How many electrons does calcium hace?

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

siniylev [52]

Answer : The reaction rate will be, $1.9\times 10^{-4}M/s$

Explanation :

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:

$2HgCl_2(aq)+C_2O_2^{4-}(aq)\rightarrow 2Cl^-(aq)+2CO_2(g)+HgCl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_2^{4-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_2^{4-}$

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 1 from 2, we get:

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

Dividing 3 from 2, we get:

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

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

Now, calculating the value of 'k' by using any expression.

Putting values in above rate law, we get:

$3.2\times 10^{-5}=k(0.164)^1(0.15)^2$

$k=8.7\times 10^{-3}M^{-2}s^{-1}$

Now we have to determine the reaction rate when the concentration of $HgCl_2$ is 0.135 M and that of $C_2O_2^{-4}$ is 0.40 M.

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

$\text{Rate}=(8.7\times 10^{-3})\times (0.135)^1\times (0.40)^2$

$\text{Rate}=1.9\times 10^{-4}M/s$

Therefore, the reaction rate will be, $1.9\times 10^{-4}M/s$

6 0

3 years ago

The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:

motikmotik

The allowable combination for the atomic orbital is n=3, l=1, $m_{l}$ =-1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

Three quantum numbers specify an atomic orbital:

- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.

- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),

- $\boldsymbol{m}_{l}$ is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus; $m_{l}$ values are integers from -l to =l(l limits ml)

For n = 3, l can have three values: 0, 1, and 2. Since $m_{l}$ values are integers from -l to 0 to +l, for l = 0 the value of $m_{l}$ cannot be -1 (l = 0 has $m_{l}$ = 0).

There are two l values that are consistent with n and $m_{l}$ values:

l=1 or 2

Therefore, the allowable condition is n=3, l=1, $m_{l}$ =-1.

To know more about quantum numbers, visit: brainly.com/question/16979660

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4 0

2 years ago

Ap chem, please help with number 31!!

Leona [35]

Answer:

this old man looks so weird ..

Explanation:

but i m sorry .....u know the reason very well

do u wanna be my friend .

3 0

3 years ago

Help I need an answer today

Ivahew [28]

Here you are! I hope it helps, and also for the ones I put a red ‘x’ it depends on how you round it.

5 0

3 years ago

Read 2 more answers

Which of the following elements are most likely to form an ionic bond

rodikova [14]

Answer:

<u>K (Potassium) and Cl (Chlorine)</u>

Explanation:

Ionic bonds are formed with metals and nonmetals

Covalent bonds are formed only with nonmetals

Calcium and potassium are both metals

Sulfur and Chlorine are both nonmetals so they don't form ionic bonds

Carbon and Oxygen are both nonmetals so they don't form ionic bonds

6 0

3 years ago