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melomori [17]
3 years ago
13

The expected value for a chemical equation is 47g of water, after an experiment you find that you have 2.58 moles of water. What

is the actual yield of your chemical reaction? *
Chemistry
1 answer:
strojnjashka [21]3 years ago
6 0
<h3><u>Answer</u>;</h3>

Actual yield = 46.44 g

<h3><u>Explanation;</u></h3>

1 mole of water = 18 g/mol

Therefore;

The experimental yield = 2.58 moles

equivalent to ; 2.58 × 18 = 46.44 g

The theoretical value is 47 g

Percentage yield = 46.44/47 × 100%

                             = 98.8%

The questions asks for actual yield = 46.44 g

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When NH3(g) reacts with N2O(g) to form N2(g) and H2O(g), 105 kcal of energy are evolved for each mole of NH3(g) that reacts. Wri
Ganezh [65]

Answer : The balanced chemical equation is,

2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal

Explanation :

Balanced chemical equation : It is defined as the number of atoms of individual elements present on the reactant side must be equal to the number of atoms of individual elements present on product side.

The given unbalanced chemical reaction is,

NH_3(g)+N_2O(g)\rightarrow N_2(g)+H_2O(g)+105kcal

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of individual elements are not balanced.

In order to balanced the chemical reaction, the coefficient 2 is put before the NH_3, the coefficient 3 is put before the N_2O\text{ and }H_2O and the coefficient 4 is put before the N_2.

The energy evolved in this reaction = 105Kcal\times 2=210Kcal

Thus, the balanced chemical reaction will be,

2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal

7 0
3 years ago
At a certain temperature the vapor pressure of pure thiophene is measured to be . Suppose a solution is prepared by mixing of th
Lesechka [4]

Answer:

0.35 atm

Explanation:

It seems the question is incomplete. But an internet search shows me these values for the question:

" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."

Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>

First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:

  • 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
  • 111 g heptane ÷ 100 g/mol = 1.11 moles heptane

Total number of moles = 1.63 + 1.11 = 2.74 moles

The<u> mole fraction of thiophene</u> is:

  • 1.63 / 2.74 = 0.59

Finally, the <u>partial pressure of thiophene vapor is</u>:

Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene

  • Partial Pressure = 0.59 * 0.60 atm
  • Pp = 0.35 atm

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