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AlekseyPX
2 years ago
5

Consider the following system at equilibrium:

Chemistry
1 answer:
alexgriva [62]2 years ago
6 0

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

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Answer is:  0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
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V(chlorine) = 3,10 l.
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Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
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c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

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2 years ago
Why is it advisable to wear long sleeves when students work in a chemistry lab? to provide warmth near the lab refrigerator to p
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2 years ago
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What is the percentage of water in the following compound? Answer using three significant figures. Sodium carbonate decahydrate,
Karo-lina-s [1.5K]
The trick for this problem is to understand atomic mass: the fact that different atoms have different masses. What we need to do is add up all the atomic masses of the compound and work out the ratio of mass of water to the mass of sodium carbonate. Atomic masses are often given for each atom in the periodic table, but you can look them up on google too.

You can do this by adding up individual atoms for each molecule, or you can shortcut and lookup the molar mass of the compound (i.e.the task already done for you). 

The molar mass of water is 18.01g/mole so for 10 moles of water we have a mass of 180.1g.


The molar mass of sodium carbonate is 106g/mole (google).

So the total mass of the sodium carbonate decahydrate compound is 180.1+106 = 286.1g, of which water would make up 180.1g, so the percentage of water is is 180.1/286.1 = 0.629, so we can round this to 63%

:)




3 0
2 years ago
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2 years ago
As illustrated, the below manometer consists of a gas vessel and an open-ended U-tube containing a nonvolatile liquid with a den
STALIN [3.7K]

Answer:

1.01atm is the pressure of the gas

Explanation:

The difference in heights in the two sides is because of the difference in  pressure of the enclosed gas and the atmospheric pressure. This difference is in mm of the nonvolatile liquid. The difference in mm Hg is:

32.3mm * (0.993g/mL / 13.6g/mL) = 2.36mmHg

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In atm:

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