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soldi70 [24.7K]
3 years ago
15

PLEASE HELP Which of the following properties of a molecule is related to temperature?

Chemistry
2 answers:
Sholpan [36]3 years ago
5 0
I think the answer is C. Speed?
I'm not 100% sure, but like 96% sure.
Hopefully, you get it right!~
crimeas [40]3 years ago
4 0
D. heat of condensation
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Please help brainiest is award.
sineoko [7]

Answer:

7688

Explanation:

3 0
3 years ago
Which of the following statements is true about the sense of smell? Men and women smell the same number of odors. The amount of
puteri [66]

Answer:

I think it is B

which I think is the correct answer

4 0
3 years ago
Read 2 more answers
In a 63.17 g sample of SO3, how many grams are sulfur?
Evgesh-ka [11]

Answer:

25.30 gram

Explanation:

No of moles = given mass / molar mass

No of moles = 63.17/80.06

0.7890 moles

Mass of sulphar = no of moles× molar mass of sulphar

Mass of sulphur = 0.7890×32.065

25.30 gram

4 0
3 years ago
This table shows the densities of a sample of substances.
Arisa [49]
When placed in a container, the heaviest (most dense) will sink to the bottom and the lightest (least dense) will rise to the top.
Therefore, Gasoline would rise to the top.
3 0
4 years ago
Helpppp pleaseee ill give brainliest
Studentka2010 [4]

Answer:

The answers are in the explanation.

Explanation:

The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:

Increasing temperature of ice from -10°C - 0°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g

Q = 2.06J/g°C*10°C*10g

Q = 206J

Change from solid to liquid:

The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:

Q = 333.55J/g*10g

Q = 3335.5J

Increasing temperature of liquid water from 0°C - 100°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g

Q = 4.18J/g°C*100°C*10g

Q = 4180J

Change from liquid to gas:

The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:

Q = 2260J/g*10g

Q = 22600J

Increasing temperature of gas water from 100°C - 120°C:

Q = S*ΔT*m

Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g

Q = 1.87J/g°C*20°C*10g

Q = 374J

Total Energy:

206J + 3335.5 J + 4180J + 22600J + 374J =

30695.5J =

30.7kJ

5 0
3 years ago
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