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lesantik [10]
3 years ago
13

The overall change in enthalpy of a reaction depends only on

Chemistry
2 answers:
Nimfa-mama [501]3 years ago
6 0
The choices can be found elsewhere and as follows:

A. the reactants and the products. 
<span>B. the path that is taken from the reactants to the products. </span>
<span>C. the reactants and the catalysts only. </span>
<span>D. the reactants, the products, and the catalysts.
</span>
I believe the correct answer is option A. The overall change in enthalpy of a reaction depends only on the reactants and the products. Hope this answers the question. Have a nice day.
serious [3.7K]3 years ago
4 0
<h3><u>Answer;</u></h3>

A. the reactants and the products.

<h3><u>Explanation</u>;</h3>
  • The enthalpy of a reaction is the heat released or absorbed by a reaction at constant pressure.
  • Enthalpy is important as it tells us the amount of heat energy present in a system. In chemical reactions it tells us the amount of heat energy lost or gained.
  • <em><u>Changes in enthalpy are determined as the reactants are converted into products. Therefore, enthalpy depends on the reactants and the products. In other words it depends on the heat of reactants and that of products.</u></em>
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Based on this;
The boiling point of bromine is a physical property of bromine.
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4 0
3 years ago
Arrange the measurements from least to greatest.
NNADVOKAT [17]
1) 1.8 micrograms(least)
2) 1.8 grams
3) 1.8 kilograms(greatest)
7 0
2 years ago
Approximately how much of the world’s oil and natural gas reserves are believed to be in the arctic? View Available Hint(s) Appr
Gnoma [55]

Answer:

1/4 or 25%

Explanation:

The Arctic region of the earth refers to that part of the earth around the north pole region. Hence, when we are talking about latitude O degrees North, the areas around this geographical location is referred to as the arctic.

Now , there is an estimated 1/4 or 25% of the world’s oil and natural gas here. Unfortunately, these are yet accessible because of the amount of ice or snow covering. With increase in technological advancements, this might be accessible in the nearest future

7 0
3 years ago
Please help!!! Is it balanced or unbalanced?
Rzqust [24]
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7 0
3 years ago
Calculate the solubility (in mol/L) of Fe(OH)3 (Ksp = 4.0 x 10^-38) in each of the following situations:
netineya [11]

Answer:

(A) 1.962x10^-10 M solubility in pure water

(B) 4.0 x 10^-33 M solubility

(C) 4.0 x 10^-27 M solubility

Explanation:

(A) Fe(OH)3 would give (Fe3+) and (3OH-)

Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

Let y = [Fe^3+]

Let 3y = [OH-]

4x10^-38 = (y)(3y)^3

4x10^-38 = 27y^4

y^4 = 4x10^-38 ÷ 27

y^4 = 1.481 x 10^-39

y = 1.962x10^-10 M solubility in pure water

(B) pH = 5.0

5.0 = - log [OH-]

-5.0 = log [OH-]

[OH-] = 10^-5.0 =  1.0 x 10^-5 M

So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

[Fe^3+][1.0 x 10^-5] = 4.0 x 10^-38

[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-5

= 4.0 x 10^-33 M solubility

(C) pH = 11.0

11.0 = - log [OH-]

-11.0 = log [OH-]

[OH-] = 10^-11.0 =  1.0 x 10^-11 M

So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

[Fe^3+][1.0 x 10^-11] = 4.0 x 10^-38

[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-11

= 4.0 x 10^-27 M solubility

6 0
3 years ago
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