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Nikolay [14]
2 years ago
8

the actual yield of a certain reaction is 38.0 g while the theoretical yield is 50.0 g calculate the percent yield

Chemistry
1 answer:
Verizon [17]2 years ago
8 0

Answer:

76%

Explanation:

Actual yield = 38.0 g

Theoretical yield = 50.0 g

Percentage yield = ?

The relationship between these quantities is given as;

Percent yield = Actual yield / Theoretical yield   *  100%

Percent yield = 38 g / 50 g    *  100 %

Percent yield = 0.76  *  100 % = 76%

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3 years ago
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Which of these substances has the lowest pH? 0.5 M HBr, pOH = 13.5 0.05 M HCl, pOH = 12.7 0.005 M KOH, pOH = 2.3
vagabundo [1.1K]
<h3><u>Answer;</u></h3>

0.5 M HBr, pOH = 13.5 ; Has the lowest pH

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

From the question;

pH = -Log [OH]

or pH = 14 - pOH

Therefore;

For 0.5 M HBr

[H+] = 0.5 M

pH = - Log [0.5]

     = 0.30

For;  pOH = 13.5

pH = 14 - pOH

     = 14 -13.5

     = 0.5

For; 0.05 M HCl

pH = - log [H+]

[H+] = 0.05

pH = - Log [0.05]

     = 1.30

For; pOH = 12.7

pH = 14 -pOH

     = 14 -12.7

     = 1.30

For;  0.005 M KOH,

pOH = - log [OH]

[OH-] = 0.005

pOH = - Log 0.005

        = 2.30

pH = 14 - 2.30

     = 11.7

For; pOH = 2.3

   pH = 14 -pOH

         = 14- 2.3

         = 11.7

6 0
3 years ago
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What isapplied in case where the base is not a OH donator
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8 0
2 years ago
If 8.74 g of CuNO3 is dissolved in water to make a 0.700 M solution, what is the volume of the solution
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Hope this helps you.

4 0
2 years ago
A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0
3 years ago
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