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Ket [755]
3 years ago
14

Consider the following reaction between mercury(II) chloride and oxalate ion.

Chemistry
1 answer:
Alina [70]3 years ago
3 0

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

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It is electrostatic in nature.

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Answer:

Requirements for a correctly written chemical equation are reactants and products, their formula and valency

Explanation:

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1 - Potassium Hydroxide - KOH

2 - Calcium Nitrate - Ca(NO_3)_2

The requirements for a correctly written chemical equation are -

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Extend your thinking: The slow decay of radioactive materials can be used to find the age of rocks, fossils, and archaeological
lisov135 [29]

The age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

<h3>What is the half-life of a radioactive material?</h3>

The half-life of a radioactive material is the time taken for half the atoms present in the material to decay or disintegrate.

The half-life, t_\frac{1}{2}, the age, t, and amount remaining, A_{r}, of a radioactive material are related by the formula below:

  • t = \frac{t_\frac{1}{2}*A_{r}}{-ln2}

Half-life of carbon-14 = 6000 years

For the Egyptian papyrus with 63% of its original carbon-14 atoms:t = \frac{6000*0.63}{-0.63} = 4000\:years

For the Aboriginal charcoal with 22% of its original carbon-14 atoms:

t = \frac{6000*0.22}{-0.63} = 13106.5\:years

For the Mayan headdress with 79% of its original carbon-14 atoms:

t = \frac{6000*0.79}{-0.63} = 2040\:years

Neanderthal skull with 3% of its original carbon-14 atoms:

t = \frac{6000*0.03}{-0.63} = 30353\:years

Therefore, the age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

Learn more about half-life at: brainly.com/question/26689704

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