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mestny [16]
3 years ago
8

A real gas behaves least like an ideal gas under the conditions of

Chemistry
1 answer:
guajiro [1.7K]3 years ago
5 0
The correct answer is option 3. A real gas behaves least like an ideal gas under the conditions of high temperature and low pressure. At this temperature and pressure, the molecules are close to each other and collisions or interactions are very likely to happen which is not an ideal gas.
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At constant P and n, T decreases at V is?
goblinko [34]

Answer:

This means that as pressure increases,the volume decreases or the volume increases as the temperature increases, and decreases as the temperature decreases

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At 19.9 degrees Celsius, we dissolve a salt crystal.
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The red dot represents the melting point of the element, which as stated is approximately 19.9 degrees Celsius and how long it took for the heat to properly completely dissolve it.

The question kind of answers itself however, is there a way to re-word it or is there a different answer you're looking for?

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3 years ago
Which of the following is not a common use of gypsum? a. wallboard/drywall for homes b. a primary ingredient for toothpaste c. s
Nataliya [291]

Answer:

pH reducing agent for acidic soils

Explanation:

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3 0
3 years ago
Which statement about Earth's core helps explain Earth's magnetic field?
Vesnalui [34]

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5 0
3 years ago
Read 2 more answers
A gas has a volume of 1.75L at -23C and 150.0kPa. At what temperature would the gas occupy 1.30L at 210.0kPa?
Katen [24]

Answer:

T2 = 260 K  

Explanation:

<em>Given data:</em>

P1 = 150.0 k Pa

T1 = (-23+ 273.15) K = 250.15 K  

V1 = 1.75 L  

P2 = 210.0 kPa  

V2 = 1.30 L

<em>To find:</em>

T2 = ?

<em>Formula:</em>

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}

<em>Calculation:</em>

T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)

T2 = 260 K  

8 0
3 years ago
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