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3 years ago

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If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?

1 answer:

s2008m [1.1K]3 years ago

7 0

The question is incomplete, the complete question is;

AgNO3(aq)+NaCl(aq)→AgCl(s)⏐↓+NaNO3(aq)

1- If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?

Answer:

1.8 g

Explanation:

Given the equation;

AgNO3(aq)+NaCl(aq)→AgCl(s)↓+NaNO3(aq)

We can see that the reaction is 1:1 Hence, 1 mole of sodium chloride yielded 1 mole of the precipitate(AgCl).

If this is so,

Number of moles of precipitate formed = 4.52g/143.32 g/mol

Number of moles of precipitate formed = 0.0315 moles

Hence, 0.0315 moles of precipitate was formed by 0.0315 moles of NaCl

Therefore;

Mass of NaCl reacted = 0.0315 moles * 58.5 g/mol = 1.8 g

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Answer:

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Explanation:

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To calculate the value of k which is a dependent variable for the above equation ; we have:

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$k = 9.030 \times 10 ^{-7}$

The time needed for 50% transformation can be determined as follows:

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$t_{0.5} =[ \dfrac{-In(1-0.4)}{9.030 \times 10^{-7}}]^{^{1/2.5}}$

= 200.00183 min

The rate of reaction for Avrami equation is:

$rate = \dfrac{1}{t_{0.5}}$

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rate = 0.00499 / min

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Explanation:

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hope this helps you

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Answer:

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Explanation:

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