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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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86.66 %  by weight of ascorbic acid in the tablet, when 0. 1220 g vitamin c tablet was dissolved in acid. This required 11. 50 ml of 0. 01740 m KIO_3 to reach the endpoint.

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A balanced chemical reaction is an equation that has equal numbers of each type of atom on both sides of the arrow.

3C_6H_80_6 +3I_2 +KIO_3 +5KI +6H^+ → 3C_6H_60_6 +6I^- +3I_2+6K^+ +3H_2O

So, the balanced chemical equation for the reaction will be:

3C_6H_8O_6 +KIO_3 → 3C_6H_6O_6 +KI + 3H_2O

That means, 3 mol of of ascorbic acid reacts with 1 mol of KIO_3

Moles of KIO_3 available is 0.0174 X 0.0115 =0.0002001 mol

Moles of ascorbic acid to be yielded should be 3 X 0.0002001 =0.0006003

So, percent mass of ascorbic acid in the tablet will be:

(0.1057 / 0.1220) X 100 %

=86.66 %

Hence, the third option is the correct answer.

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1 year ago
Cell is to a tissue as brick is to _______.<br> Cement<br> Clay<br> Bricklayer<br> wall
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Which of the following correctly describes a compound?
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A compound is a pure substance formed by the chemical combination of two or more different elements.


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3 years ago
Why are xrays used to probe the crystal structure of a material
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8 0
2 years ago
Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
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Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

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    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

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1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

7 0
3 years ago
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