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Amiraneli [1.4K]
4 years ago
15

If you increase the number of hours that you practice, you can score more goals in soccer.

Chemistry
2 answers:
vazorg [7]4 years ago
7 0

Answer:

Yes

Explanation:

Hypothesis: If i increase the number of hours i practice, then i will score more goals in soccer.

Set up. I would have two people who don't play soccer. One will be practicing for one hours every day, and the other will be practicing three hours every day. Based on my hypothesis, the person that is training three hours every day will score more goals than the person training one hour a day.

anyanavicka [17]4 years ago
6 0

Answer:

A.Hypothesis

Explanation:

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The magnesium ribbon, <u>D. It forms a material to cast the tool mark</u>.

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When a magnesium ribbon is burnt in the presence of oxygen it gives out strong light and heat is produced. Apart from it, it leads to the production of substance called as magnesium oxide which is formed as the product due to the reaction of magnesium with the oxygen present in the air.

Tool marks are the mark which is created by tools while using them. In order to identify or locate them castes made up of magnesium oxide is utilized. When this is pasted on the suspected area, the tool mark of the suspected tool gets pasted on it.

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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

7 0
3 years ago
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