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atroni [7]
2 years ago
14

How many atoms are in 9.3moles of lithium?

Chemistry
2 answers:
vladimir1956 [14]2 years ago
3 0

Answer:

3.45 moles Li  contains  2.08 × 10 (to the power of)24  atoms .

Explanation:

The relationship between atoms and moles is:

1  mole atoms =

6.022 × 10 (to the power of)23

atoms

In order to determine how many atoms occupy a given number of moles, multiply the given moles by  

6.022 × 10 (to the power of)23

atoms/mole

.

In the case of 3.45 moles lithium (Li):

3.45 mol Li × 6.022 × 10 (to the power of)23  atoms Li/ 1 mol Li =

2.08 × 10 (to the power of)24

atoms Li  rounded to three

zvonat [6]2 years ago
3 0

Answer: 3.45 moles Li contains 2.08 × 10 (to the power of)24 atoms

Explanation:

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which one of the following is an endothermic process?1.ice melting 2.water freezing 3.a boiling soup 4.the reaction of HCl and N
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<span>From the above given choices :
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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

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I believe this topic is quantitative chemistry but I need the equation to work this out sorry

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