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3241004551 [841]
2 years ago
12

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

Chemistry
1 answer:
Westkost [7]2 years ago
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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How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*
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<u>Answer:</u> 6.57 L of solution can be made.

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}} .....(1)

Given values:

Molarity of LiBr = 3.5 M

Moles of LiBr = 23 moles

Putting values in equation 1, we get:

3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L

Hence, 6.57 L of solution can be made.

5 0
3 years ago
Why the moon is the barycenter of earth?
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4 0
3 years ago
What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

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3 years ago
How many oxygen atoms are in the product of this balanced chemical equation
svlad2 [7]

Answer:

two oxygen atoms

There are two oxygen atoms in the reactants and two atoms of oxygen in the product.

Explanation:

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2 years ago
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