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Lunna [17]
3 years ago
13

Why do chemical compounds dissolve more quickly in hot solvent than in cold solvent?

Chemistry
1 answer:
Oksana_A [137]3 years ago
8 0
Chemical compounds<span> tend to </span>dissolve more quickly in hot solvent than in cold solvent, t<span>he </span>solvent<span> molecules have </span>more<span> energy to pull the solute particles apart. Give an example of two liquids that are miscible.

https://www.studyblue.com/notes/note/n/chemistry/deck/938054
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What is 0.00550 g converted to mg
Mila [183]

Answer: it is 5.5 mg

Explanation:

you have to multiply the mass value by 1000

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3 years ago
The crystallization of salt from evaporating seawater a. is an example of precipitation from a solution. b. is, in mineralogic t
Ad libitum [116K]

Answer:

is an example of precipitation from a solution.

Explanation:

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Óxido cuprico+amoniaco-------Nitrógeno +cobre +agua(redox)
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Answer:

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2 years ago
How many liters of radon gas would be in 3.43 moles at standard temperature and pressure (273 K and 100 kPa)?
MissTica

Answer: Option B. 76.83L

Explanation:

1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.

If 1 mole of Radon = 22.4L

Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L

5 0
3 years ago
"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"
Salsk061 [2.6K]

Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

8 0
3 years ago
Read 2 more answers
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