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3 years ago

Fill in the left side of this equilibrium constant equation for the reaction of acetic acid HCH3CO2 with water.

Chemistry

1 answer:

Luda [366]3 years ago

5 0

Answer:

The left side equation is:

$Ka = \frac{[CH_{3}COO^{-}] [H_{3}O^{+}]}{[HCH_{3}COO] [H_{2}O]}$

Explanation:

For the reaction of acetic acid HCH3CO2 with the water, The equilibrium constant equation is

$HCH_{3}COO + H_{2}O \rightleftharpoons CH_{3}COO^{-} + H_{3}O$

The left side of this equilibrium constant equation will be written as shown below:

$Ka = \frac{[CH_{3}COO^{-}] [H_{3}O^{+}]}{[HCH_{3}COO] [H_{2}O]}$

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Which of the following aqueous solutions will have the lowest freezing point? a. 0.75 m (NH4)3PO4 b. 1.0 m CaSO4 c. 1.0 m LiCl d

Nataly [62]

Answer: a. 0.75m (NH4)3PO4 will have the lowest freezing point .

Explanation: Freezing point decreases as the concentration of the solute substance increases. Assuming the same solvent for all of them, for instance water.

∆T= i.Kf.b

∆T= freezing point depression

i= vant Hoff factor

Kf= molality

Assuming water to be the solvent for all Kf=1.86°C/M

VANT HOFF FACTORS :

For (NH4)3PO4

This has 3 ionic bonding and 1 covalent bonding making it 4 bond

Therefore i=4

For CaSO4

This has 1 metallic bond and 1 covalent bond making it 2 bond.

Therefore i=2

For LiCl

This has 1 metallic bond and 1 non metallic bond making it 2 bond.

Therefore i=2

For CH3OH

This has only 1 covalent bond.

Therefore i=1

MOLALITY:

(NH4)3PO4 = 0.75M

CaSO4= 1.0M

LiCl= 1.0M

CH3OH= 1.5M

FREEZING POINT DEPRESSION:

For (NH4)3PO4

∆T= 4×0.75×1.86=5.58°C

For CaSO4

∆T= 2×1.0×1.86=3.72°C

For LiCl

∆T = 2×1.0×1.86= 3.72°C

For CH3OH

∆T= 1×1.5×1.86=2.79°C

REMEMBER THE HIGHER THE FREEZING POINT DEPRESSION THE LOWER THE FREEZING POINT.

FREEZING POINT DEPRESSION IS THE CHANGE IN THE FREEZING POINT PROPORTIONAL TO THE AMOUNT OF SOLUTE ADDED THE THE SOLUTION.

THEREFORE THE ONE WITH THE LOWEST FREEZING POINT IS (NH4)3PO4

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3 years ago

Which change is most likely to occur when a molecule of h2 and a molecule i2 collide with proper orientation and sufficient ener

Alekssandra [29.7K]

Answer: Formation of hydrogen iodide will occur.

Explanation:

When molecule of hydrogen gas $(H_2)$ and molecule of iodine gas $(I_2)$ collide with proper orientation and with sufficient amount of energy. This collision of the two different molecules of gases will result in the formation of hydrogen iodide.

The balanced chemical equation of the reaction will written as:

$H_2(g)+I_2(g)\righarrow 2HI(g)$

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4 years ago

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mr Goodwill [35]

Answer:

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3 years ago

How many grams of dry nh4cl need to be added to 2.00 l of a 0.600 m solution of ammonia, nh3, to prepare a buffer solution that

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Nh4cl = 117.165 gram ( solution is attached )

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Help please!!! My science homework is due in 2 days and I need help!

sertanlavr [38]

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