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frez [133]
3 years ago
7

A typical fast food meal consists of a burger, fries, and a soft-drink and contains 58.0 grams of fat, 39.0 grams of protein, an

d 177 grams of carbohydrate. If jogging burns 950.0 kJ/hour, how many minutes would it take to completely burn off the meal? The respective fuel values for protein, fat, and carbohydrate are 17, 38, and 17 kJ/g, respectively.
Chemistry
1 answer:
Burka [1]3 years ago
5 0

Answer:

371.12 minutes would it take to completely burn off the meal.

Explanation:

Amount of fat in the food = 58.0 g

The respective fuel values for fat = 38 kJ/g

Amount of protein = 39.0 g

The respective fuel values for protein = 17 kJ/g

Amount of carbohydrates in the food = 177 g

The respective fuel values for carbohydrates = 17 kJ/g

Total energy derived from the meal = Q

Q=58.0 g\times 38 kJ/g+39.0 g\times 17 kJ/g+177 g\times 17 kJ/g

Q = 5876 kJ

Energy used while jogging for 1 hour = 950.0 kJ

Energy used while jogging for 60 minutes = 950.0 kJ

Let the time require to burn 5876 kJ of energy be t.

So, \frac{950.0 kJ}{60 min}=\frac{5876 kJ}{t}

\frac{5876 kJ\times 60 min}{950.0 kJ}=371.12 min

371.12 minutes would it take to completely burn off the meal.

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2. A balloon full of air has a volume of 2.75 L at a temperature of 18°C. What is the balloon's volume
Sphinxa [80]

Answer:

3.01 L

Explanation:

V

1

: 2.75L

T

1

:  

18

∘

C

V

2

: ?

T

2

:  

45

∘

C

If you know your gas laws, you have to utilize a certain gas law called Charles' Law:

V

1

T

1

=

V

2

T

2

V

1

is the initial volume,  

T

1

is initial temperature,  

V

2

is final volume,  

T

2

is final temperature.

Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.

Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:

2.75

291

=

V

2

318

By cross multiplying, we have...

291

V

2

= 874.5

Dividing both sides by 291 to isolate  

V

2

, we get...

V

2

= 3.005...

In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.

Therefore,  

V

2

= 3.01 L

5 0
4 years ago
A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.
Alexxandr [17]

<u>Answer:</u> The empirical and molecular formula of the compound is C_4H_4O and C_8H_8O_2 respectively

<u>Explanation:</u>

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = \frac{0.313}{0.078}=4.01\approx 4

For Hydrogen  = \frac{0.316}{0.078}=4.05\approx 4

For Oxygen  = \frac{0.078}{0.078}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is C_4H_4O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

n=\frac{130g/mol}{68g/mol}=1.9\approx 2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2

Hence, the empirical and molecular formula of the compound is C_4H_4O and C_8H_8O_2 respectively

4 0
3 years ago
2) Gay-Lussac's law shows a direct relationship between temperature and
shusha [124]

Answer:

The correct option is (b) "pressure".

Explanation:

Gay-Lussac's law states that the pressure of an ideal gas is directly proportional to its temperature at constant mass and volume.

Mathematically, Gay-Lussac's law is as follows :

P=kT

or

\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

Hence, the correct option is (b) "pressure".

4 0
3 years ago
How many bonding domains around the central atom and nonbonding domains around the central atom are their in PF4Cl?
Greeley [361]

Answer:

NH2- has two pairs of bonding and two pairs of non-bonding electrons participated in the formation of a molecule. The central nitrogen atom has two pairs of non-bonding electrons cause repulsion on both bonding pairs which pushes the bonds closer to each other. So, NH2- has a bent (angular) molecular geometry.

Explanation:

7 0
3 years ago
What is the percent composition of hydrogen in NH4HCO3?
Readme [11.4K]
To find the Percent Composition of an atom, you use this formula:
Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent. 
So we get: 
\frac{5}{79} * 100
So you get about: 
0.063 * 100
6.3

So, the percent composition of Hydrogen in NH4HCO3 is 6.3%

Hope this Helps! :D


7 0
3 years ago
Read 2 more answers
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