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Y_Kistochka [10]
3 years ago
6

Is there a qualitative difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process

? If so, explain those differences.​
Chemistry
1 answer:
Vesna [10]3 years ago
3 0

Answer:

No

Explanation:

given that, enthalpy is a state function, that means it depends only on the initial and final states,  there is no difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process, when the cooling or heating process finish in a change of phase.

It does not  matter which way we take to cool or heat the substances the Enthalpy of this process will be the same.

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Mademuasel [1]
In the atom model, the poppy seeds are placed in different places to represent the different position of electrons.
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3 years ago
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The strongest forces of attraction occur between molecules of(1) HCl (3) HBr(2) HF (4) HI
Alexxandr [17]
It would likely be HF. It displays hydrogen bonding. 
4 0
3 years ago
Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1
bekas [8.4K]

Explanation:

2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant of reaction = K=154

Concentration of NO = [NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M

Concentration of bromine gas = [Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M

Concentration of NOBr gas = [Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M

The reaction quotient is given as:

Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}

Q=328.06

Q>K

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced.  False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

8 0
3 years ago
what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C
svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

\frac{V}{T}=k

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

\frac{V1}{T1}=\frac{V2}{T2}

In this case:

  • V1= 0.730 L
  • T1= 28 °C= 301 °K (0°C= 273°K)
  • V2= ?
  • T2= 0°C= 273 °K

Replacing:

\frac{0.730 L}{301K}=\frac{V2}{273K}

Solving:

V2=273K*\frac{0.730L}{301K}

V2=0.662 L

<u><em>The volume of the air is 0.662 L</em></u>

6 0
3 years ago
If a compound has a log kow value of 6.5, what would be its predicted concentration (in ppm) in the fat of fish that swim in wat
blsea [12.9K]
Easy ans I don't know .......
5 0
3 years ago
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