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morpeh [17]
3 years ago
15

Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a

luminum container of mass 0.325 kg filled with oxygen. This container is placed in 1.75 kg of water in an aluminum calorimeter cup of mass 0.624 kg at an initial temperature of 15.0°C. The oxygen–candy mixture in the small container (a "bomb calorimeter") is ignited, and the final temperature of the whole system is 53.5°C.
Chemistry
1 answer:
GrogVix [38]3 years ago
8 0

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

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The alkali metals play virtually the same general chemical role in all their reactions. (b) How is it based on atomic properties
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6 0
2 years ago
What is the pressure, in atmospheres, of a 0.108-mol sample of helium gas at a temperature of 20.0°C if its volume is 0.505 L?
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Answer:

P = 5.14ATM

Explanation:

Number of moles = 0.108moles

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Volume V = 0.505L.

Pressure (P) = ?

R = 0.082J/mol.K

From ideal gas equation,

PV = nRT

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What is the molarity of a solution made by dissolving 8.60 g of a solid with a
Dima020 [189]

Answer:

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Explanation:

We'll begin by calculating the number of mole of the solid. This can be obtained as follow:

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Mole of solid =?

Mole = mass / molar mass

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Next, we shall convert 280 mL to litre (L). This can be obtained as follow:

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Therefore,

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Thus, 280 mL is equivalent to 0.28 L.

Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:

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Thus, the molarity of the solution is 1.43 M.

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