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Lana71 [14]
3 years ago
14

While conducting a lab experiment, Ali calculated that 1.20 E6 Joules of heat were needed to melt 18.5 kilograms of an unknown s

ubstance at its melting point. What is the latent heat of fusion of the substance
Chemistry
1 answer:
nordsb [41]3 years ago
7 0

Answer:

Latent heat of fusion of the substance is 6.49\times 10^{4}J/kg

Explanation:

Latent heat of fusion denotes amount of energy (heat) per unit mass required to melt a solid material at constant temperature and pressure i.e. at it's melting point

Here amount of heat required = 1.20\times 10^{6}J

Mass of unknown substance being melted = 18.5 kg

So, latent heat of fusion of the substance = (required heat energy to melt)/(mass of the unknown substance) = \frac{1.20\times 10^{6}}{18.5}J/kg=6.49\times 10^{4}J/kg

So, latent heat of fusion of the substance is 6.49\times 10^{4}J/kg

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Which statement correctly describes the relationship between reactant and yield? The actual yield is calculated from the amount
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This is google's answer for the last question

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5 0
3 years ago
Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc
jeka94

Answer:

T_f = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

T_i = 21.5°C

C_{calorimeter} = 6.15 kJ/°C

T_f = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate ×  \frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

=  0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter =  C_{calorimeter} (T_f-T_i} )

21.84 kJ =6.15 kJ/°C * (T_f-21,5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - 132.225 kJ

21.84 KJ + 132.225 kJ = (6.15 kJ/^0C * T_f)

154.065 kJ = (6.15 kJ/^0C * T_f)

T_f = \frac{154.065kJ}{6.15kJ/^0C}

T_f =25.05°C

4 0
4 years ago
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