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Mashcka [7]
3 years ago
9

if you supply 3600 kJ of heat, how many grams of ice at 0°C can be melted, heated to its boiling point? (Make M your X in equati

ons) (figure out which q you need)
Chemistry
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

=1.36kg

Explanation:

First heat is absorbed to melt the ice- latent heat of fusion of ice then heat is absorbed to raise the temperature to 373K

The specific heat capacity of water is 4.187kJ/kg/K while  the latent heat of fusion of ice is 2230kJ/kg/K.

Letting the mass of the ice to be =x

x×2230kJ/kg/K + x×4.187kJ/kg/K×100K= 3600kJ

2230x+418.7x=3600

2647.8x=3600

x=1.36 kg

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Determine whether a precipitate will form when 0.96g of Na2CO3 is combined with 0.20g BaBr2 in a 10 L solution.
Mashcka [7]

Answer:

Qsp > Ksp, BaCO3 will precipitate

Explanation:

The equation of the reaction is;

Na2CO3 + BaBr2 -------> 2NaBr + BaCO3

Since BaCO3 may form a precipitate we can determine the Qsp of the system.

Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles

concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M

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Qsp = 6.1 * 10^-8

But, Ksp for BaCO3 is 5.1*10^-9.

Since Qsp > Ksp, BaCO3 will precipitate

3 0
3 years ago
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