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Alik [6]
3 years ago
10

A 12.5 g sample of granite initially at 82.0 oc is immersed into 25.0 g of water that is initially at 22.0 oc. what is the final

temperature of both substances when they reach thermal equilibrium? (specific heat capacity of water = 4.18 j/g.oc and specific heat capacity of granite = 0.790 j/g.oc)
Chemistry
1 answer:
Paraphin [41]3 years ago
7 0

The amount of heat lost by granite is equal to the amount of heat gained by water. Therefore their change in enthalpies must be equal. The opposite in sign means that one is gaining while the other is losing

ΔH granite = - ΔH water

ΔH is the change in enthalpy experienced by a closed object as it undergoes change in energy. This is expressed mathematically as,

ΔH = m Cp (T2 – T1)

Given this information, we can say that:

12.5 g * 0.790 J / g ˚C * (T2 – 82 ˚C) = - 25.0 g * 4.18 J / g ˚C * (T2 – 22 ˚C)

9.875 (T2 – 82) = 104.5 (22 – T2)

9.875 T2 – 809.75 = 2299 – 104.5 T2

114.375 T2 = 3108.75

T2 = 27.18 ˚C

The temperature of 2 objects after reaching thermal equilibrium is 27.18 ˚<span>C.</span>

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Wine goes bad soon after opening because the ethanol (CH,CH,OH) in it reacts with oxygen gas (0.) from the air to form water (11
daser333 [38]

<u>Answer:</u> The mass of water produced is 1.8 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

Given mass of ethanol = 4.59 g

Molar mass of ethanol = 46.07 g/mol

Putting values in equation 1, we get:

\text{Moles of ethanol}=\frac{4.59g}{46.07g/mol}=0.0996mol

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CH_3CH_2OH+O_2\rightarrow CH_3COOH+H_2O

By Stoichiometry of the reaction;

1 mole of ethanol produces 1 mole of water.

So, 0.0996 moles of ethanol will produce = \frac{1}{1}\times 0.0996=0.0996mol of water.

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

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Putting values in equation 1, we get:

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3 years ago
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Explanation:

The net equation will be as follows.

          K(s) + Cl_{2}(g) \rightarrow KCl(s)

So, we are required to find \Delta H_{formation} for this reaction.

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Step 1:  Convert K from solid state to gaseous state

          K(s) \rightarrow K(g),    \Delta H_{1} = 89 kJ

Step 2:  Ionization of gaseous K

           K(g) \rightarrow K^{+}(g) + e^{-},    H_{2} = 418 KJ

Step 3:  Dissociation of Cl_{2} gas into chlorine atom .

            \frac{1}{2} Cl_{2}(g) \rightarrow Cl(g),   \Delta H_{3} = \frac{244}{2} = 122 KJ

Step 4: Iozination of chlorine atom.

              Cl(g) + e^{-} \rightarro Cl^{-}(g),      H_{4} = -349 KJ

Step 5:  Add K^{+} ion and Cl^{-} ion formed above to get KCl .

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Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

      \Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}

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                  = - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

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