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3 years ago

Write a balanced net ionic equation for the following reaction. Explain how you arrived at this answer. What did you cancel out?

Chemistry

1 answer:

Alex787 [66]3 years ago

4 0

The net ionic equation is

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

Explanation

AgNO₃ (aq) + KCl (aq)→ AgCl(s) +KNO₃(aq)

from above molecular equation break all soluble electrolyte into ions

Ag⁺(aq) +NO₃⁻ (aq) + K⁺(aq) +Cl⁻(aq) → AgCl (s) + K⁺(aq) + No₃⁻(aq)

cancel the spectator ions in both side of equation =K⁺ and NO₃⁻ ions

The net ionic equation is therefore

= Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

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Read 2 more answers

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

Alina [70]

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

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:

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

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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