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2 years ago

Use the diagram below to answer the question.

Chemistry

2 answers:

oksian1 [2.3K]2 years ago

5 0

Answer: Every action force results in an equal and opposite reaction force.

Explanation: newtons law

kifflom [539]2 years ago

3 0

Every action force results in an equal and opposite reaction force

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Suppose you carry out a titration involving 2.50 molar NaOH and an unknown concentration of HBr. To bring the reaction to its en

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Answer:

Explanation:

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2 years ago

In a solution of salt and water which compound is the solute?

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3 years ago

Which of the following are properties of solution?

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3 years ago

When an atom gives off energy in the form of light, a. it becomes excited. b. it becomes more stable. c. it returns to ground st

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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:

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

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

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

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2 years ago