2 years ago

Calculate the molar solubility of pbi2 in the presence of 0.10 m nai.

Chemistry

1 answer:

lys-0071 [83]2 years ago

5 0

The equation representing the solubility of $PbI_{2}$ :

$PbI_{2}(aq)Pb^{2+}(aq) + 2 I^{-}(aq)$

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Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N2(g)+3H2(g)⟶2NH3(g) Ass

yaroslaw [1]

Answer:

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Explanation:

The reaction is

$N_{2}(g)+3H_{2}(g) --> 2NH_{3}(g)$

As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia

So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen

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this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia

the moles of ammonia produced = 0.28 moles

Here the nitrogen is limiting reagent.

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I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti

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<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
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