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ludmilkaskok [199]
2 years ago
9

What are the partial pressures of NH3 and H2S at equilibrium, that is, what are the values of PNH3 and PH2S, respectively?Enter

the partial pressure of ammonia followed by the partial pressure of hydrogen sulfide numerically in atmospheres separated by a comma.
Chemistry
1 answer:
mylen [45]2 years ago
6 0

Answer:

Ammonium bisulfide, NH4HS , forms ammonia, NH3 , and hydrogen sulfide, H2S , through the reaction NH4HS(s)⇄NH3(g)+H2S(g) This reaction has a Kp value of 0.120 at 25°C .

An empty 5.00-L flask is charged with 0.300 g of pure H2S(g) , at 25°C

Partial pressure of NH3 = 0.325 atm

Partial pressure of H2S  = 0.368 atm

Explanation:

Temperature = 25 °C = 298 K

Volume = 5 L

Mass of H2S = 0.30 g

Moles of H2S = 0.30 / 34.08

= 0.0088

Using PV = nRT

Initial pressure of H2S = 0.0088 * 0.0821 * 298 / 5

= 0.043 atm

Kp = 0.120

(a). NH4HS(s) ⇄ NH3(g) + H2S(g)

Initial - 0.0 0.043

Change - + x + x

Equilibrium - x 0.043 + x

Kp = P(NH3) * P(H2S)

0.120 = x (0.043 + x)

x = 0.325 atm

Hence, at equilibrium:

Partial pressure of NH3 = 0.325 atm

Partial pressure of H2S = 0.043 + 0.325 = 0.368 atm

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M_{A}V_{A}=M_{B}V_{B}\\(0.295)V_{A}=(0.107)(24.8)\\V_{A}=\frac{(0.107)(24.8)}{(0.295)} \approx \boxed{9.00 \text{ mL}}

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What is the empirical formula of a compound that contains 75% Ag and 25% Cl by mass
lesya [120]

Answer:

The answer to your question is     AgCl

Explanation:

Data

Silver = Ag = 75%

Chlorine = Cl = 25%

1.- Convert percent numbers to grams

Silver = 75 g

Chlorine = 25 g

2.- Calculate the moles of each element

Mass number Ag = 108 g

Mass number Cl = 35.5 g

                            108 g of Ag ------------------  1 mol

                              75 g of Ag ------------------   x

                                x = (75 x 1) / 108

                                x = 75 / 108

                                x = 0.69 moles

                            35.5 g of Cl ------------------ 1 mol

                             25 g of Cl    -----------------   x

                                    x = (25 x 1) / 35.5

                                    x = 0.70 moles

3.- Divide by the lowest number of moles

Ag = 0.69/0.69 = 1

Cl = 0.70 / 0.69 = 1.02 ≈ 1

4.- Write the empirical formula

                                                 Ag₁ Cl₁  = AgCl

5 0
2 years ago
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