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12345 [234]
3 years ago
8

Indicate the type of solute-solvent interaction that should be most important in each of the following solutions.a. LiCl in wate

rb. NF3 in acetonitrile (CH3CN)c. CCl4 in benzene (C6H6)d. methylamine (CH3NH2) in watere. Dispersion forcesf. Dipole-dipole forcesg. Ion-dipole forcesh. Hydrogen bonding
Chemistry
1 answer:
topjm [15]3 years ago
5 0

Explanation:

a. LiCl is an ionic molecule whereas water is a polar molecule with net dipole moment in it. There LiCl in water would have an ion-dipole force of interaction.

b. Both NF3 and CH3CN have dipole moment in them, since both are polar molecule. Hence, there would be dipole-dipole interaction.

c. Here both CCl4 and benzene are non polar molecules therefore, they have London dispersion force of interaction.

d. In methylamine and water both have hydrogen bonding in them. The nitrogen of CH3NH2 forms hydrogen bond with water.

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In terms of disorder and energy, systems in nature have a tendency to undergo changes toward
Simora [160]

Answer:

Option C is correct

Explanation:

Entropy is defined as the randomness in a system. In nature, a system tends to have high entropy. The higher the entropy , the spontaneous the reactions would be. As a system loses energy, it becomes more ordered and less random. Hence, as the reaction proceeds in a system, the system  has higher energy/entropy and less order.  

Hence, option c is correct

8 0
4 years ago
A mixture of 14.0 grams of H2, 84.0 grams of N2, and 64.0 grams of O2 are placed in a flask. The partial pressure of the O2 is 7
Hitman42 [59]

Answer:

P_{tot}=465.27torr

Explanation:

Hello there!

In this case, according to the given information, it will be possible for us to use the Dalton's law, in order to solve this problem. However, we first need to calculate the mole fraction of oxygen by firstly calculating the moles of each gas:

n_{H_2}=\frac{14.0g}{2.02g/mol} =6.93mol\\\\n_{N_2}=\frac{84.0g}{28.02g/mol}=3.00mol\\\\n_{O_2}=\frac{64.0}{16.00g/mol}  =2.00mol

Next, we calculate such mole fraction as follows:

x_{O_2}=\frac{2}{6.93+3+2} =0.168

Then, given the following equation:

P_{O_2}=P_{tot}*x_{O_2}

So we solve for the total pressure as follows:

P_{tot}=\frac{P_{O_2}}{x_{O_2}} \\\\P_{tot}=\frac{78.00torr}{0.168} \\\\P_{tot}=465.27torr

Regards!

6 0
3 years ago
After an experiment, scientists write a ___ which summarizes their experiment and results
Rudiy27
After an experiment, scientists write a Conclusion which summarizes their experiment and results.
6 0
3 years ago
Which list below contains only renewable resources​
gulaghasi [49]

      - Tidal Energy

     -  Biomass Energy

      - Solar Energy

      - Wind Energy

      - Hydropower

       -Ocean Thermal Energy

       - Biogas

       -Geothermal Energy

       -Air

       -water

        -wood

Explanation:

  • renewable resources are the natural resources which will help to replace the portion got deplete. These resources will fill up the gaps by any recurring processes or any natural process in a finite time.
  • Most of the renewable resources are part of the earth’s ecosphere and considered as part of the earth's natural environment.
  • water is technically a renewable resource because it can use over and over but the sustainability of this renewable resource is a question.

5 0
3 years ago
Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
3 years ago
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