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Alisiya [41]
3 years ago
13

The amount of energy needed to heat 6.2 g of a substance from 50.0°C to 80.0°C is 27.4 J. What is the specific heat capacity of

this sample?
Chemistry
1 answer:
taurus [48]3 years ago
7 0

Answer:

The heat capacity for the sample is 0.913 J/°C

Explanation:

This is the formula for heat capacity that help us to solve this:

Q / (Final T° - Initial T°) = c . m

where m is mass and c, the specific heat of the substance

27.4 J / (80°C - 50°C) = c . 6.2 g

[27.4 J / (80°C - 50°C)] / 6.2 g = c

27.4 J / 30°C . 1/6.2g = c

0.147 J/g°C = c

Therefore, the heat capacity is 0.913 J/°C

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Answer:

Increasing Surface Area

Explanation:

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Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).
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The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE  for C≡O  = 1072 kJ/mol

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BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

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ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

To learn more about enthalpy here

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