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3 years ago

Na2S2O3 + AgBr NaBr + Na3ſAg(S203)2] What is

Chemistry

1 answer:

gtnhenbr [62]3 years ago

8 0

Answer:

Mass of NaBr produced = 23.67 g

Explanation:

Given data:

Mass of AgBr = 42.7 g

Mass of NaBr produced = ?

Solution:

Chemical equation:

2Na₂S₂O₃ + AgBr → NaBr + Na₃(Ag(S₂O₃)₂

Number of moles of AgBr:

Number of moles = mass/molar mass

Number of moles = 42.7 g/ 187.7 g/mol

Number of moles = 0.23 mol

now we will compare the moles of AgBr with NaBr.

AgBr : NaBr

1 : 1

0.23 : 0.23

Mass of NaBr:

Mass = number of moles × molar mass

Mass = 0.23 mol × 102.89 g/mol

Mass = 23.67 g

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Answer : The correct rate law for the reaction is,

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Expression for rate law for second observation:

$0.0069=k(0.020)^a(0.010)^b$ ....(2)

Expression for rate law for third observation:

$0.0098=k(0.020)^a(0.020)^b$ ....(3)

Expression for rate law for fourth observation:

$0.027=k(0.040)^a(0.040)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1$

Dividing 2 from 3, we get:

$\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}$

Calculation used :

$1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}$

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$\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}$

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Answer:

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