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Evgen [1.6K]
3 years ago
8

At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1·M−1s−1 : →2SO3g+2SO2gO2g Suppo

se a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
Chemistry
1 answer:
RoseWind [281]3 years ago
4 0

Answer:

[SO_3]=0.25M

Explanation:

Hello there!

In this case, since the integrated rate law for a second-order reaction is:

[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}

Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:

[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\

[SO_3]=0.25M

Best regards!

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4 years ago
If 42.68 ml of 0.43 M KOH is used to neutralize 18.40 ml of H, SO4, calculate
Temka [501]

The concentration of the sulfuric acid : 0.499 M

The net ionic equation

2 OH ⁻(aq]+ 2 H ⁺ (aq] → 2 H ₂O (l]

<h3>Further explanation</h3>

Given

42.68 ml of 0.43 M KOH

18.40 ml of H2SO4

Required

the concentration of the sulfuric acid

the net ionic equation

Solution

Acid-base titration formula  

Ma. Va. na = Mb. Vb. nb  

Ma, Mb = acid base concentration  

Va, Vb = acid base volume  

na, nb = acid base valence  

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The net ionic equation

Reaction

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2 K ⁺ (aq]+ 2 OH ⁻ (aq] + 2 H ⁺ (aq] + SO ₄²⁻(aq] → 2 K ⁺ (aq] + SO ₄²⁻(aq] + 2 H ₂ O (l]

canceled the spectator ions :

2 OH ⁻(aq]+ 2 H ⁺ (aq] → 2 H ₂O (l]

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4 years ago
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