What is the net ionic equation for 2AgNO3 + 2NaOH = Ag2O + 2NaNO3 + H2O

Chemistry

1 answer:

soldi70 [24.7K]3 years ago

7 0

Answer:

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

Explanation:

Step 1: RxN

2AgNO₃ + 2NaOH → Ag₂O + 2NaNO₃ + H₂O

Step 2: Define states of matter

2AgNO₃ (aq) + 2NaOH (aq) → Ag₂O (s) + 2NaNO₃ (aq) + H₂O (l)

Step 3: Total Ionic Equation

2Ag⁺ (aq) + 2NO₃⁻ (aq) + 2Na⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + 2Na⁺ (aq) + 2NO₃⁻ (aq) + H₂O (l)

Step 4: Cancel out spectator ions

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

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Convert 7.50 grams of glucose C6H12O6 to moles

Lera25 [3.4K]

Answer:

The number of mole is 0.04167mole

Explanation:

To convert gram to mole, we need to calculate the molecular weight of the compound

C6H12O6

C - 12

H - 1

O - 16

Molecular weight = 6 * 12 + 1 *12 + 6 * 16

= 72 + 12 + 96

= 180g/mol

To covert gram to mole

Therefore,

= 7.50g/ 180g/mol

= 0.04167 mole of glucose

3 0

3 years ago

Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

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Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$