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taurus [48]
3 years ago
14

if 8.15 ml of water is placed in a graduated cylinder the mass of the cylinder = 56.98g and the combined mass of both the cylind

er of both the cylinder + water is 65.11g What is the mass of the water? what is the density of the water?
Chemistry
1 answer:
Allushta [10]3 years ago
4 0

1. The mass of the water is 8.13 g.

2. The density of the water is 1 g/mL

Density is related to mass and volume by the following equation:

<h3>Density = mass / volume </h3>

<h3>1. Determination of the mass of water. </h3>

Mass of cylinder = 56.98 g

Mass of cylinder + water = 65.11 g

<h3>Mass of water =? </h3>

<h3>Mass of water = (Mass of cylinder + water) – (Mass of cylinder) </h3>

Mass of water = 65.11 – 56.98

<h3>Mass of water = 8.13 g</h3>

Thus, the mass of the water is 8.13 g

<h3>2. Determination of the density of water. </h3>

Mass of water = 8.13 g

Volume of water = 8.15 mL

<h3>Density of water =? </h3>

Density = mass / volume

Density = 8.13 / 8.15

Density of water = 0.998 g/mL

<h3>Density of water ≈ 1 g/mL </h3>

Therefore, the density of the water is 1 g/mL

Learn more: brainly.com/question/24679715

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Explanation:

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Step 3: Convert "t" to hours

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Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

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As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

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Therefore, change in pH will be calculated as follows.

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                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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