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Artist 52 [7]
3 years ago
7

A curve showing the relationship between temperature and time for a given amount of liquid heated a constant rate is a _________

_ curve. a. heating c. liquid b. molar d. energy
Chemistry
2 answers:
svlad2 [7]3 years ago
3 0
Answer: A curve showing the relationship between temperature and time for a given amount of liquid heated at constant rate is a heating curve.

This is the option a. heating.

The increase of temperatue as result of heating may be ploted to show the changes. This increase of temperature related with the heat depends of the heat capacity of the substance. For a fixed amount of heat, substances with high heat capacity exhibit a lower increase in temperature than substances with low heat capacity.
Pavel [41]3 years ago
3 0
The answer you are looking for is: A: Heating Curve. Good Luck!
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Inessa [10]

Answer: b

Explanation:

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4 years ago
Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h
OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

<u>Step 1:</u> Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

<u>Step 2:</u> Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

<u>Step 3:</u> Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q =  16650 J

<u> </u>

<u>Step 4</u>: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

 ⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0  = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

<u>Step 5:</u> Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

8 0
3 years ago
What type of front in which warm air mass is cut off from the ground by cool air beneath it?
vodomira [7]

Answer:

Occluded Front

Explanation:

"Occluded Front Forms when a warm air mass gets caught between two cold air masses. The warm air mass rises as the cool air masses push and meet in the middle. The temperature drops as the warm air mass is occluded, or “cut off,” from the ground and pushed upward."

  - www.eduplace.com › science › hmxs › pdf

7 0
3 years ago
a heliox tank contains 32% helium and 68% oxygen. the total pressure in the tank is 395 kPa. What is the partial pressure of oxy
trapecia [35]

Answer:

Pp O2 = 82.944 KPa

Explanation:

heliox tank:

∴ %wt He = 32%

∴ %wt O2 = 68%

∴ Pt = 395 KPa

⇒ Pp O2 = ?

assuming a mix of ideal gases at the temperature and volumen of the mix:

∴ Pi = RTni/V

∴ Pt = RTnt/V

⇒ Pi/Pt = ni/nt = Xi

⇒ Pi = (Xi)*(Pt)

∴ Xi: molar fraction (ni/nt)

⇒ 0.68 = mass O2/mass mix

assuming mass mix = 100 g

⇒ mass O2 = 68 g

∴ molar mass O2 = 32 g/mol

⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2

⇒ mass He = 32 g

∴ molar mass He = 4.0026 g/mol

⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He

⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles

molar fraction O2:

⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099

⇒ Pp O2 = (X O2)(Pt)

⇒ Pp O2 = (0.2099)(395 KPa)

⇒ Pp O2 = 82.944 KPa

6 0
3 years ago
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