Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it
If the pressure of a 2.00 L sample of gas is 50.0 kPa, what pressure does the gas exert if its volume is decreased to 20.0 mL? W
2 answers:
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Answer:
Explanation:
Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant.
All zero’s between integers are always significant.
All zero’s after the decimal point are always significant.
All zero’s preceding the first integers are never significant.
Thus has three significant figures
<u>Answer:</u> The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.
<u>Explanation:</u>
We are given:
Initial concentration of sulfur dioxide = 0.500 M
Initial concentration of nitrogen dioxide = 0.500 M
Initial concentration of sulfur trioxide = 0.00500 M
Initial concentration of nitrogen monoxide = 0.00500 M
The chemical reaction follows:
<u>Initial:</u> 0.500 0.500 0.005 0.005
<u>At eqllm:</u> 0.500-x 0.500-x 0.005+x 0.005+x
The expression of equilibrium constant for the above reaction follows:
We are given:
Putting values in above equation, we get:
Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration
So, equilibrium concentration of sulfur dioxide =
Equilibrium concentration of nitrogen dioxide =
Equilibrium concentration of sulfur trioxide =
Equilibrium concentration of nitrogen monoxide =
Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.