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algol13
3 years ago
8

What does supersaturated mean?

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
6 0

Explanation:

There is more solute than the solvent can hold (C).

You might be interested in
If you have a 1500 g aluminum pot, how much heat energy is needed to raise its temperature by 100°C?
Nataly [62]

The heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ.

The heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

Explanation:

The heat energy required to raise the temperature of any body can be obtained from the specific heat formula. As this formula states that the heat energy required to raise the temperature of the body is directly proportional to the product of mass of the body, specific heat capacity of the material and temperature change experienced by the material.

So in this problem, the mass of the aluminium is given as m = 1500 g, the specific heat of the aluminium is 0.900 J/g °C. Then as it is stated that the temperature is raised by 100 °C, so the pots are heat to increase by 100 °C from its initial temperature. This means the difference in temperature will be 100°C (ΔT = 100°C).

Then, the heat energy required to raise the temperature will be

q = m*c*del T = 1500 * 0.900 * 100 = 135000 = 135 kJ

Thus, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100 °C is 135 kJ.

Similarly, the mass of copper pot is given as 1500 g, the specific heat capacity of copper is 0.385 and the difference in temperature is 100  °C.

Then, the heat energy required to raise its temperature will be

q = m*c*del T = 1500 * 0.385 * 100 = 57750 = 57.75 kJ

And the heat energy required to raise the temperature of 1500 g of copper pot by 100°C is 57.75 kJ.

So, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ. And the heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

8 0
3 years ago
≡↔∛ππΔ≅×π⊄∴<br>khvjh j hjgvvvvvvvvvvvvvh jh
cupoosta [38]

its 40

I just took the test right now to get this answer

7 0
4 years ago
In the reaction K2CrO4 (aq) + PbCl2 (aq) 2KCl (aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction
tankabanditka [31]
The answer is 24 grams                       
6 0
3 years ago
Radioactive americium-241 is used in household smoke detectors and in bone mineral analysis.
Varvara68 [4.7K]

Answer:

Electrons- 95

Protons- 95

Neutrons-146

Explanation:

An atoms is made up of three fundamental particles; electrons, protons and neutrons,

Americium belongs to the f block in the periodic table. It is an actinide element.

An atom of Am-241 contains 95 protons, 95 electrons and 146 neutrons.

4 0
3 years ago
(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S
Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

    Question 1:

  • <u>H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)</u>

    Question 2:

  • <u>0.201 M</u>

Explanation:

<u>Question 1:</u>

The<em> neutralization</em> reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) <u>Word equation:</u>

  • sulfuric acid + potassium hydroxide → potassium sulfate + water

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

<u>2) Skeleton equation (unbalanced)</u>

  • H₂SO₄ + KOH → K₂SO₄ + H₂O

<u>#) Balanced chemical equation (including phases)</u>

  • H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

<u>Question 2:</u>

<u>1) Mol ratio:</u>

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

  • 1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

<u>2) Moles of H₂SO₄:</u>

  • V = 0.750 liter
  • M = 0.480 mol/liter
  • M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

<u>3) Moles of KOH:</u>

  • V = 0.700 liter
  • M = 0.290 mol/liter
  • M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

<u>4) Determine the limiting reagent:</u>

a) Stoichiometric ratio:

   1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

   0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄  is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

<u>5) Amount of H₂SO₄ that reacts:</u>

  • Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

         x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

         x = 0.203 / 2 = 0.0677 mol of H₂SO₄

<u>6) Concentration of H₂SO₄ remaining:</u>

  • Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

  • Total volume = 0.700 liter + 0.750 liter = 1.450 liter

  • Concetration = M

        M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0
3 years ago
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