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2 years ago

Calculate the percentage of zinc I zncl2

Chemistry

1 answer:

Murrr4er [49]2 years ago

4 0

Answer:

47.8 %

Explanation:

Ar of Zn = 65

Mr of ZnCl2 = 65 + (2 * 35.5)

= 65+ 71

= 136

Percentage Composition

= ( Ar of element × No of atom in formula ) / Mr of compound * 100 %

% composition of Zn = ( 65*1 )/ 136 × 100 %

= 47.7941...

= 47.8 % ( 3 s.f )

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NikAS [45]

Answer:

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2 years ago

Which coefficients correctly balance the formula equation nh4no2(s)-> n2(g) + h2o(l)?

horrorfan [7]

NH4NO2 > N2 + 2 H2O

8 0

3 years ago

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

3 years ago

What is the number of orbitals in an atom of radon

yan [13]

86 elections so 43 orbitals

6 0

3 years ago

Why do you think some of the indicators used in experiment 2 were different than the ones used in experiment 1?

pentagon [3]

During selection of indicator. We choose an indicator which have pH range equivalent to the pH change of reaction to give better result and better observation.

So there are some different indicator are used in table 2 as compared to the table 1.

- Alizarin and phenolphthalein are basic indicator and their pH range is more than 8 so they are used in table 2

6 0

3 years ago