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ankoles
2 months ago
13

A careless student forgets to label one of their reagent containers. They know it contains one of the following solutions: 0.1 M

NaCl 0.1 M Ca(NO3)2 0.1 M KCH3COO The student decides to add some 0.1 M Pb(NO3)2 to a sample of the reagent in question. Upon stirring, a white precipitate formed. Which of the possible solutions is present in the unlabeled container based on this observation?
A. 0.1 M KCH3COO
B. 0.1 M Ca(NO3)2
C. 0.1 M NaCl
Chemistry
1 answer:
castortr0y [3K]2 months ago
4 0

Sagot:

0.1 M NaCl

Paliwanag:

Ang tanong na ito ay nagpapaalala sa atin ng mga patakaran sa solubility. Alalahanin natin na ang lahat ng chlorides ay natutunaw maliban sa mga ng lead, mercury II at silver na hindi natutunaw sa tubig.

Ang sumusunod na reaksyon ay mangyayari na humahantong sa pagbuo ng isang precipitate;

Pb(NO3)2(aq) + 2NaCl(aq) -------> 2NaNO3(aq) + PbCl2(s)

Ang puting precipitate na nabuo ay  PbCl2.

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Answer:

To break a single I-I bond, the wavelength of light required is 7.92 × 10⁻⁷ m

Explanation:

The energy needed to break one mole of iodine-iodine single bonds is 151 KJ

The energy necessary to rupture one iodine-iodine bond is calculated as (151 KJ/mol) / 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

or

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Formula:

E = hc / λ    

Where h is Planck's constant    = 6.626 × 10⁻³⁴ js

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λ   = hc / E

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λ   = 19.878 × 10⁻²⁶ j.m / 2.51 × 10⁻¹⁹ J

λ   = 7.92 × 10⁻⁷ m

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