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zaharov [31]
3 years ago
8

The hot dogs you ate at the barbecue last week were 75% fat-free by weight, had 275 calories, and weighed 110 g. a) What percent

age of these hot dogs are fat? b) How many grams of fat does each have? e) What is the percentage of Calories from fat? (Hint: Fat has approximately 9 Cal/g.)
Chemistry
1 answer:
vovangra [49]3 years ago
6 0

Answer:

a) 25%

b) 27.5 g

c) 90%

Explanation:

a) 75% fat-free by weight means 25% of the weight is made by fat.

b) 110 g ___ 100%

      x    ___ 25%

          x = 27.5g

Each hot dog has 27.5g of fat.

c) 9 cal ___ 1 g fat

      y    ___ 27.5 g fat

          y = 247.5 cal

275 cal ___ 100%

247.5 cal ___ z

     z = 90%

90 % of the calories come from fat.

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Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

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a) The ideal gas is experimenting an isocoric process and the following relationship is used:

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n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

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The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

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P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

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V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

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