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Mariulka [41]
2 years ago
14

What volume of .15M NaOH could be produced with only 5 grams of NaOH and water?

Chemistry
1 answer:
fenix001 [56]2 years ago
4 0

The volume of 0.15M NaOH that could be produced with only 5 grams of NaOH and water is 0.83L.

<h3>How to calculate volume?</h3>

The volume of a substance can be calculated using the following expression:

Molarity = number of moles ÷ volume

The number of moles of NaOH can be calculated by dividing the mass of NaOH by its molar mass.

Molar mass of NaOH = 40g/mol

n = 5/40

n = 0.125moles

Volume of NaOH = 0.125mol ÷ 0.15M

Volume = 0.83L

Therefore, the volume of 0.15M NaOH that could be produced with only 5 grams of NaOH and water is 0.83L.

Learn more about volume at: brainly.com/question/905400

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Answer:

The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.

Explanation:

Given data:

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Percent abundance of Si-28 = 92.21%

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Percent abundance of Si-29 = 4.70%

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Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass)  / 100

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Average atomic mass = 2808.86 / 100

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The ionization equation for the given acid is written as:

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Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

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Equilibrium expression for the above equation would be:

\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}

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From given info, equilibrium concentration of the acid is 0.12

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We know that, pH=-log[H^+]

pH = -log(0.00465)

pH = 2.33

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