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3 years ago

What is the net ionic equation for AgNO3 (aq) + Na2SO4 (aq)

Chemistry

3 answers:

Alik [6]3 years ago

8 0

The net ionic equation is

2Ag^+(aq) + SO4^2-(aq) → Ag2SO4 (s)

<h3><u>explanation</u></h3>

write the chemical equation for the reaction

2AgNO3 (aq) + Na2SO4 (aq)→ Ag2SO4 (s) + 2 NaNO3 (aq)

write the ionic equation

2Ag^+(aq) + 2 NO3^- (aq) +2 Na^+(aq) + SO4^2-(aq) → Ag2SO4 (s) +2 Na^+(aq) + 2NO3^-

cancel the spectator ions (NO3^- and Na^+)

the net ionic equation is therefore

= 2Ag ^+(aq) + SO4^2-(aq) → Ag2SO4(s)

Doss [256]3 years ago

7 0

The net ionic equation is

2Ag^+(aq) + SO4^2-(aq) → Ag2SO4 (s)

<h3><u>explanation</u></h3>

write the chemical equation for the reaction

2AgNO3 (aq) + Na2SO4 (aq)→ Ag2SO4 (s) + 2 NaNO3 (aq)

write the ionic equation

2Ag^+(aq) + 2 NO3^- (aq) +2 Na^+(aq) + SO4^2-(aq) → Ag2SO4 (s) +2 Na^+(aq) + 2NO3^-

cancel the spectator ions (NO3^- and Na^+)

the net ionic equation is therefore

= 2Ag ^+(aq) + SO4^2-(aq) → Ag2SO4(s)

Marat540 [252]3 years ago

7 0

The net ionic equation is

2Ag^+(aq) + SO4^2-(aq) → Ag2SO4 (s)

<h3><u>explanation</u></h3>

write the chemical equation for the reaction

2AgNO3 (aq) + Na2SO4 (aq)→ Ag2SO4 (s) + 2 NaNO3 (aq)

write the ionic equation

2Ag^+(aq) + 2 NO3^- (aq) +2 Na^+(aq) + SO4^2-(aq) → Ag2SO4 (s) +2 Na^+(aq) + 2NO3^-

cancel the spectator ions (NO3^- and Na^+)

the net ionic equation is therefore

= 2Ag ^+(aq) + SO4^2-(aq) → Ag2SO4(s)

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3 years ago

Read 2 more answers

the chloride of a metal M contains 47.25% of metal 1.0 gram of metal would be displaced from a compound by 0.88 gram of another

GenaCL600 [577]

Answer:

The equivalent weight of M is approximately 31.8 g

The equivalent weight of N is approximately 27.98 g

Explanation:

The given parameters are;

The percentage of the the metal M in in the chloride = 47.25%

Where by the chemical formula for the metal chloride is MClₓ, we have;

47.25% of the mass of MClₓ = Mass of M = W

Therefore, we have;

$\dfrac{0.4725}{W} = \dfrac{1}{W + 35.5 \cdot x}$

0.4725 × (W + 35.5·x) = W

0.4725·W + 0.4725×35.5×x = W

W - 0.4725·W = 16.77·x

0.5275·W = 16.77·x

W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M

The equivalent weight of M = 31.799 ≈ 31.8 g

Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g

The equivalent weight of N = 27.98 g.

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