All categories

3 years ago

A chemical change in which a single compound is broken down into two or more simpler products.

Chemistry

2 answers:

Sergeu [11.5K]3 years ago

8 0

Answer:

decomposition reaction

Explanation:

andreyandreev [35.5K]3 years ago

5 0

Answer:

Decomposition or cracking

Explanation:

Decomposition reaction is a chemical change in which a single compound is broken down into two or more simpler products.

For example;

A → B + C

The driving force of such reaction is the high positive heat of formation of the compound which indicates that they are highly unstable.

Some stable compounds also decompose when subjected to high temperature and pressure.

You might be interested in

Help please I don’t know the answer

Nostrana [21]

The answer is 3 It is a body of knowledge gained using inquiry and experimentation. Hope this helped!

6 0

3 years ago

Read 2 more answers

The ionization constant (Ka) of HF is 6.7 X 10 ^-4. Which of the following is true in a 0.1 M solution of this acid?

spin [16.1K]

The ionization equation is:

HF ⇄ H(+) + F(-)

The ionization constant is Ka = [H(+)] * [H(-)] / [HF]

=> [H(+)] * [F(-)] = Ka * [HF]

Given that Ka < 1

[H(+)] * [F(-)] < [HF]

Which is [HF] > [H(+)] * [F(-)] the option a. fo the list of choices.

8 0

3 years ago

Read 2 more answers

Which term describes “one or more cells that carry out all of the processes needed to sustain life”?

Sergio039 [100]

Answer:

B. Organism

Explanation:

7 0

3 years ago

Read 2 more answers

How many liters of 1.5 M potassium permanganate could be made if 152 g of the solute are available?

Misha Larkins [42]

Answer:

0.64 L

Explanation:

Recall that

n= CV where n=m/M

Hence:

m/M= CV

m= given mass of solute =152g

M= molar mass of solute

C= concentration of solute in molL-1 = 1.5M

V= volume of solute =????

Molar mass of potassium permanganate= 158.034 g/mol

Thus;

152 g/158.034 gmol-1= 1.5M × V

V= 0.96/1.5

V= 0.64 L

6 0

3 years ago

Read 2 more answers

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago