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almond37 [142]
3 years ago
7

"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"

Chemistry
1 answer:
Andru [333]3 years ago
3 0

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]

<em>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</em>

<em />

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.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

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

  • <em>H-H equation:</em>

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>

<em />

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How many atoms are in 57.5 liters of xenon gas at STP
TiliK225 [7]

Answer:

1.55 × 10²⁴ atoms Xe

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 57.5 L Xe at STP

[Solve] atoms Xe

<u>Step 2: Identify Conversions</u>

[STP] 22.4 L = 1 mol

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 57.5 \ L \ Xe(\frac{1 \ mol \ Xe}{22.4 \ L \ Xe})(\frac{6.022 \cdot 10^{23} \ atoms \ Xe}{1 \ mol \ Xe})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 1.54583 \cdot 10^{24} \ atoms \ Xe

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.54583 × 10²⁴ atoms Xe ≈ 1.55 × 10²⁴ atoms Xe

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Which inorganic substance taken up by plants is also taken in and used directly by animals, including humans?
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Answer: water

water can not be organic...but is is took by humans and plants because theur life dpends on it

Explanation:

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tamaranim1 [39]

Answer:

The answer is 0.023 moles of phosphorus

Explanation:

The 15-15-15 fertilizer is a fertilizer of great versatility, made with nitrogen, phosphorus and potassium, which makes it one of the fertilizers most used for fertilizer in the sowing plant, thus covering the crop requirements from planting. .

This fertilizer consists of 14.25% phosphorus pentoxide (P2O5). Therefore, we have to remove 14.25% at 10 grams of 15-15-15 fertilizer to calculate the moles of phosphorus. As follows:

Grams of P2O5 = 10 g x 0.1425 = 1.425 g

We calculate the molecular weight of phosphorus. We use the periodic table:

Phosphorus molecular weight = 2 x 30.97 = 61.94 g/mol

Now we calculate the moles of phosphorus in the fertilizer:

Phosphorus moles = 1,425 g/61.94 g/mol = 0.023 moles

8 0
2 years ago
in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?
dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of C_3H_5N_3O_9

\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol

Now we have to calculate the moles of CO_2,O_2,N_2\text{ and }H_2O

The balanced chemical reaction is:

4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)

From the balanced chemical reaction we conclude that,

As, 4 moles of C_3H_5N_3O_9 react to give 12 moles of CO_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{12}{4}=3 moles of CO_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 1 moles of O_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{1}{4}=0.25 moles of O_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 6 moles of N_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{6}{4}=1.5 moles of N_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 10 moles of H_2O

So, 1 moles of C_3H_5N_3O_9 react to give \frac{10}{4}=2.5 moles of H_2O

Now we have to calculate the mole fraction of water.

\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}

\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

p_{H_2O}=X_{H_2O}\times p_T

where,

p_{H_2O} = partial pressure of water vapor gas  = ?

p_T = total pressure of gas  = 58 atm

X_{H_2O} = mole fraction of water vapor gas  = 0.345

Now put all the given values in the above formula, we get:

p_{H_2O}=X_{H_2O}\times p_T

p_{H_2O}=0.345\times 58atm=20.01atm

Therefore, the partial pressure of the water vapor is, 20.01 atm

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