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Kay [80]
3 years ago
11

A beaker containing 80 grams of lead(ii) nitrate, pb(no3)2, in 100 grams of water has a temperature of 30 ºc. approximately how

many grams of the salt are undissolved, on the bottom of the beaker?
Chemistry
1 answer:
sukhopar [10]3 years ago
4 0

Answer:

14 g.

Explanation:

  • From the figure attached:

<em>the solubility of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC is </em><em>(66 g).</em>

When beaker containing 80 grams of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC.

<em>∴ The grams of the salt are undissolved, on the bottom of the beaker are </em><em>(14 g).</em>

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