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liraira [26]
3 years ago
7

1. Write the word equation for the reaction occurring between lead nitrate and sodium chloride. 2. From your answer in 1 write a

balance chemical equation showing all state symbols. 3. Then write the ionic equation for the react ion occurring.
Chemistry
1 answer:
ziro4ka [17]3 years ago
8 0

Explanation:

Word Equation

Lead Nitrate + Sodium Chloride --> Lead Chloride + Sodium Nitrate

Balanced Chemical Equation

2 NaCl (aq) + Pb(NO3)2 (aq) → PbCl2 (s) + 2 NaNO3 (aq)

Ionic Equation

In an ionic equation, the aqueous species in the reaction are broken up into ions.

2Na⁺ + 2Cl⁻  + Pb²⁺  + 2NO₃⁻  → PbCl2 (s)  + 2Na⁺ +  2NO₃⁻

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7 0
3 years ago
Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.
STALIN [3.7K]

Answer: Solution A : [H_3O^+]=0.300\times 10^{-7}M

Solution B : [OH^-]=0.107\times 10^{-5}M

Solution C : [OH^-]=0.177\times 10^{-10}M

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

pH=-\log[H_3O^+]

pOH=-log[OH^-}

pH+pOH=14

[H_3O^+][OH^-]=10^{-14}

a. Solution A: [OH^-]=3.33\times 10^{-7}M

[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M

b. Solution B : [H_3O^+]=9.33\times 10^{-9}M

[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M

c. Solution C : [H_3O^+]=5.65\times 10^{-4}M

[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M

7 0
2 years ago
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