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kykrilka [37]
3 years ago
7

A solution is saturated in both nitrogen gas and potassium bromide at 750C. When the solution is cooled to room temperature, wha

t is most likely to happen? Why? (5 pts)
a. Some nitrogen gas bubbles out of solution.

b. Some potassium bromide precipitates out of solution.

c. Some nitrogen gas bubbles out of solution and some potassium bromide precipitates out of solution.

d. Nothing happens.
Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
8 0

Some Potassium bromide precipitates out of solution.

Option B.

<h3><u>Explanation:</u></h3>

Solubility is defined as the tendency of a substance to get mixed into a solvent at a particular temperature and pressure. The amount of solubility is defined as the amount of substance in grams which will make a saturated solution of 100ml at a temperature and pressure.

Potassium bromide is a salt and nitrogen is a gas. The solubility of the salts generally increase with temperature in water, and decreases with decrease in temperature. So in case of potassium bromide, the solubility of the salt will decrease, leaving some precipitate in the room temperature.

While in case of gases, the solubility of them do decrease with increase in temperature. So at room temperature, solubility of nitrogen will be more than that in 750°C. So no gas will bubble off.

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Substance in which all atoms are alike
Troyanec [42]

Answer:

That substance is an element.

Explanation:

Atoms in a substance have the same identity, that substance is an element. in other words if all atoms are alike then it is an element.

Ive gotten this question asked to me before.

8 0
4 years ago
In order for a hypothesis to become a theory, scientists MUST _____. A. do many experiments B. find a well-defined question C. m
Dovator [93]

the answer is A do many experiments.

8 0
3 years ago
Read 2 more answers
A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit
nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = <em>1.750x10⁻³ mol aniline</em>

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing  C₆H₅NH₃⁺.  C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = <em>2.63x10⁻⁵ </em><em>(1)</em>

<em />

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ =  [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

<em>pH = 3.70</em>

<em></em>

7 0
3 years ago
The concentration of the KCN solution given in Part A corresponds to a mass percent of 0.473 %. What mass of a 0.473 % KCN solut
tiny-mole [99]
If the Ka of HCN = 5.0 x 10^-10
Since
(Ka) (Ka) - 1 x 10^ -14
then
the Kb of its conjugate base (CN-) = 2.0 X 10^-5

since
pH + pOH = 14
when the pH = 10.00
then
the pOH = 4.00
& the OH-
would then equal 1.0 X 10^-4

NaCN as a base does a hydrolysis in water:
CN- & water --> HCN & OH-
notice that equal amounts of OH- & HCN are formed

Kb = [HCN] [OH-] / [CN-]

2.0 X 10^-5 = [1.0 X 10^-4] [1.0 X 10^-4] / [CN-]

[CN-] =(1.0 X 10^-8) / (2.0 X 10^-5)

[CN-] = (5.0 X 10^-4)

that's 0.00050 Molar
which is 0.00050 moles in each liter of aqueous KCN solution
which is
0.00025 moles KCN in 500. mL of aqueous KCN solution

use molar mass of KCN, to find grams:
(0.00025 moles KCN) (65.12 grams KCN / mole) = 0.01628 grams of KCn

which is 16.3 mg of KCN
& rounded to the 2 sig figs which are showing in the Ka of HCN , "5.0" X 10^-10
your answer would be
16 mg of KCN

sorry even after making a correction in calcs , I don't get one of your answers.
the only way that I could get one of them is to pretend that yours was a 1 sig fig problem,
in which case your 16 mg would round off to 20 mg.
but you have 3 sig figs in "500. ml", & 2 sig figs in both the "pH of 10.00."
& The Ka of HCN = "5.0 x 10^-10."

it does however take 12 mg of NaCN, to make 500. mL of aqueous solution pH of 10.00. the molar mass of NaCN has the smaller molar mass of 49.00 grams per mole.
maybe they meant NaCN, but wrote KCN instead.

I hope i answered this correctly for you.
8 0
4 years ago
The student dissolves the entire impure sample of in enough distilled water to make of solution. Then the student measures the a
GuDViN [60]

Answer:

Check explanation

Explanation:.

NOTE: kindly check for attached file/picture for the graph.

From the graph of absorbance against concentration from the question. We can see that the 0.3 mark absorbance is equivalent to 0.15 M. So, the concentration of CuSO4 is 0.15 M.

The concentration can also be calculated using the Beer-lambert equation for absorbance. The equation is given below;

A= ɛ×C×l --------------------------------------------------------------------------------------(1).

Where A= absorbance, ɛ= molar absorptivity, C= concentration and l= length.

Therefore, the concentration,C will now be; C= A/ ɛ×l. -------------------------------------------------------------------------(2).

Assuming the length,l is 1cm.

Hence, C= 0.300/ ɛ×1.

C= (0.300/ ɛ) M.

4 0
3 years ago
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