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natita [175]
3 years ago
7

A 155g sample of an unknown substance was; heated from 25.0°c to 40°c. In the process, the substance absorbed 2085 J of energy.

What is the specific heat of the substance?
Chemistry
1 answer:
Archy [21]3 years ago
8 0

Answer:

0.897 J/g°C

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass (M) of substance = 155g

Initial temperature (T1) = 25.0°C

Final temperature (T2) = 40°C

Change is temperature (ΔT) = T2 – T1 = 40°C – 25.0°C = 15°C

Heat Absorbed (Q) = 2085 J

Specific heat capacity (C) of the substance =?

Step 2:

Determination of the specify heat capacity of the substance.

Applying the equation: Q = MCΔT, the specific heat capacity of the substance can be obtained as follow:

Q = MCΔT

C = Q/MΔT

C = 2085 / (155 x 15)

C = 0.897 J/g°C

Therefore, the specific heat capacity of the substance is 0.897 J/g°C

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What is the ratio of lactic acid (Ka = 1.37x^10-4) to lactate in a solution with pH =4.29
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Henderson–Hasselbalch equation is given as,

                                         pH  =  pKa  +  log [A⁻] / [HA]   -------- (1)

Solution:

Convert Ka into pKa,

                                         pKa  =  -log Ka

                                         pKa  =  -log 1.37 × 10⁻⁴

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Putting value of pKa and pH in eq.1,

                                         4.29  =  3.863 + log [lactate] / [lactic acid]

Or,

                   log [lactate] / [lactic acid]  =  4.29 - 3.863

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Taking Anti log,

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6 0
3 years ago
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Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

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Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

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=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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