density charge on the oxygen.

4. Therefore, when comparing H₂Se and H₂O the one with stronger intermolecular forces has a higher boiling point. That's why the boiling point of H₂O is much higher than the boiling point of H₂Se.

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4 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

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

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

3 years ago

Please answer, with explanation. Thanks!

nadya68 [22]

Answer:

Explanation:

a = 40.1 g of Ca