How can you balance a chemical equation?

Chemistry

1 answer:

sergey [27]3 years ago

7 0

Answer:

When you balance a chemical equation, you change coefficients. You never change subscripts. A coefficient is a whole number multiplier. To balance a chemical equation, you add these whole number multipliers (coefficients) to make sure that there are the same number of atoms on each side of the arrow.

Explanation:

You might be interested in

During the combustion of methane (CH4), one mole of methane gas combines with two moles of oxygen gas (O2) to produce one mole o

astraxan [27]

Answer:

CH4(g) +2O2____ CO2(g) +2H2O (l)

3 0

3 years ago

What is the charge of oxygen in Na2O?

Nuetrik [128]

In Na2O, what is the oxidation state of oxygen? In Na2O oxidation state of Na is 1+

3 0

3 years ago

Read 2 more answers

An instrument that is used to measure air pressure is a ?

Zolol [24]

A barometer measures atmospheric pressure.

8 0

3 years ago

In general, what causes a sea breeze?

ella [17]

Answer:

The usual cause of sea breeze is the difference in specific heat capacity between land and water Land heats and cools more quickly than water. What is the cause of a sea breeze? The air over the land is warmed and rises, to a greater temperature than that over the sea.

Explanation:

Your answer is C ! <3

7 0

3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$