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Sedbober [7]
3 years ago
8

Both hydrogen sulfide (H2S) and ammonia (NH3) have strong, unpleasant odors. Which gas has the higher effusion rate? If you open

ed a container of each gas in a corner of a large room, which odor would you detect first on the other side of the room? Assume the temperature is constant. Explain your answer.
Chemistry
1 answer:
Artist 52 [7]3 years ago
3 0

Answer:

yooooo D kardi

Explanation:

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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0
3 years ago
If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha
NARA [144]
There are 2.32 x 10^6 kg sulfuric acid in the rainfall. 

Solution: 
We can find the volume of the solution by the product of 1.00 in and 1800 miles2: 
     1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m 
     1.00 in * 1 m / 39.3701 in = 0.0254 m  
     Volume = 4.662 x 10^9 m^2 * 0.0254 m
                  = 1.184 x 10^8 m^3 * 1000 L / 1 m3
                  = 1.184 x 10^11 Liters 

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70: 
     [H+] = 10^-pH = 10^-3.7 = 0.000200 M 
     [H2SO4] = 0.000100 M  
 
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid: 
     1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4 

We can now calculate for the mass of sulfuric acid in the rainfall: 
     mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
                               = 2.32 x 10^9 g * 1 kg / 1000 g
                               = 2.32 x 10^6 kg H2SO4
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