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3 years ago

Can a solution with undissolved solute be supersaturated

1 answer:

5 0

Answer: A supersaturated solution will not contain undissolved solute because the undissolved solute will be indicative of saturated solution.

Explanation:

A supersaturated solution is the one that consists of more than the maximum concentration of the solute in the solvent that is being dissolved at a given temperature. A saturated solution is the one in which the maximum concentration of solute has been dissolved in the solvent and no additional solute can be dissolved further.

According to the given statement, a solution with undissolved solute is a saturated solution rather a supersaturated solution.

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lol click other and say sumin like I dont watch the news I only watch netflix.

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3 years ago

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Four students presented different analogies to describe the formation of an ionic bond.Student A: A tug of war between seven mid

MissTica

Answer:

The answer is A

Explanation:

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3 years ago

When 1.104 grams of iron metal are mixed with 26.023 grams of hydrochloric acid in a coffee cup calorimeter, the temperature ris

Alika [10]

Answer:

a) An exothermic reaction, will release heat. No heat will be absorbed.

b) 903.71 J of heat released

c) reaction is exothermic and ∆H will be negative.

d) ΔHreaction = 818.6 J/g

e) ΔHreaction = 45.6 kJ/mol

f) ΔHreaction = 91.2 kJ

Explanation:

<u>Step 1:</u> Data given

Mass of iron = 1.104 grams

Mass of hydrochloric acid = 26.023 grams

Initial temperature = 25.2°C

Final temperature = 33.5 °C

Temperature change = 8.3 °C

<u>Step 2:</u> The balanced equation

2Fe(s)+6HCl(aq) → 2FeCl3 (aq)+3H2 (g)

(A) Determine the amount of heat (in J) absorbed by the reaction mixture.

Since we have a rise of temperature, this means the reaction is exothermic.

An exothermic reaction, will release heat. No heat will be absorbed.

(B) How much heat (in J) was released by the reaction that occurred?

q = mC∆T

with q = heat released (in J)

with m = the mass (in grams)

with c = the specific heat capacity (in J/g°C)

with ∆T = The change in temperature (in °C)

q = (26.023g)*(4.184 J/g°C)*(8.3 °C) = 903.71 J of heat released

(C) Is this reaction exothermic or endothermic? Is ΔHreaction positive or negative?

Since we have a rise of temperature, this means the reaction is exothermic.

There is heat released so ∆H will be negative.

(D) Under constant pressure conditions (as used in this experiment), the heat released by the reaction equals the reaction enthalpy, qreleased = ΔHreaction. Determine ΔHreaction in Joules per gram of metal used (J/g).

ΔHreaction = 903.71 J/1.104 g = 818.6 J/g

(E) Determine ΔHreaction in kilojoules per mole of metal used (kJ/mol)

Number of moles of iron =1.104 grams / 55.845 g/mol = 0.0198 moles

ΔHreaction = 903.71 J / 0.0198 moles = 45641.9 J/mol = 45.6 kJ/mol

(F) Determine ΔHreaction in kilojoules per mole for the balanced reaction equation provided

Since we have 2moles of Fe in the balanced reaction;

ΔHreaction = 45.6 kJ/mol * 2 mol = 91.2 kJ

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3 years ago

What is a nonpolar bond?

serious [3.7K]

A bond between 2 nonmetal atoms that have the same electronegativity and therefore have equal sharing of the bonding electron pairExample: In H-H each H atom has an electronegativity value of 2.1, therefore the covalent bond between them is considered nonpolar. Nonpolar covalent bonds, with equal sharing of the bond electrons, arise when the electronegativities of the two atoms are equal.

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3 years ago

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Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

(b)Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = 0 ; total change in entropy = 2.881 J/K

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$

Final volume = $V_{2}$ = $2V_{1}$

(a) Change in entropy of the system Δ $S_{sys}$ = $nRIn\frac{V_{2} }{V_{1} }$

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

Δ $S_{sys}$ = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding Δ $S_{sur}$ = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ $S_{sys}$ = 2.881 J/K

Since surrounding does not change in this process Δ $S_{sur}$ = 0.

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 0 = 2.88 J/K

Δ $S_{sys}$ = 0

Since heat energy is not transferred from the system to the surrounding

Δ $S_{sur}$ = 0

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 0

6 0

3 years ago