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SVETLANKA909090 [29]
4 years ago
8

Which statement is true?

Chemistry
2 answers:
MissTica4 years ago
8 0

Answer:

The correct answer is in an exothermic reaction the energy of the product is less than the energy of the reactants.

Explanation:

Exothermic reaction is a type of reaction that generates heat.As a result in case of an exothermic reaction the energy of the reactant is more than the energy of the product.

       That"s why the enthalpy change in an exothermic reaction is always positive .

lorasvet [3.4K]4 years ago
7 0

The correct answer is C. In an exothermic reaction, the energy of the products is less than the energy of the reactants.

Explanation:

A chemical reaction is a process during which two or more substances interact and form new products. In these, the first substances are known as the reactants while the substances at the end of the reaction are the products. Moreover, reactions often involve a change in the energy of substances this known as enthalpy.

In the case of exothermic reactions in which energy or heat is released, the enthalpy is negative, this means, the energy in the products is less than in the reactants because during the chemical reaction substances release energy. This makes option C correct. On the other hand, in endothermic reactions, energy is absorbed, and therefore it is expected energy in the products is more than the one in the reactants.

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Answer:

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The correct answer for the question that is being presented above is this one: "<span>16.728 g."</span>

Given that 
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This means that 5.66 kJ of heat is released when 1 mole of NH3 solidifies 

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<span>1 mole of NH3 = 17 g </span>
So, 0.984 moles of NH3 = 17 X 0.984 = 16.728 g
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3 years ago
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2 years ago
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The rate constant k for a certain reaction is measured at two different temperatures:
bogdanovich [222]

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are K_1  and K_2  at temperature T_1  and T_2

By using Arrhenius equation at two different two different temperature,

Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K  ;K_1=4.8*10^8;T_2=273+280=553K  ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;  

By putting value of R=2 cal/mole.K

E_a=5510.265cal/mole;

By rounding off upto 2 significant figure;

E_a=5.5Kcal/mole;

8 0
4 years ago
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