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ira [324]
2 years ago
13

During a reaction, the enthalpy of formation of an intermediate is 34 kJ/mol.

Chemistry
1 answer:
galina1969 [7]2 years ago
5 0

Since the enthalpy can be calculated from the heat of formation, the enthalpy is 136 kJ/mol.

<h3>What is enthalpy?</h3>

The enthalpy of a reaction is the heat that is lost or gained in that reaction. We know that the enthalpy can be calculated from the heat of  formation.

Thus, we can obtain the enthalpy of the reaction as 4 * 34 kJ/mol = 136 kJ/mol.

Learn more about enthalpy:brainly.com/question/13996238

#SPJ1

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We have gravity here on earth while in space there’s none
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2 years ago
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How many atoms are in a sample of 175 grams of sodium (Na)? The answer needs to be <br> a x 10^b
ValentinkaMS [17]
Atomic mass Sodium ( Na ) = 22.98 u.m.a

22.98 g ----------------- 6.02x10²³ atoms
175 g ------------------- ?? atoms

175 x ( 6.02x10²³) / 22.98 =

4.58x10²⁴ atoms of Na

hope this helps!

4 0
3 years ago
What was the chemical purpose of heating the MnSO4⋅H2O? Why is this procedure important?
Alla [95]

The chemical purpose of heating the MnSO4⋅H2O is to eliminate water with the formation of new compounds.

<h3>Dehydration reactions</h3>

Dehydration reaction is a type of chemical reaction that involves the removal of water which would eventually lead to the formation of a new compound.

The molecule of MnSO4⋅H2O contains a molecule of water(H2O). After heating of the molecule, water is lost giving rise to the dry compound MnSO4.

Other examples of dehydration reactions are:

  • Reactions that produce acid anhydrides.

  • Reactions that involve the production of polymers.

  • Reaction of sucrose with concentrated sulfuric acid

Therefore, the chemical purpose of heating the MnSO4⋅H2O is to eliminate water with the formation of new compounds.

Learn more about dehydration here:

brainly.com/question/12261974

8 0
2 years ago
The exchange rate for one United States dollar (US$1.00) is two dollars and seventy cents in Eastern Caribbean currency (EC$2.70
love history [14]

Answer:

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8 0
3 years ago
You are asked to prepare 500. mL 0.150 M acetate buffer at pH 5.10 using only pure acetic acid ( MW = 60.05 g/mol, p K a = 4.76
Amanda [17]

Answer:

4.504g of acetic acid

Explanation:

The acetic acid in reaction with NaOH produce acetate ion, thus:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

<em>That means the moles of acetate buffer comes, in the first, from the acetic acid</em>

As you need 500mL (0,500L) of a 0.150M acetate buffer, moles are:

0.500L × (0.150mol / 1L) = <em>0.075 moles of acetate</em>. That is:

0.075mol = [CH₃COO⁻] + [CH₃COOH]

Thus, grams of acetic acid you need to prepare the buffer are:

0.075 moles acetic acid × (60.05g / 1mol) = <em>4.504g of acetic acid</em>

6 0
3 years ago
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