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kaheart [24]
3 years ago
12

The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.085 M, how long would it take u

ntil the concentration decreases to 0.055 M
Chemistry
1 answer:
olasank [31]3 years ago
5 0

Answer: It will take 8.2 minutes until the concentration decreases to 0.055 M

Explanation:

The time after which 99.9% reactions gets completed is 40 minutes

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{13min}=0.053min^{-1}

b)  Time taken for 0.085 M to decrease to 0.055 M

t=\frac{2.303}{0.053}\log\frac{0.085}{0.055}

t=8.2min

Thus it will take 8.2 minutes until the concentration decreases to 0.055 M

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A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

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\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

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\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

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<h3><u>Answer;</u></h3>

<em>D. Have different structures and different functions</em>

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<h3><u>Explanation;</u></h3>
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