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bija089 [108]
3 years ago
6

What happens to molecules when a substance is cooled?​

Chemistry
2 answers:
igor_vitrenko [27]3 years ago
8 0

The molecules are constantly moving, but lowering the temperature slows this movement. Because movement is slow enough, intermolecular forces can overcome the movement and take form, creating more bonds. This results in a gas turning into liquid or a liquid turning to solid, as more bonds are formed.

marissa [1.9K]3 years ago
3 0

Answer:

causes a substance to change from a liquid to a solid.

Explanation:

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g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the
weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0
3 years ago
The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at e
Sunny_sXe [5.5K]

Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

           N_{2}O_{4} \rightarrow 2NO_{2}

Let us assume that x concentration of N_{2}O_{4} is present at the initial stage. Therefore, according to the ICE table,

                    N_{2}O_{4} \rightarrow 2NO_{2}

Initial :               x                   0

Change :       - 0.1        2 \times 0.1

Equilibrium : (x - 0.1)             0.2

Now, expression for K_{p} of this reaction is as follows.

     K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

Putting the given values into the above formula as follows.

          K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

                 2 = \frac{(0.2)^{2}}{(x - 0.1)}

                2 \times (x - 0.1) = (0.2)^{2}

                            x = 0.12

This means that P_{N_{2}O_{4}} = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

6 0
3 years ago
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Density equals Mass/Volume. The equation for this problem would be 1.62 kg/205 mL. If it’s easier, this can be converted to 1620 g/205 mL. The answer would be 7.902 (rounded to the nearest thousandth). Good luck!
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I need help asap please :) it says to “identify the reactants.”
Lelechka [254]

The reactants are A: Carbon Dioxide and water

5 0
3 years ago
Calculate the number of grams of C2H6 in 4.6 miles of C2H6
Bezzdna [24]
M (C2H6) = n × M = 4.6 × 30 = 13.8 g.
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