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pychu [463]
3 years ago
7

Each sentence describes potential and kinetic energy at various locations on a slide. Complete each statement by selecting the p

osition of the object on the slide
Chemistry
2 answers:
babunello [35]3 years ago
8 0

Answer:

C

A

B

Explanation:

marshall27 [118]3 years ago
7 0

The question is incomplete, the complete question is;

Each sentence describes potential and kinetic energy at various locations on a slide. Complete each statement by selecting the position of the object on the slide.

An object at position has all kinetic energy. A B or C

An object at position has all potential energy. A B or C

An object at position has about half potential energy and half kinetic energy. A B or C

Answer:

An object at position C has all kinetic energy.

An object at position A has all potential energy.

An object at position B has about half potential energy and half kinetic energy.

Explanation:

Lets take a good look at the image.

At position A, the object is stationary and we know that the energy possessed by an an object by virtue of its position is known as potential energy. Hence at A, the energy possessed by the body is all potential.

At position B, the total mechanical energy of the object becomes potential energy + kinetic energy as the object is now in motion.

At position C, the object is still in motion and the energy is all kinetic.

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You put two open cans of soda on the table. You took one can from the refrigerator and the other from your car. When you open th
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Read 2 more answers
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)} ⇄ 2NO_{(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)} andO_{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

3 0
2 years ago
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