All categories

3 years ago

What is the balanced equation for KOH+CO2—>K2CO3+H2O

Chemistry

1 answer:

insens350 [35]3 years ago

3 0

The balanced equation :

2KOH+CO₂⇒K₂CO₃+H₂O

<h3>Further explanation</h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

Reaction

KOH+CO₂⇒K₂CO₃+H₂O

give coeffiecient

aKOH+bCO₂⇒K₂CO₃+cH₂O

make equation

K, left=a, right=2⇒a=2

H, left=a, right=2c⇒a=2c⇒2=2c⇒c=1

O, left=a+2b, right=3+c⇒a+2b=3+c⇒2+2b=3+1⇒2b=2⇒b=1

the equation becomes :

2KOH+CO₂⇒K₂CO₃+H₂O

You might be interested in

Substances like neon , which is a gas at room temperature and pressure, can often be liquified or solidified only at very low te

marshall27 [118]

Answer:

Condenses at 27.25K.

Freezes at 24.65K.

Explanation:

In order to solve this above question, there is is need to make use of the following equation. The main idea here is to convert degree celsius to Kelvin. Hence,

0°C + 273.15 = 273.15K---------------------(1).

Therefore, we will make use of the above equation (1) and slot in the values for at degree celsius at which it condenses and at degree celsius at which it freezes.

So, we have at temperature at which it condenses:

-245.9°C + 273.15 = 27.25K.

Also, we have at temperature at which it freezes.

-248.5°C + 273.15 = 24.65K.

8 0

3 years ago

Which statments regarding the henderson-hasselbalch equation are true?

ziro4ka [17]

Complete question is;

Which statements regarding the Henderson-Hasselbalch equation are true?

1. If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined.

2. At pH = pKa for an acid, [conjugate base] = [acid] in solution.

3. At pH > pKa for an acid, the acid will be mostly ionized.

4. At pH < pKa for an acid, the acid will be mostly ionized.

A. All of the listed statements are true. B. 1, 2, and 3 are true.

C. 2, 3, and 4 are true.

D. 1, 2, and 4 are true.

Answer:

B. 1, 2, and 3 are true.

Explanation:

The formula for the Henderson-Hasselbalch equation is:

pH = pka + log₁₀([A^(-)]/[HA])

Where;

PH is acidity of solution

ka is acid dissociation constant

A^(-) is concentration of conjugate base

HA is concentration of Acid

- For statement 1; If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined;

pH = pka + log₁₀([A^(-)]/[HA])

pH - pka = log₁₀([A^(-)]/[HA])

10^(pH - pka) = ([A^(-)]/[HA])

Since we can find the ratio as seen, then the statement is true

- For statement 2: At pH = pKa for an acid, [conjugate base] = [acid] in solution;

We will substitute pH for pKa;

pH = pH + log₁₀([A^(-)]/[HA])

This give;

0 = log₁₀([A^(-)]/[HA])

10^(0) = [A^(-)]/[HA]

1 = [A^(-)]/[HA]

Thus; [A^(-)] = [HA]

Thus, the statement is true

- For statement 3: At pH > pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is greater than 0 and thus;

10^(pH - pKa) is greater than 1.

Thus;

[A^(-)]/[HA] > 1

[A^(-)] > [HA]

So more acid is ionized than base.

So the statement is true.

- For statement 4: At pH < pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is less than 0 and thus;

10^(pH - pKa) is less than 1.

Thus;

[A^(-)]/[HA] < 1

[A^(-)] < [HA]

So we have more base ionized than acid. So statement is false

7 0

3 years ago

A chemist needs to create a series of standard Cu2+(aq) solutions for an absorbance experiment. For the first standard, he uses

morpeh [17]

Answer:

0.01 M

Explanation:

The chemist is performing a serial dilution in order tyo obtain the calibration curve for the instrument.

First we must obtain the concentration of the solution in the 250ml flask from

C1V1 = C2V2

Where;

C1 = concentration of the stock solution

V1 = volume of the stock solution

C2 = concentration of the diluted solution

V2= volume of the diluted solution

2.61 × 10 = C2 × 250

C2 = 2.61 × 10/250

C2 = 0.1 M

Hence for solution in 100ml flask;

0.1 × 10 = C2 × 100

C2 = 0.1 × 10/100

C2 = 0.01 M

6 0

3 years ago

What is the entry qualification of optician

IRISSAK [1]

Answer:

Opticians typically have a high school diploma or equivalent and receive some form of on-the-job training. Some opticians enter the occupation with an associate's degree or a certificate from a community college or technical school. About half of the states require opticians to be licensed.

5 0

3 years ago

Someone please help me with the second question ONLY

wlad13 [49]

Answer:

Explanation:

this experiment could not be replicated, because there are no specific measurements. the details would not be the same because it does not include the type of paper used nor the diameter, width or length. A guess of the measurements would be the only way to replicate this experiment but other then that, no you cannot.

4 0

3 years ago