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Vera_Pavlovna [14]
1 year ago
10

A sample of a monoprotic acid (ha) weighing 0.384 g is dissolved in water and the solution is titrated with aqueous naoh. if 30.

0 ml of 0.100 m naoh is required to reach the equivalence point, what is the molar mass of ha?
Chemistry
1 answer:
posledela1 year ago
3 0
The formula for the monoprotic acid is taken as HA, reaction with base is as follows;
HA + NaOH ---> NaA + H₂O
Stoichiometry of acid to base is 1:1
At the neutralisation point, number of HA moles = number of base moles
Number of NaOH moles reacted = 0.100M / 1000 mL /L x 30.0 mL = 0.003 mol
Therefore number of HA moles reacted = 0.003 mol
the mass of acid 0.384 g
Therefore molar mass - 0.384 g/ 0.003 mol = 128 g/mol
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Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of CH_4 will be,

2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=29.7kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

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Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

\Delta H=\frac{q}{n}

or,

q=\Delta H\times n

where,

\Delta H = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

Thus, the total heat will be, -2251 kJ

4 0
2 years ago
The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured a
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Answer:

See explanation

Explanation:

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4 0
2 years ago
Read 2 more answers
I NEED HELP ASAP, WILL MARK BRAINLEST!
Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial        SiCl₄                   O₂                    SiO₂

 1           120 g                  240 g              38.2 g

 2           75 g                   50 g                25.2 g

<u>Percentage yield for trial 1</u>

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

 molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

 120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

 For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

 .706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

 molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

 .706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

 actual/theoretical x 100

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<u>Leftover reactant for trial 1</u>

We know oxygen is the excess reactant.

Mass to moles:

 molar mass O₂ = 32.00 g/mol

 240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

 7.5 - .706 = 6.8 moles left over

Moles to mass:

 6.8 mol O₂ x 32.00g/mol =<u> 217.4 g O₂</u>

<u />

<u>Percentage yield for trial 2</u>

Mass to moles:

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 75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

 For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

 .441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

 molar mass SiO₂: 60.09 g/mol

 .441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

 actual/theoretical x 100

 25.2 / 26.5 = .950 = <u>95.0% yield</u>

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

6 0
2 years ago
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