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MAVERICK [17]
3 years ago
6

A researcher studying the nutritional value of a new candy places a 4.90 g sample of the candy inside a bomb calorimeter and com

busts it in excess oxygen. The observed temperature increase is 2.08 ∘C. If the heat capacity of the calorimeter is 33.50 kJ⋅K−1, how many nutritional Calories are there per gram of the candy?
Chemistry
1 answer:
snow_lady [41]3 years ago
8 0

Answer:

449730.879 cal/g

Explanation:

Given data:

Mass of sample = 4.9 g

Change in temperature  = 2.08 °C  (275.23 k)

Heat capacity of calorimeter = 33.50 KJ . K⁻¹

Solution:

C(candy) = Q/m

Q = C (calorimeter) × ΔT

C(candy) = C (calorimeter) × ΔT / m

C(candy) =  33.50 KJ . K⁻¹ × 275.23 K / 4.90 g

C(candy) = 9220.205 KJ / 4.90 g

C(candy) =  1881.674 KJ / g

It is known that,

1 KJ /g = 239.006 cal/g

1881.674 × 239.006 = 449730.879 cal/g

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Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

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