The rate of energy production is referred to as Light Mass Power

"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"

Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKa NH₃/NH₄⁺

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

Moles NH₃

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

H-H equation:

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 = log [0.400 moles] / [NH₄Cl]

0.0087 = [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

3 0

3 years ago

In terms of entropy and energy, systems in nature tend to undergo changes toward

andrew-mc [135]

Number 2 lower entropy and higher entropy

7 0

3 years ago

Is there a qualitative difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process

Vesna [10]

Answer:

No

Explanation:

given that, enthalpy is a state function, that means it depends only on the initial and final states, there is no difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process, when the cooling or heating process finish in a change of phase.

It does not matter which way we take to cool or heat the substances the Enthalpy of this process will be the same.

3 0

3 years ago

) If you started with 2.3182 g of 3-nitrophthalic acid and had a 93% yield of 3-nitrophthalhydrazide, how many grams of 3-nitrop

MArishka [77]

Answer: 1.8124 g of 3-nitrophthalhydrazide were recovered.

Explanation:

The balanced chemical reaction will be :

$C_8H_5NO_6+N_2H_4\rightarrow C_8H_5N_3O_4+2H_2O$

moles of 3-nitrophthalic acid = $\frac{\text {given mass}}{\text {molar mass}}=\frac{2.3182g}{211.13g/mol}=0.0110mol$

As 1 mole of 3-nitrophthalic acid gives = 1 mole of 3-nitrophthalhydrazide

0.0110 moles of 3-nitrophthalic acid gives = $\frac{1}{1}\times 0.0110=0.0110$ mole of 3-nitrophthalhydrazide

mass of 3-nitrophthalhydrazide = $moles\times {\text {molar mass}}=0.0110mol\times 177.16g/mol=1.9488g$

As the percentage yield is 93% , the mass of 3-nitrophthalhydrazide recovered = $\frac{1.9488\times 93}{100}=1.8124g$

Therefore 1.8124 g of 3-nitrophthalhydrazide were recovered.

4 0

3 years ago

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My name is Ann [436]

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6 0

3 years ago