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Gelneren [198K]
2 years ago
13

1. Which of the following shows the process of freezing?'

Chemistry
2 answers:
melamori03 [73]2 years ago
8 0

Answer:

A. candle wax cooling on the table

Explanation:

Candle wax cooling on a table is an example of freezing because the atoms in the candle wax start to slow down as the wax cools. This is considered "freezing" because the slower an atom is vibrating then the colder it is and of course the faster an atom is vibrating then the hotter it is! :)

larisa [96]2 years ago
6 0
A. because the wax was hot meaning it’s molecules are moving fast and then cooling down which slows down the molecules. this is the same thing that happens when water changes it’s state to ice
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The rate constant for a certain reaction is measured at two different temperatures:
Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

k_1 = rate constant at temperature T_1 = 2.3\times 10^8

k_2 = rate constant at temperature T_2 = 4.8\times 10^8

E_a= activation energy = ?

R= gas constant = 8.314 J/kmol

T_1 = temperature = 280.0^0C=(273+280)=553K

T_2 = temperature = 376.0^0C=(273+376)=649K

Putting in the values ::

ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

E_a=22689.8J/mol

The activation energy Ea for this reaction is 22689.8 J/mol

3 0
3 years ago
What is the volume of 70.0 g of ether if the density of ether is 0.70 g/mL?
frosja888 [35]
M=70.0 g
p=0.70 g/mL

v=m/p

v=70.0/0.70=100.00 mL
8 0
3 years ago
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kirill [66]
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8 0
2 years ago
Cobalt-63 has a half-life of 5.3 years. If a pellet that has been in storage for 15.9 years contains 40.0g of Cobalt-63, how muc
Cerrena [4.2K]

Answer:

320 g  

Step-by-step explanation:

The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of               Fraction         Mass

half-lives   t/yr   Remaining   Remaining/g

      0        0              1

      1         5.3           ½

     2        10.6           ¼

     3        15.9           ⅛                 40.0

     4        21.2           ¹/₁₆

We see that 40.0 g remain after three half-lives.

This is one-eighth of the original mass.

The mass of the original sample was 8 × 40 g = 320 g

5 0
3 years ago
IF 3.25 mol of argon gas occupies a volume of 100. L at a particular temperature and
Serga [27]

Answer:

435.38 L

Explanation:

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

Initial mole (n₁) = 3.25 mole

Initial volume (V₁) = 100 L

Final mole (n₂) = 14.15 mole

Final volume (V₂) =?

The final volume occupied by the gas can be obtained as follow:

V₁/n₁ = V₂/n₂

100 / 3.25 = V₂ / 14.15

Cross multiply

3.25 × V₂ = 100 × 14.15

3.25 × V₂ = 1415

Divide both side by 3.25

V₂ = 1415 / 3.25

V₂ = 435.38 L

Thus, the final volume of the gas is 435.38 L

3 0
3 years ago
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