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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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almond37 [142]

Answer:

At lower temperatures the solubility of the KCl decreased and recrystallization occurred

Explanation:

Solubility refers to the amount of solute that dissolves in one litre of water at a particular temperature. Solubility is a property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. The resulting solution is called a saturated solution.

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Let's return to the question, at elevated temperatures, we can see that the KCl is very soluble in water as evidenced by the clear solution obtained at high temperature. However, as the temperature decreases, the solubility of KCl also decreases accordingly and recrystallization of the solute occurred hence the formation of a cloudy solution and the settling of some solid at the bottom of the flask.

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4 years ago
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Digiron [165]
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4 0
3 years ago
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Kryger [21]

Answer:

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Explanation:

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3 years ago
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How many milliters of a 0.171 solution contains 1.00 g of NaCl
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