Silver chloride breaks down to form silver and chlorine in the presence of light. What type of reaction is this?

Chemistry

1 answer:

Vitek1552 [10]3 years ago

7 0

This is the decomposition reaction, namely the reaction of photolysis.

2AgCl = 2Ag + Cl₂

You might be interested in

1) How many Joules of energy are released when 75g of water is heated

Pie

Answer: 23,199 J

Explanation:

8 0

1 year ago

What is the degree of oxidation of a simple substance

marissa [1.9K]

Answer: The oxidation state of a free element (uncombined element) is zero. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. For example, Cl– has an oxidation state of -1. When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2.

hope this helps........ Stay safe and have a Merry Christmas!!!!!!!!! :D Explanation:

8 0

2 years ago

What is a change that will not affect the pressure in a container ? A.increasing the volume of the fluid B . Changing the materi

ExtremeBDS [4]

Answer:

B . Changing the material that the fluids container is made of

Explanation:

Changing the material of the container does not affect the pressure in a container whereas increasing the volume, changing the weight of the fluid, and heating/cooling the fluid will all change the pressure.

4 0

3 years ago

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.600 M. N2(g)+O2(g)↽−−⇀2NO(g) If

cestrela7 [59]

Answer:

0.84M

Explanation:

Hello,

At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:

$K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9$

Now, the new equilibrium condition, taking into account the change x, becomes:

$9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}$

Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:

$\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}$