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IgorLugansk [536]
3 years ago
9

List two properties of mixtures

Chemistry
1 answer:
Volgvan3 years ago
3 0
Composition, and a c<span>hemical reaction.</span>
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Help me i don’t know what to put here pls thank you!
taurus [48]
I’m pretty sure that the answer is life, they all have life
4 0
3 years ago
If I have 31 liters of helium in a balloon at 9°C and increase the temperature of the balloon to 28°C, what will the new volume
Ede4ka [16]

Answer: The new volume of the balloon is 33 liters while T2 is 301K

Explanation:

Initial volume of helium V1 = 31 liters

Initial temperature T1 = 9°C

Convert temperature in Celsius to Kelvin

( 9°C + 273°C = 280K)

Final temperature T2 = 28°C

( 28°C + 273°C = 301K)

Final volume V2 = ?

According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.

Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2

31/280 = V2/301

To get the value of V2, cross multiply

V2 = (31 x 301) / 280

V2 = 9331 / 280

V2 = 33.325 liters (approximated to 33liters)

Thus, the new volume of the balloon is 33 liters while T2 is 301K

8 0
4 years ago
Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
How many moles would there be in 1.5x10^24 molecules of water?
Nookie1986 [14]

Answer:

5.85 moles

Explanation:

Moles in (3.52 X 10^24) molecules of water. 3.52 x 10^24 molecules x `1mol/6.02 x 10^23 molecules = 5.85 moles of H2O.

3 0
3 years ago
Which evidence suggests that tulip populations may be affected by a warming climate?
zalisa [80]
B. I say be because she is watering the plants early so nothing happened to her
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3 years ago
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