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Paha777 [63]
2 years ago
9

Questions

Chemistry
1 answer:
Ulleksa [173]2 years ago
6 0

Answers:

See below  

Step-by-step explanation:

1. Most food energy

(a) Pringles

Heat from Pringles + heat absorbed by water = 0

m₁ΔH  + m₂CΔT = 0

1.984ΔH + 100 × 4.184 × 18 = 0

1.984ΔH + 7530 = 0

ΔH = -7530/1.984 = -3800 J/g

(b) Cheetos

0.884ΔH + 418.4 × 13 = 0

ΔH = -5400/0.884 = -6200 J/g

Cheetos give you more food energy per gram.

(c) Snickers

Food energy = 215 Cal/28 g × 4184 J/1 Cal = 32 000 J/g

The food energy from Cheetos is much less than that from a Snickers bar

2. Experimental uncertainty

The experimental values are almost certainly too low.

Your burning food is heating up the air around it, so much of the heat of combustion is lost to the atmosphere.

3. Percent efficiency

Experimental food energy = 3800 J/g

Actual food energy = 150 Cal/28 g × 4184 J/1 Cal = 22 000 J/g

% Efficiency = Experimental value/Actual value × 100 %

                    = 3800/22 000 × 100 %

                    = 17 %

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3 0
3 years ago
Read 2 more answers
Weighted averages, help please
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Answer:

Average atomic mass of Cl = 35.48 amu

Average atomic mass of B  = 10.8  amu

Average atomic mass of Ag  = 107.96 amu

Explanation:

1)

Given data:

Abundance of Cl-35 = 75.8%

Abundance of Cl-37 = 24.2%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (75.8×35)+(24.2×37) /100

Average atomic mass =  2653 + 895.4 / 100

Average atomic mass  = 3548.4 / 100

Average atomic mass  = 35.48 amu

2)

Given data:

Abundance of B-10 = 19.8%

Abundance of B-11 = 80.2%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (19.8×10)+(80.2×11) /100

Average atomic mass = 198  +882.2  / 100

Average atomic mass  = 1080.2 / 100

Average atomic mass  = 10.8  amu

3)

Given data:

Abundance of Ag-107 = 52%

Abundance of Ag-109 = 48%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (52×107)+(48×109) /100

Average atomic mass =  5564 +5232  / 100

Average atomic mass  = 10796 / 100

Average atomic mass  = 107.96 amu

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