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lbvjy [14]
3 years ago
9

The theoretical yield of ammonia in an industrial synthesis was 550 kg, but only 480 kg was obtained. What was the percentage yi

eld of the reaction?
Chemistry
1 answer:
Klio2033 [76]3 years ago
7 0

We know that when calculating percent yield, we use the equation:


Percent yield=\frac{Actual yield (A)}{Theoretical yield (T)}


Since the quantities that we are given in the question are equal, we can just directly divide them to find percent yield:


Percentyield=\frac{480kg}{550kg} =0.8727*100=87.27


So now we know that the percent yield of the synthesis is 87.27%.

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Fluorine is the element or atom of the greatest electronegativity. Electronegativity would increase as we move left to right of the periodic table. 

 
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Atoms have been traditionally viewed as being composed of three different types of particles: protons, neutrons, and electrons.
sattari [20]

Answer:

See explanation

Explanation:

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True or false:The nucleus is the largest part of the atom and takes up most space
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5 0
3 years ago
How many atoms are in 22 grams of copper metal?
strojnjashka [21]
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6 0
3 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
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