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victus00 [196]
3 years ago
6

Which one of the following sets of units is appropriate for a second-order rate constant?A. s-1B. mol L-1 s-1C. L mol-1 s-1D. mo

l2 L-2 s-1E. L2 mol-2 s-1
Chemistry
1 answer:
Korolek [52]3 years ago
6 0

Answer:

E

Explanation:

A second order reaction is one in which the reaction rate is proportional to the concentrations of the 2 reactants. We only consider the reactants whose concentration are changing however.

For example, in the case:

A + B ----> C + D , provided that both A and B undergo concentration changes, the rate is proportional to the product of their concentrations.

Now, like we know, the unit of the reaction rate is per second I.e 1/s or simply s^-1 . Also it is important to note that the concentration of the reactants is exprested in units of mol/L. We can now mathematically calculate the unit of the rate constant.

Mathematically,

R = k × [A] [B]

Hence,

S^-1 = k × (mol/L)(mol/L)

S^

s^-1 = k × (mol/L)^2

K = s^-1 ÷ (mol/L)^2

K = s^-1 × L^2 × mol^-2.

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In order to satisfy charge conservation and lepton number conservation the other products must be neutron.

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A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i
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Answer:

MM_{acid}=140.1g/mol

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}

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n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol

Then, by using the mass of the acid, we compute its molar mass:

MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol

Regards.

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Explanation:

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