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12345 [234]
3 years ago
15

The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr

ium that occurs in an aqueous solution of trimethylamine:
Chemistry
1 answer:
GREYUIT [131]3 years ago
6 0

Answer:

The K_b expression for the weak base equilibrium is:

K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}

Explanation:

(CH_3)_3N(aq)+H_2O(l)\rightlefharpoons (CH_3)_3NH^++OH^-(aq)

The expression of the equilibrium constant of base K_c can be given as:

K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}

]K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}

As we know, water is pure solvent, we can put [H_2O]=1

K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}

So, the the K_b expression for the weak base equilibrium  is:

K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}

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The statement that best explains why magnesium and chlorine combine in a 1:2 ratio is; Magnesium has two valence electrons, and chlorine can accept one electron in its outer shell.

The number of electrons that an atom of an element has in its outermost shell determines the chemical formula of the compounds formed by atoms such elements.

Magnesium is in group 2, as such it has two electrons in its outermost shell while chlorine in group 17 only accepts one electron in its outermost shell. This one electron will give chlorine an inert gas configuration while the loss of two electrons give magnesium an inert gas configuration.

Therefore; The compound MgCl2 is formed in the ratio of 1:2 because Magnesium has two valence electrons, and chlorine can accept one electron in its outer shell.

Learn more: brainly.com/question/11527546

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How many protons and electrons in Arsenic ion
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How many bonding and ione pairs does HCN have?
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3 years ago
Calculate the heat required to melt 7.35 g of benzene at its normal melting point. Heat of fusion (benzene) = 9.92 kJ/mol Heat =
grigory [225]

Answer:

The heat required to melt 7.35 g of benzene at its normal melting point is 934.8 Joules.

The heat required to vaporize 7.35 g of benzene at its normal melting point is 2,893 Joules.

Explanation:

Mass of benzene = 7.35 g

Moles of benzene = \frac{7.35 g}{78 g/mol}=0.09423 mol

Heat fusion of benzene,\Delta H_{fus} = 9.92 kJ/mol

1) Heat required to melt 7.35 g of benzene at its normal melting point = Q

Q=\Delta H_{fus}\times 0.09423 mol

=9.92 kJ/mol\times 0.09423 mol=0.9348 kJ=934.8 J

(1 kJ = 1000 J)

2) Heat vaporization of benzene,\Delta H_{vap} = 30.7 kJ/mol

Heat required to vaporize 7.35 g of benzene at its normal melting point = Q

Q=\Delta H_{Vap}\times 0.09423 mol

=30.7 kJ/mol\times 0.09423 mol=2.893 kJ=2,993 J

(1 kJ = 1000 J)

3 0
4 years ago
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