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3 years ago

Describe how you would prepare a supersaturated solution.

Chemistry

2 answers:

earnstyle [38]3 years ago

5 0

Answer:

To prepare a supersaturated solution, the added amount must be higher than the solubility for the given volume of solvent

Explanation:

Hello,

This could be answered by knowing that all the solutes have a property called solubility which accounts for the maximum amount of it that can be thoroughly dissolved into a specific solvent. Thus, to prepare a supersaturated solution, the added amount must be higher than the solubility for the given volume of solvent at a specific temperature. For example, at 20°C, 45.8g of aluminium chloride are completely dissolved into 100 mL of water, so at that amount, the solution will be saturated, thus, if one adds more than 45.8g the solution will start being supersaturated.

Best regards.

Anastasy [175]3 years ago

4 0

To make a supersaturated solution<span>, make a saturated </span>solution<span> of sugar by adding 360 grams of sugar to 100 mL of water at 80 degrees Celsius. When the water cools back down to 25 degrees, that 360 grams of sugar will still be dissolved even though the water </span>should<span> only dissolve 210 grams of sugar.</span>

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Define atmpospheric pressure

krok68 [10]

Answer:

that pressure is called atmospheric pressure or air pressure. It is the force exerted on a surface by the air above is as gravity pulls it to earth. atmospheric pressure is commonly measured with a barometer. In a barometer , a column of mercury is a glass tube rises ot falls as the weight of the atmospheric changes

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3 years ago

During an experiment, the percent yield of calcium chloride from a reaction was 85.22%. Theoretically, the expected amount shoul

devlian [24]

The formula to find yield is

(Actual Yield)/(Theorectical Yield) x100

Just do the math.

85.22% x 113 = 96.2986

Convert it to 3 significant figures

Ans: 96.3g

7 0

3 years ago

Read 2 more answers

The data below shows the change in concentration of dinitrogen pentoxide over time, at 330 K, according to the following process

tensa zangetsu [6.8K]

Answer: a) $1.7\times 10^{-4}$

b) $3.4\times 10^{-4}$

Explanation:

The reaction is :

$2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)$

Rate = Rate of disappearance of $N_2O_5$ = Rate of appearance of $NO_2$

Rate = $-\frac{d[N_2O_5]}{2dt}$ = $\frac{d[NO_2]}{4dt}$

Rate of disappearance of $N_2O_5$ = $\frac{\text {change in concentration}}{time}$ = $\frac{0.100-0.066}{200-0}=1.7\times 10^{-4}$

a) Rate of disappearance of $N_2O_5$ = $-\frac{d[N_2O_5]}{2dt}$

Rate of appearance of $NO_2$ = $\frac{d[NO_2]}{4dt}$

b) Rate of appearance of $NO_2$ = $\frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}$

8 0

2 years ago

What is the molarity of a HCl solution, if 28.6 mL of a 0.175 m NaOH solution is needed to neutralize a 25.0 mL sample of the HC

Katen [24]

The molarity of the HCl solution needed to neutralize 28.6 mL of a 0.175 M NaOH solution is 0.2002 M

We'll begin by writing the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 1

The mole ratio of the base, NaOH (nB) = 1

From the question given above, the following data were obtained:

Volume of base, NaOH (Vb) = 28.6 mL

Molarity of base, NaOH (Mb) = 0.175 M

Volume of acid, HCl (Va) = 25 mL

<h3>Molarity of acid, HCl (Ma) = ?</h3>

The molarity of the acid, HCl can be obtained as follow:

MaVa / MbVb = nA / nB

(Ma × 25) / (0.175 × 28.6) = 1

(Ma × 25) / 5.005 = 1

Cross multiply

Ma × 25 = 5.005

Divide both side by 25

Ma = 5.005 / 25

Therefore, the molarity of the acid, HCl needed for the reaction is 0.2002 M

Learn more: brainly.com/question/25573711

7 0

3 years ago

A 455 g piece of bopper tubing is heated to 89.5 C and placed in an insulated vessel containing 159 g of water at 22.8 C. Assumi

postnew [5]

Answer:

The final temperature of the setup = 36.6°C

Explanation:

Let the final temperature of the setup be T

Heat lost by the copper tubing = Heat gained by water and the vessel

Heat lost by the copper tubing = mC ΔT = 455 × 0.387 × (89.5 - T) = (15759.61 - 176.1T) J

Heat gained by water = mC ΔT = 159 × 4.186 × (T - 22.8) = (665.6T - 15175.1) J

Heat gained by vessel = c ΔT = 10 × (T - 22.8) = (10T - 228) J

Heat lost by the copper tubing = Heat gained by water and the vessel

(15759.61 - 176.1T) = (665.6T - 15175.1) + (10T - 228)

851.7 T = 31162.71

T = 36.6°C

7 0

3 years ago