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2 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:

1 answer:

GREYUIT [131]2 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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A chemistry student needs of isopropenylbenzene for an experiment. He has available of a w/w solution of isopropenylbenzene in a

Lera25 [3.4K]

Question:

A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answer:

The answer to the question is as follows

The mass of solution the student should use is 23.42 g.

Explanation:

To solve the question we note the following

A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution

Therefore we have 10 g of isopropenylbenzene contained in

100 g * 10 g/ 42.7 g = 23.42 g of solution

Available solution = 120 g

Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.

8 0

3 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

2 years ago

A rock dropped in a graduated cylinder raises the level of water from 20 mL to 35 mL. The rock has a mass of 45 g. What is the d

Brut [27]

I think 15 I might be wrong

7 0

3 years ago

If the pressure on a gas sample is tripled and the absolute temperature is quadrupled, by what factor will the volume of the sam

antoniya [11.8K]

Answer:

b. 4/3

Explanation:

Given data

Initial pressure: P₁

Initial volume: V₁

Initial temperature: T₁

Final pressure: P₂ = 3 P₁

Final volume: V₂

Final temperature: T₂ = 4 T₁

We can find by what factor will the volume of the sample change using the combined gas law.

$\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}}\\\frac{P_{1}.V_{1}}{T_{1}} =\frac{3P_{1}.V_{2}}{4T_{1}}\\V_{1} =\frac{3V_{2}}{4}\\V_{2}=4/3 V_{1}$

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2 years ago

solmaris [256]

Answer:

В. No, because the mass of the reactants is less than the mass of the products.

Explanation:

Chemical equation:

NaBr + Cl₂ → 2NaCl + Br₂

The given equation is not balanced because number of moles of sodium and bromine atoms are less on reactant side while more on the product side.

There are one mole of sodium and one mole of bromine atom on left side of equation while on right side there are 2 moles of bromine and 2 moles of sodium atom are present. The number of moles of chlorine atoms are balanced.

Balanced chemical equation:

2NaBr + Cl₂ → 2NaCl + Br₂

Now equation is balanced. Number of moles of sodium , chlorine and bromine atoms are equal on both side.

5 0

2 years ago