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AlexFokin [52]
3 years ago
12

Which compound decreases in solubility as the temperature of the solution is increased from 10c to 50c

Chemistry
1 answer:
Marysya12 [62]3 years ago
5 0

Answer:

<u><em></em></u>

  • <u><em>2)NH</em></u><em><u>₃</u></em>

Explanation:

The answer choices of this question are:

  • 1)NH₄Cl
  • 2)NH₃
  • 3)NaNO₃
  • 4)NaCl
<h2 /><h2>Solution</h2>

The key to answer this question is to realize that <em>NH₃</em> is a gas at the given temperatures.

The solutilibty of all gases always decreases when the temperature increases. Thus, the solubility of NH₃ will decrease when the temperature rises from 10ºC to 50ºC.

The reason for this behavior of gases is that the temperature and the kinetic energy of the particles are directly proportional. Thus, as the temperature incrases the kinetic energy of the particles increases.

As the kinetic energy of the molecules of gas in the liquid solution increases, their speeds also increase, meaning that more molecules will escape from the solution to the gas phase, leaving the soluton with less dissolved gas molecules.

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A swimming pool is filled with 160,000,000 mL of water, how many gallons of water are in the pool. Please show your work, and re
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4 years ago
Which process represents a chemical change?(1) melting of ice(2) corrosion of copper(3) evaporation of water(4) crystallization
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(2) Corrosion of copper.
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3 0
4 years ago
Pyridinium is a weak acid having a pKa of 5.2. How much pyridine (the conjugate base of pyridinium) must be added to an aqueous
Whitepunk [10]

Answer:

Amount of pyridine required = 0.0316 M

Explanation:

pH of a buffer solution is calculated by using Henderson - Hasselbalch equation.

pH=pK_a+log\frac{[Conjugate\ base]}{[weak\ acid]}

Pyridinium is a weak acid and in the presence of its conjugate base, it acts as buffer.

Henderson - Hasselbalch equation for pyridine/pyridinium buffer is as follows:

pH=pK_a+log\frac{[Py]}{PyH^+]}

pH = 4.7

pK_a=5.2

PyH^+ (Pyridinium)=0.100 M

Substitute the values in the formula

pH=pK_a+log\frac{[Py]}{PyH^+]}\\4.7=5.2 log\frac{[Py]}{0.100}

4.7-5.2=log\frac{[Py]}{0.100} \\-0.5=log\frac{[Py]}{0.100}\\\frac{[Py]}{0.100}=antilog -0.5\\\frac{[Py]}{0.100}=0.316

\frac{[Py]}{0.100} =0.316

[Py]=0.0316\ M

Amount of pyridine required = 0.0316 M

7 0
3 years ago
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