All categories

2 years ago

What is the ph of a solution labeled .30 trimethylamine k for trimethylamine is 7.42 x 10^4

Chemistry

1 answer:

valentina_108 [34]2 years ago

5 0

Answer:the pH is 12

Explanation:

First We need to understand the structure of trimethylamine

Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline

C3H9N+ H2O --> C3H9NH + OH-

$k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}$

Then:

The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine

[OH-]= $\sqrt{ 0.3*7.42x10^{-4}}$

[OH-]=0.01

pH=14-(-log[OH-])

pH=12

You might be interested in

Calculate the grams of solute in each of the following solution: 278 mL of 0.038 M Fe2(SO4)3

Goryan [66]

Answer: 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Now put all the given values in the formula of molality, we get

$0.038M=\frac{n}{0.278L}$

$n=0.0105mol$

mass of $Fe_2(SO_4)_3$ = $moles\times {\text {Molar Mass}}=0.0105\times 399.88g/mol=4.22g$

Thus 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

3 0

3 years ago

Tetrachloromethane, CCl4 is produced from the substitution reaction between methane and chlorine gas. If the rate of formation o

Korolek [52]

The rate of disappearance of chlorine gas : 0.2 mol/dm³

<h3>Further explanation</h3>

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

For reaction :

$\tt aA+bB\rightarrow cC+dD$

The rate reaction :

$\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}$

Reaction for formation CCl₄ :

<em />

From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³

Rate of formation of CCl₄ = reaction rate x coefficient of CCCl₄

0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³

The rate of disappearance of chlorine gas (Cl₂) :

Rate of disappearance of Cl₂ = reaction rate x coefficient of Cl₂

Rate of disappearance of Cl₂ = 0.05 x 4 = 0.2 mol/dm³

4 0

2 years ago

The lowest layer of Earths atmosphere is the__

alina1380 [7]

Answer:

Troposphere

Explanation:

The troposphere is the lowest layer in the atmosphere, and where all the weather occurs. After the troposphere, there's the stratosphere, mesosphere, thermosphere, and exosphere.

I hope this helps!

7 0

2 years ago

Read 2 more answers

31. calculate the amount of energy released as heat by the freezing of 13.3 g of a liquid substance, given that the substance ha

spin [16.1K]

From the calculations, the heat of fusion of the substance is 0.73 kJ

<h3>What is is the heat of freezing?</h3>

The heat of freezing is the energy released when the substance is converted from liquid to solid.

Now we know that the molar mass of the substance is 82.9 g/mol hence the number of moles of the substance is; 13.3 g /82.9 g/mol = 0.16 moles

Now the heat of fusion shall be;

H = 4.60 kj/mol * 0.16 moles

H = 0.73 kJ

Learn more about freezing:brainly.com/question/3121416

#SPJ4

5 0

1 year ago

7.0 mol Mn reacts with 5.0 mol

Elenna [48]

The moles of MnO formed with the reaction of 5 moles of Mn is 3 moles.

<h3>What is a limiting reagent?</h3><h3 />

In a chemical reaction, the reagent that is present in a lesser quantity and governs the rate of the reaction is termed as the limiting reagent.

In the reaction of the formation of MnO, according to the stoichiometric law 2 moles of Mn reacts with 1 moles of Oxygen.

Thus, the moles of oxygen consumed by 7 moles of Mn is:

2 moles Mn = 1 mole O₂

7 moles Mn = 3.5 moles O₂

The available moles of O₂ = 5 mol.

The remaining moles of O₂ = 5-3.5 moles

The remaining moles of O₂ = 1.5 moles

The reaction of 5 moles Mn requires 2.5 moles of O₂, whereas the available moles of oxygen is 1.5 moles. Thus, oxygen serves as the limiting reagent.

The moles of MnO formed with the reaction of 5 moles Mn and 1.5 moles O₂

1 mole O₂ = 2 moles MnO

1.5 moles O₂ = 2 * 1.5 moles MnO

1.5 moles O₂ = 3 moles MnO

Thus, the moles of MnO formed with the reaction of 5 moles of Mn with the available oxygen is 3 moles.

Learn more about stoichiometric law, here:

brainly.com/question/14465605

#SPJ1

6 0

1 year ago

Read 2 more answers