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3 years ago

If the theoretical yield of a reaction is 0.110 g and the actual yield is 0.104 g , what is the percent yield?

1 answer:

3 0

Theoretical Yield is an Ideal yield with 100 % conversion of reactant to product. It is in fact a paper work.

While,

Actual Yield is the yield which is obtained experimentally. It is always less than theoretical yield because it is not possible to have 100% conversion of reactants into products. Even some amount of product is lost while handling it during the process.

Percentage Yield is Calculated as,

%age Yield = Actual Yield / Theoretical Yield × 100

Data Given:
Actual Yield = 0.104 g

Theoretical Yield = 0.110 g

Putting Values,

%age Yield = 0.104 g / 0.110 g × 100

%age Yield = 94.54 %

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a quantity of zinc reacts with sulfuric acid and produces 0.10g of hydrogen. how many particles of zinc are required socratic.or

alekssr [168]

Answer:

grams of zinc required = 3.24 g

Particles of zinc required = $2.99\times 10^{22}$

Explanation:

Given that:-

Mass of hydrogen gas produced = 0.10 g

Calculation of the moles of $H_2$ as:-

Mass = 0.10 g

Molar mass of $H_2$ = 2.016 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.10\ g}{2.016\ g/mol}$

$Moles= 0.0496\ mol$

According to the reaction shown below:-

$Zn+H_2SO_4\rightarrow ZnSO_4+H_2$

1 mole of hydrogen gas is produced when 1 mole of zinc reacts

So,

0.0496 mole of hydrogen gas is produced when 0.0496 mole of zinc reacts

Moles of Zinc required = 0.0496 moles

Molar mass of zinc = 65.38 g/mol

mass= Moles*Molar mass = $0.0496\ moles\times 65.38\ g/mol=3.24\ g$

Also, 1 mole of Zinc contains $6.023\times 10^{23}$ particles

0.0496 mole of Zinc contains $0.0496\times 6.023\times 10^{23}$ particles

Particles of zinc required = $2.99\times 10^{22}$

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3 years ago

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) The Δ G ° ′ of the reaction is − 9.130 kJ ⋅ mol − 1 . Calculate the equil

Liula [17]

Answer:

K = 39.85

ΔG= -6.9 kJ/mol

Explanation:

Step 1: Data given

The ΔG°′ of the reaction is − 9.130 kJ/mol

Temperature = 25.0 °C = 298 K

Body temperature = 37.0 °C = 310K

the concentration of A is 1.9 M

the concentration of B is 0.80 M

Step 2: The reaction

A (aq) ⇌ B (aq)

Step 3:

ΔG° = -RT ln K

⇒with ΔG° = standard Gibbs free energy change = − 9.130 kJ/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T = the temperature = 298 K

⇒with K = the equilibrium constant = TO BE DETERMINED

− 9130 J/mol = - 8.314 * 298 * ln K

ln K = 3.685

K = e^3.685

K = 39.85

Step 4: The reaction at body temperature

ΔG= ΔG° + RT ln [B]/[A]

⇒with ΔG° = Gibbs free energy change

⇒with ΔG° = standard Gibbs free energy change = − 9130 J/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T = the temperature = 310 K

⇒with [A] = 1.9 M

⇒with [B] = 0.80 M

ΔG= -9130 J/mol + 8.314 J/mol*K * 310 K * ln (1.9/0.80)

ΔG= -9130 J/mol + 2229.4J/mol

ΔG=-6900.6 J/mol = -6.9 kJ/mol

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Answer:

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Explanation:

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