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saveliy_v [14]
3 years ago
7

Select the correct Answers.

Chemistry
2 answers:
Mandarinka [93]3 years ago
8 0

Answer:

3. The temperatures of the two substances equalize.

Explanation:

  • As two objects at different temperatures are placed in contact, heat is transferred from the warmer to the cooler object until the temperature of the two objects be the same.
  • The amount of heat that is transferred from the warmer object is equal to the amount of heat that is transferred into the cooler object.
  • This is in agreement with the law of conservation of energy.
  • <em>So, the right choice is: 3. The temperatures of the two substances equalize. </em>

<em></em>

Vanyuwa [196]3 years ago
7 0

Answer:

A and C

Explanation:

All temperatures end up the same if placed next to the other, and the heat is going out of the carbon dioxide into the water.

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John Dalton

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If 100 yards = 1 football field, what is the length, in meters, of 4.9<br> football fields?
Veseljchak [2.6K]
* Answer: 448.056 meters *
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1 yard = .9144 meters (according to Google)
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4 0
3 years ago
There are over 100 elements, but only
elena-14-01-66 [18.8K]

Answer:

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He_Helium

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C_Carbon

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F_Fluorine

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6 0
3 years ago
Chemistry - please help!!​
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4 0
2 years ago
How many moles of methanol must be added to 4.50 kg of water to lower its freezing point to -11.0 ∘c? for each mole of solute, t
DochEvi [55]

<u>Given:</u>

Mass of solvent water = 4.50 kg

Freezing point of the solution = -11 C

Freezing point depression constant = 1.86 C/m

<u>To determine:</u>

Moles of methanol to be added

<u>Explanation:</u>

The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:

ΔTf = kf*m

where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution

ΔTf = 0 C - (-11.0 C) = 11.0 C

m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg

11.0 = 1.86 * moles of methanol/4.50

moles of methanol = 26.613 moles

Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.



3 0
3 years ago
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