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ICE Princess25 [194]
2 years ago
9

1. How much heat is required to melt 25.0 g of ice at 0°C?

Chemistry
1 answer:
tigry1 [53]2 years ago
8 0

Answer: The heat required to melt 25.0 g of ice at 0^0C is 8350 Joules

Explanation:

Heat of Fusion tells us how much energy is needed to convert 1g of a solid to a liquid at the same temperature.

Q=m\times L

Q = Heat absorbed = ?  

m = mass of ice = 25.0 g

L = Latent heat of fusion of ice = 334 J/g

Putting in the values, we get:

Q=25.0g\times 334J/g=8350J

Thus heat required to melt 25.0 g of ice at 0^0C is 8350 Joules

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This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

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M_2=\frac{M_1V_1}{V_2}.

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\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

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Answer: 1.87 atm

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Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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P_1 = initial pressure of gas = 2.50 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 26.4 ml

V_2 = final volume of gas = 36.2 ml

T_1 = initial temperature of gas = 2.5^oC=273+2.5=275.5K

T_2 = final temperature of gas = 10^oC=273+10=283K

Now put all the given values in the above equation, we get:

\frac{2.50\times 26.4}{275.5}=\frac{P_2\times 36.2}{283}

P_2=1.87atm

The new pressure is 1.87 atm by using combined gas law.

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