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vovangra [49]
3 years ago
15

Read this for the next two problems: When a 125 g obsidian rock is placed into a graduated

Chemistry
1 answer:
alekssr [168]3 years ago
3 0

Answer:

29. Option (b) 32 mL

30. Option (d) 3.9 g/mL

31. Option (d) 13.195 g

Explanation:

29. Determination of the volume of the rock.

Volume of water = 15 mL

Volume of water + Rock = 47 mL

Volume of Rock =.?

Volume of Rock = (Volume of water + Rock) – (Volume of water)

Volume of Rock = 47 – 15

Volume of Rock = 32 mL

30. Determination of the density of the rock.

Density of a substance is simply defined as the mass of the substance per unit volume of the substance. It can be represented mathematically as:

Density (D) = mass (m) /volume (V)

D = m/V

With the above formula, the density of the rock can be obtained as follow:

Mass (m) of Rock = 125 g

Volume (V) of Rock = 32 mL

Density (D) =?

D = m/V

D = 3.9 g/mL

Therefore, the density of the rock is 3.9 g/mL.

31. Determination of the sum of 3.7 g and 9.495 g.

The sum of 3.7 g and 9.495 g can be obtained as follow:

3.7 g + 9.495 g = 13.195 g

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Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).

pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.

LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:

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