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taurus [48]
3 years ago
15

a piece of food is burned in a calorimeter that contains 200.0g of water. If the temperature of the water rose from 65.0°C to 83

.0°C, how much heat energy in joules was contained in the food? Specific heat if water is 4.184 j/g°C.
Chemistry
1 answer:
Flauer [41]3 years ago
4 0

Answer: 15062.4 Joules

Explanation:

The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of food = 200.0g

C = 4.184 j/g°C

Φ = (Final temperature - Initial temperature)

= 83.0°C - 65.0°C = 18°C

Then, Q = MCΦ

Q = 200.0g x 4.184 j/g°C x 18°C

Q = 15062.4 J

Thus, 15062.4 joules of heat energy was contained in the food.

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The free energy (ΔG) measures the total energy that is presented in a thermodynamic system that is available to produce useful work, especially at thermal machines. In a reaction, the value of the variation of it indicates if the process is spontaneous or nonspontaneous because the free energy intends to decrease, so, if ΔG < 0, the reaction is spontaneous.

The standard value is measured at 25°C, 298 K, and the value of free energy varies with the temperature. It can be calculated by the standard-free energy of formation (G°f), and will be:

ΔG = ∑n*G°f products - ∑n*G°f reactants, where n is the coefficient of the substance in the balanced reaction.

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saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<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)

  • <u>For B_5H_9 :</u>

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Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

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