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3 years ago

Which of the following statements is always TRUE for the freezing of water?

Chemistry

1 answer:

Rom4ik [11]3 years ago

3 0

ΔG > 0 is always true for the freezing of water.

Explanation:

The freezing of water is only spontaneous when the temperature is fairly small. Over 273 K, the higher value of TΔS causes the sign of ΔG to be positive, and there is no freezing point.
The entropy decreases as water freezes. This does not infringe the Thermodynamics second law. The second law doesn't suggest entropy will never diminish anywhere.
Entropy will decline elsewhere, provided it increases by at least as much elsewhere.

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Which is and element A) Salad B) Sugar C) Water D) Potassium

Semenov [28]

Answer:

the only element above is potassium

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4 years ago

a sample of argon gas is cooled and its volume went from 380. mL to 250. mL. If its final temperature was -55.0C, what was its o

inn [45]

Answer:

The initial temperature was 58.4°C

Explanation:

Given the following data:

initial volume = V₁ = 380 mL = 0.38 L

final volume = V₂ = 250 mL = 0.25 L

final temperature = T₂ = -55°C = 218 K

According to Charles's law, the volume of a gas is directly proportional to the temperature (in Kelvin). The mathematical expression is:

V₁/T₁= V₂/T₂

So, we calculate the initial temperature (V₁) as follows:

T₁ = T₂/V₂ x V₁ = 218 K/(0.25 L) x 0.38 L = 331.36 K ≅ 331.4 K

Finally, we convert the initial temperature from K to °C:

T₁= 331.4 K - 273 = 58.4°C

6 0

3 years ago

Water’s heat of vaporization is 2,257 joules/gram. How much energy is released when 11.2 grams of water vapor condenses into liq

atroni [7]

Heat of vaporization is the heat required to vaporize unit mass of a substance at it's boiling point. The heat of vaporization or condensation of water is 2257 J/g.

It means that 1 g of water involves 2257 J energy during condensation or vaporization at its boiling point (100 C)

Given mass of water vapor that is condensing = 11.2 g

Calculating the heat released when 11.2 g of water vapor condenses in to liquid water at 100 $^{0}C$ :

$11.2 g *2257\frac{J}{g} = 25,278 J$

Rounding the final answer 25,278 J to three significant figures we get, 25,300 J

Therefore the correct answer is E. 25,300 J

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4 years ago

a cell was taken from dolly the sheep and used to make a geneticly identical copy of her the copy lived to be 6 years old what i

nevsk [136]

Explanation:

dolly was cloned I guess

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3 years ago

Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Which ion would NOT be present in

Allisa [31]

Answer:

The question is incomplete and confusing.

In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.

In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.

See below the details.

Explanation:

Which compound will not form ions?

1. Write the balanced molecular equation:

Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

2. Write the ionizations for the ionic aqueous compounds:

Pb(NO₃)₂(aq) → Pb⁺²(aq) + 2NO₃⁻(aq)

2KI(aq) → 2K⁺(aq) + 2I⁻(aq)

2KNO₃(aq) → 2K⁺(aq) + 2NO₃⁻(aq)

3. Write the complete ionic equation:

Pb⁺²(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)

Hence, since PbI₂(s) does not ionize, but stays in solid form, it will not form ions.

All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.

It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.

3 0

3 years ago