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Zina [86]
3 years ago
9

The quantity of energy released or absorbed as heat during a chemical reaction is called the 28. a. temperature. c. entropy d. f

ree energy enthalpy of reaction.
Chemistry
1 answer:
Tatiana [17]3 years ago
5 0

Answer: enthalpy of reaction.

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like ^0C and K

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

Free energy is the amount of energy that can be converted into useful work.

Enthalpy of the reaction is the difference between the energy of products and the energy of reactants. it is either the heat released or absorbed during the reaction. It is either positive or negative.

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Which two systems in the human body work together to eliminate carbon dioxide from the body?
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Cardiovascular and circulatory

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2 years ago
Based on the chemical bonds present in this molecule, we would classify it as a(n)
MrMuchimi

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8 0
3 years ago
Read 2 more answers
n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph
olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.</em>

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = <em>0.111M</em>

<em></em>

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = <em>0.2501 g of acetic acid</em>

Now, assuming density of solution as 1.00g/mL, 37.54mL weights <em>37.54g</em>.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = <em>0.666% (w/w)</em>

8 0
3 years ago
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