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schepotkina [342]
4 years ago
12

551 cal of heat is added to 5.00 g ice at –20.0 °c. what is the final temperature of the water?

Chemistry
1 answer:
hram777 [196]4 years ago
5 0
This item can be answered using the equation,
  
               h = mcp(dT) + mHv

where h is the heat, m is the mass of the substance, cp is the specific heat, and dT is the temperature difference, Hv is the latent heat of fusion. Substituting the known values from the given above,

            551 cal = (5 g)(1 cal/g°C)(T - -20) + (5 g)(80 cal/g)

The value of T from the equation is 10.2.

Answer: 10.2°C
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The total amount that Staci had in hand is $200.00.

The ticket at that time costs $87.96.

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Now that she has paid ticket, she will deduct the ticket from the cash she has in hand, so as to know the amount she is remaining with.

Tha is, $200-$87.96 = $112.04.

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3 0
3 years ago
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kykrilka [37]

Answer:

a)23.2 L

b)68.3kPa

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b)

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P1= 205×10^3Pa

V2= 12.0L

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c)

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V2= 26.0L

P2= P1V1/V2=1×196.0/26.0

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d)

V1= 40.0L

P1= 12.7×10^3Pa

V2=???

P2= 8.4×103Pa

V2= P1V1/P2= 12.7×10^3×40.0/8.4×103

V2=60.5L

e)

V1= 100mL

P1= 1atm

V2= 60mL

P2=???

P2= P1V1/V2= 1×100/60

P2= 1.67 atm

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Answer:

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Answer:

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