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ss7ja [257]
3 years ago
15

The following initial rate data are for the reaction of mercury(II) chloride with oxalate ion: 2 HgCl2 + C2O42- 2 Cl- + Hg2Cl2 +

2 CO2
Experiment [HgCl2] [C2O42-]o, M Initial Rate
1 0.124 0.115 1.61E-5
2 0.248 0.015 3.23E-5
3 0.124 0.229 6.40E-5
4 0.248 0.229 1.28E-4
1) Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n.
2) From these data, the rate constant is ____ M-2s-1?
Chemistry
1 answer:
Sever21 [200]3 years ago
3 0

Answer:

Explanation:

2 HgCl₂ + C₂O₄²⁻   =  2 Cl⁻ + Hg₂Cl₂ + 2CO₂

1 )

Rate of reaction = k[HgCl_2]^m[C_2O_4^{-2}]^n

             [HgCl₂]        [C₂O₄²⁻ ]           Rate  

1 .              .124             .115               1.61 x 10⁻⁵

2 .             .248             .115             3.23 x 10⁻⁵

3 .              .124             .229              6.4 x 10⁻⁵

4 .              .248             .229            1.28 x 10⁻⁴

comparing 1 and 3 , when concentration of HgCl₂ remains constant and concentration of C₂O₄²⁻  becomes twice , rate becomes 4 times so rate is proportional to square of concentration of C₂O₄²⁻  .

Hence n = 2

comparing 1 and 2 , when concentration of HgCl₂ becomes twice  and concentration of C₂O₄²⁻  remains constant  , rate becomes 2 times so rate is proportional to simply  concentration of C₂O₄²⁻  .

Hence m = 1

Putting the data of  1 in the rate equation found

 1.61 x 10⁻⁵ = k x .124 x  .115²

k = 11.3 x 10⁻⁴ M⁻² s⁻¹

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A 10.99 g sample of NaBr contains 22.34% Na by mass.

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Answer:

Choice A: Approximately 5.23 \times 10^{-29} joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the \rm 5.81\times 10^{-29} kilograms of mass.

E = m \cdot c^{2},

where

  • E stands for energy,
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The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of c?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}.

Apply the mass-energy equivalence:

\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}.

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

\rm 1\; J = 1\; N \times 1\; m.

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

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In other words, the mass defect here is also \rm 5.23\times 10^{-12}\; J.

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