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My name is Ann [436]
2 years ago
14

According to the cooling curve, what is the approximate freezing point of the substance?

Chemistry
2 answers:
Dmitry_Shevchenko [17]2 years ago
6 0

Correct answer: B

Cooling curve is the plot of temperature versus time as the sample is allowed to cool. In a cooling curve, we start at a temperature greater than the boiling point. At this temperature, the sample is in gaseous state. At the boiling point, there is no change in temperature as the gaseous and liquid states are in equilibrium. As the temperature reduces further, the liquid starts to condense and at the melting point of the sample the liquid undergoes phase transition to solid state. At the melting temperature, a second plateau is observed as the temperature remains unchanged. At temperatures below the melting point, the sample exists as a solid.

So from the curve, the second plateau is observed at around -111^{0}C. This point represents the phase transition from liquid to solid state.

Vilka [71]2 years ago
5 0

Answer:

Its B broooooooo

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A sample of neon occupies a volume of 752 ml at 25 0
lyudmila [28]
V_{initial} = 752\:mL
T_{initial} = 25.0^0C
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → T_{initial} = 298.0\:K
V_{final} = ? (in\:milliliters)
T_{final} = 50.0^0C
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → T_{final} = 323.0\:K

By the first Law of Charles and Gay-Lussac, we have: 
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }

Solving:
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }
\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }
Product of extremes equals product of means:
298.0* V_{f} = 752*323.0
298.0 V_{f} = 242896
V_{f} = \frac{242896}{298.0}
\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark
5 0
3 years ago
Suggest why some manufacturers of car and jewellery would use metal alloys instead of pure metals?
nika2105 [10]
I think using metal alloys would be better than using the pure metal would cut down on the amount of gases that are released in the air.
6 0
3 years ago
If 25.0 ml of seawater has a mass of 25.895 g and contains 1.295 g of solute, what is the mass/mass percent concentration of sol
Romashka-Z-Leto [24]
The  %  mass/mass concentration  of   solute  in the  seawater  sample  is  calculated  as  below


% mass =  mass  of the  solute /mass of the solvent(sea water)  x100
mass  of  the  solute =1.295  g
mass  of the  solvent(sea   water_)  =  25.895 g

there  the % mass =  1.295/25.895  x100 =  5.001 %  
8 0
2 years ago
HURRY PLEASE HELP <br><br> 5 x 10^4 ÷ 2.5 x 10^2 =
weqwewe [10]

Answer:

1600

Explanation:

5×10^4÷2.5×10^2

(5×10^4)

(10^4)

(5×40)

(200)

(200÷2.5)

(80)

(80×10^2)

(10^2)

(20)

(80×20)

Answer is 1600.

Sorry if it's not correct.

7 0
3 years ago
Fill in the blanks to describe the work of Ben Kennedy, a volcanologist who works in New Zealand. To learn more about why his pu
AnnyKZ [126]

Answer:

1)- foam

2)- bubbles

3)- tension

Explanation:

5 0
2 years ago
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