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elena55 [62]
3 years ago
13

using your knowledge of colligative properties explain whether sodium chloride or calcium chloride would be a more effective sub

stance to melt the ice on a slick sidewalk. I already know that it is Calcium chloride but I don't have the answer completely worked out. can someone help explain it please? thank you
Chemistry
2 answers:
Vsevolod [243]3 years ago
8 0

Answer:

Calcium chloride

Explanation:

Colligative  property depends upon number of particles. The colligative property is independent of the nature of solute present in a solution.

More the number of particles larger the colligative property.

The following are colligative property

a) depression in freezing point

b) elevation in boiling point

c) relative lowering of vapor pressure

d) osmotic pressure

Thus if we wish to melt the we will take calcium chloride as it will give more number of ions per molecules as compared to sodium chloride.

Romashka [77]3 years ago
6 0
Calcium chloride dissociates into more ions than sodium chloride.
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Explanation:

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The question is incomplete. Her eis the complete question.

Steam reforming methane  (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

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CH_{4} + H_{2}O ⇒ CO_{} + 3H_{2}

To calculate the concentration equilibrium constant, first calculate the molarity (\frac{mol}{L}) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

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<u>Molarity of H20</u>:

H2O = \frac{10}{125} = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = \frac{18}{125} = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = \frac{0.144}{3} = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }

K_{c} = \frac{0.048.0.144^{3} }{0.112.0.032}

K_{c} = 2.10^{-2}

The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

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