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Alexus [3.1K]
3 years ago
5

A precipitate is a solid that sometimes forms when two liquids combine. A true B False

Chemistry
2 answers:
lisov135 [29]3 years ago
6 0
<h2>Question:</h2>

A precipitate is a solid that sometimes forms when two liquids combine.

<h2>Choosing:</h2>

A. True

B. False

<h2>Answer:</h2>

<u>A</u><u>.</u><u> </u><u>True</u>

<h3><u>#READINGHELPSWITHLEARNING</u><u> </u></h3><h3><u>#CARRYONLEARNING</u><u> </u></h3><h3><u>#STUDYWELL</u><u> </u></h3>
r-ruslan [8.4K]3 years ago
3 0
<h2>Question:</h2>

A precipitate is a solid that sometimes forms when two liquids combine.

<h2>Answer:</h2>

<u>A</u><u>.</u><u> </u><u>True</u><u> </u>

<h2>Explanation:</h2>
  • <u>Because</u><u> </u><u>the</u><u> </u><u>Precipitate</u><u> </u><u>it's</u><u> </u><u>forms</u><u> </u><u>solid</u><u> </u><u>when</u><u> </u><u>two</u><u> </u><u>liquids</u><u> </u><u>combine</u><u> </u><u>to</u><u> </u><u>precipitate</u><u>.</u><u> </u>

<h2><u>#CARRYONLEARNING</u><u> </u></h2><h2><u>#STUDYWELL</u><u> </u></h2>
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A gas has a volume of 5.00 L at 0°C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to ea
Diano4ka-milaya [45]

Answer:

A = -213.09°C

B = 15014.85 °C

C = -268.37°C

Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = 0°C  (273 K)

Final volume = 1100 mL, 280 L, 87.5 mL

Final temperature = ?

Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

Final volume =  87.5 mL/1000 = 0.0875 L

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

4.78 K - 273 = -268.37°C

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