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Lelu [443]
2 years ago
9

A student dissolved 3.50g of copper (II) nitrate in water and mixed it with a solution of sodium carbonate. The student recovere

d 1.89g of copper carbonate precipitate. If the theoretical yield is 2.35g, what is the percent yield
Chemistry
1 answer:
Arisa [49]2 years ago
3 0

Answer: If the theoretical yield is 2.35g then the percent yield is 80.4%

Explanation:

Given: Actual yield = 1.89 g

Theoretical yield = 2.35 g

Formula used to calculate the percentage yield is as follows.

Percent yield = \frac{actual yield}{theoretical yield} \times 100

Substitute the values into above formula as follows.

Percent yield = \frac{actual yield}{theoretical yield} \times 100\\= \frac{1.89}{2.35} \times 100\\= 80.4 percent

Thus, we can conclude that if the theoretical yield is 2.35g then the percent yield is 80.4%

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Bubba falls out of a plane flying over his peanut field. What will bubba speed be after falling for 8 seconds?
chubhunter [2.5K]

Variables we know:

t = 8 seconds

Vi = 0 m/s

g = -9.81

Δy = ?

Vf = ?

Equation we will be using to solve for Vf: Vf = Vi + gt

Steps to solve:

Vf = (0) + (-9.81)(8)

Vf = -78.48 m/s

Hope this helps!! :)

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2 years ago
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Most metals allow electricity to flow freely through them. So, most metals are.....
algol [13]

Answer:

Conductors

Explanation:

Metals that are conductors let electric currents flow freely. Insulators have a resistance of a charge to flow through them.

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Answer:

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Explanation:

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2 years ago
True or false?
Virty [35]
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3 years ago
An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

7 0
3 years ago
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