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Eddi Din [679]
3 years ago
13

write the symbols for the ions that form when potassium and iodine react to form the ionic compound potassium iodide.

Chemistry
2 answers:
irakobra [83]3 years ago
5 0

Answer:

I^{-}

K^{+}

Explanation:

As iodine has a higher electronegativity value and 7 valence electrons, it takes the valence electron of potassium to complete 8 electrons in its valence shell thus forming the anion I^{-} and, as consequence of losing one electron, the K atoms forms the cation K^{+}

The formula of potassium iodide is KI

Bingel [31]3 years ago
3 0
I believe it isK+  + I-=KI
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For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
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Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

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:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

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

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

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Answer: The molarity of KBr in the final solution is 1.42M

Explanation:

We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.

We will first calculate the number of moles of KBr in the individual sample before mixing together

In the first sample:

Volume (V) = 35.0 mL

Concentration (C) = 1.00M

Number of moles (n) = C × V

n = (35.0mL × 1.00M)

n= 35.0mmol

For the second sample

V = 60.0 mL

C = 0.600 M

n = (60.0 mL × 0.600 M)

n = 36.0mmol

Therefore, we have (35.0 + 36.0)mmol in the final solution

Number of moles of KBr in final solution (n) = 71.0mmol

Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.

Therefore,

Final volume of solution (V) = 50mL

Number of moles of KBr in final solution (n) = 71.0mmol

From

C = n / V

C= 71.0mmol/50mL

C = 1.42M

Therefore, the molarity of KBr in the final solution is 1.42M

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