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Komok [63]
2 years ago
6

In which direction will the following equilibrium shift when solid sodium fluoride is added to an aqueous solution of acetic aci

d? CH3COOH â CH3COOâ + H+ A. The equilibrium shifts to the right, to form more products. B. The equilibrium shifts to the left, to form more reactants. C. There is no change; the system is still at equilibrium.
Chemistry
1 answer:
Sedbober [7]2 years ago
6 0

Answer:

A. The equilibrium shifts to the right, to form more products.

Explanation:

When sodium fluoride, NaF, is added to an aqueous solution, some HF is produced:

NaF + H₂O ⇄ OH⁻ + Na⁺ + HF.

That is, some H⁺ reacts decreasing its concentration.

Now, the equilibrium of the acetic acid is:

CH3COOH ⇄ CH3COO⁻ + H⁺

As the concentration of H⁺ decreases:

A. The equilibrium shifts to the right, to form more products.

In order to restore the initial equilibrium.

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A low-pressure weather system comes into the city of Denver. The atmospheric pressure is 693 mmHg.693 mmHg. If 78.0%78.0% of dry
scoundrel [369]

Answer:

540.54 mmHg

Explanation:

We know that the partial pressure of a substance is defined as; Mole fraction * total pressure.

If the total amount of gases in the atmosphere is 100%, the mole fraction of nitrogen gas is now

78/100 = 0.78

Thus, partial pressure of nitrogen gas = 0.78 * 693 = 540.54 mmHg

6 0
2 years ago
If water was used to rinse the burette used to dispense the hydrochloric acid solution, how would this affect
Rufina [12.5K]

Answer:

The concentration of hydrochloric acid would be estimated to be less.

Explanation:

This is because, the hydroxyl ions from water react with the hydrogen ions from the hydrochloric acid, hence decreasing the moles of hydrogen ions which lowers the acidic strength of Hydrochloric acid.

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3 years ago
Draw the diagram of 4-propylnonane
andriy [413]

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4 0
3 years ago
How many moles are in 12.5 g ammonium chloride
77julia77 [94]

Answer:

0.233682161601125

Explanation:

3 0
3 years ago
A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what
KIM [24]
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water 

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
      = (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)- 
      = 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
       = (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
       = 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C

Answer: 28.4 °C

3 0
3 years ago
Read 2 more answers
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