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Alexxx [7]
2 years ago
9

During a reaction, the enthalpy of formation of an intermediate is 34 kJ/mol. During the reaction, 4 moles of the intermediate a

re produced. What is the enthalpy value for this step of the reaction?​
Chemistry
1 answer:
Lapatulllka [165]2 years ago
7 0

The enthalpy change for the formation step of this intermediate is 136 kJ/mol.

<h3>What is the enthalpy?</h3>

The enthalpy of reaction can be obtained from the enthalpies of formation of the species involved. The enthalpy of reaction is the heat evolved or absorbed in a reaction.

Thus, the enthalpy change for the formation step of this intermediate is 4 *  34 kJ/mol = 136 kJ/mol.

Learn more about enthalpy change:brainly.com/question/4526346

#SPJ1

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An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit
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<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of triprotic acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL

Putting values in above equation, we get:

3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

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3 years ago
Write a balanced equation for the complete combustion of each compound. a. formaldehyde (CH2O(g)) b. heptane (C7H17(l)) c. benze
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Answer:

a. formaldehyde (CH₂O(g)): CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

b. heptane (C₇H₁₇(l)): 4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l))​: C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

Explanation:

In a reaction of combustion, a hydrocarbon compound (composed of C, H and O) reacts with oxygen gas (O₂). The <u>complete</u> combustion of a hydrocarbon - such as formaldehyde, heptane, and benzene - produces carbon dioxide (CO₂) and water (H₂O).  

Thus, we write the reactants and products for each combustion reaction and then we balance the atoms: C, H, and O.

a. formaldehyde (CH₂O(g)):

CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

The chemical equation is already balanced with the coefficients 1: we have the same number of C atoms (1), H (2) and O (3) on both sides of the equation.

b. heptane (C₇H₁₇(l)):

C₇H₁₇(l) + O₂(g) → CO₂(g) + H₂O(g)

Here we have to write a coefficient 28 for CO₂ to balance the C atoms in the products side, and for C₇H₁₇ we write 28/4 = 7. With similar reasoning we found the coefficients for O₂ and H₂O:

4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l))​:

C₆H₆(l) +  O₂(g) → CO₂(g) +  H₂O(g)

First, we write a coefficient of 6 for CO₂ to balance the C atoms. Then, we have to balance H atoms: we write a coefficient 3 in H₂O. Now, we have 12 + 3 = 15 atoms of O on the reactants side. So, we write a half of these number of atoms in the coefficient for O₂: 15/2. We obtain the balanced equation:

C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

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