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Archy [21]
2 years ago
12

Round off your answer to one decimal place.

Chemistry
1 answer:
8090 [49]2 years ago
8 0

Answer: 10.4

Explanation:

ph is just as follows:

pH = -log[H_{3}O^{+}]

So plugging in our concentration we get:

-log(4.45*10^{-11})=10.35163999...

Then rounding to one decimal place takes us to pH = 10.4

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The correct answer is b :)
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A permanent magnet can affect:
ruslelena [56]

A. both permanent magnets and electromagnets.

Explanation:

A permanent magnet can affect and attract any other permanent magnet and even electromagnet.

They also affect any magnetic materials especially metals that can be magnetized.

In the vicinity of such substances, an attractive or repulsive force sets in and they both interact in the presence of the force field in place.

Permanent magnets cannot magnetize non-magnets.

An electromagnet is a magnet produced by the passage of electric current through a wire wound round a metallic core.

learn more:

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3 0
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
4 years ago
Ch2(c(ch3) 2 iupac name​
Slav-nsk [51]

Answer:

hope this helps please like and mark as brainliest

5 0
3 years ago
Heart, 5 stars, and Brainiest to first right answer!
mezya [45]

Answer:

Give person above me brainliest

Explanation:

7 0
2 years ago
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