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alexandr402 [8]
3 years ago
8

Given that the density of Hg(l) at 0°C is about 14 g mL-­1, which of the following is closest to the volume of one mole of Hg(l)

at this temperature?
Chemistry
1 answer:
Mila [183]3 years ago
8 0

Answer: 14mL

Explanation:

From the question, we've been told that the that the density of Hg(l) at 0°C is about 14 g mL-­1.

14g = 1mL

We should note that atomic weight of mercury = 200.59.

Therefore, 1 mol Hg = 200.59 g

Therefore, the answer that is closest to the volume of one mole of Hg(l) at this temperature will be:

= 200.59/14

= 14.3 mL

= 14mL approximately

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H3PO4 + Ca(OH)2 → Ca(H2PO4)2 + H2O
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Given question is incomplete. The complete question is as follows.

Balance the following equation:

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