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lisabon 2012 [21]
3 years ago
15

The label on a box of granola indicates that it contains 13 grams of added sugars, and 240 Calories per serving. What percent of

Calories in a serving comes from added sugars?
Chemistry
1 answer:
Afina-wow [57]3 years ago
6 0

Answer:

22 percent of Calories in a serving comes from added sugars

Explanation:

Sugars mentioned in the food products are usually processed sugars, chemically called fructose (which is a combination of glucose and sucrose). Sugars gives four Calories of energy per gram. So, 13 grams of sugar will give thirteen times of four, which is 52 Calories.

The percentage can be calculated from the following formula:

Percentage= (\frac{Calories from sugars}{TotalCalories perserving})(100)

Percentage= (\frac{52}{240})(100)\\ \\ Percentage= 22

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How many moles of nitrogen are in 2.0×10−2mole of quinine?
galben [10]

Answer: 4 x 10 ∧-2  moles of nitrogen.

Explanation:

The chemical formular for quinine is ; C20 H24 N2 O2

As can be seen from the chemical formular;

1 mole of quinine contains 2 moles of Nitrogen

Thus; 2.0 x 10 ∧-2 moles of quinine would contain

         2.0 x 10 ∧-2 x 2 = 4 x 10 ∧-2  moles of nitrogen.

 Therefore 4 x 10 ∧-2  moles of nitrogen are in 2.0×10−2mole of quinine        

7 0
3 years ago
Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o
konstantin123 [22]

Answer:

- From octane: m_{CO_2}=1.54x10^5gCO_2

- From ethanol: m_{CO_2}=9.57x10^4gCO_2

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O

Thus, the produced mass of carbon dioxide is:

m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2

Now, for ethanol:

C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O

m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2

Best regards.

3 0
3 years ago
Please help I need a answer to both
irga5000 [103]

Hi , I’m really sorry because I just have the first question... so here it is :

So it is 0.52 mol .

Have a nice day .

3 0
2 years ago
A 45.0-gram sample of copper metal was heated from 20.0°C to 100.0°C. Calculate the heat absorbed, in kJ, by the metal.
s2008m [1.1K]

Answer:

1.386 KJ

Explanation:

From the question given above, the following data were obtained:

Mass (M) of copper = 45 g

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

Heat absorbed (Q) =..?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

Change in temperature (ΔT) =?

ΔT = T2 – T1

ΔT = 100 – 20

ΔT = 80 °C

Next, we shall determine the heat absorbed by the sample of copper as follow:

Mass (M) of copper = 45 g

Change in temperature (ΔT) = 80 °C

Specific heat capacity (C) of copper = 0.385 J/gºC

Heat absorbed (Q) =..?

Q = MCΔT

Q = 45 × 0.385 × 80

Q = 1386 J

Finally, we shall convert 1386 J to KJ. This can be obtained as follow:

1000 J = 1 KJ

Therefore,

1386 J = 1386 J × 1 KJ /1000 J

1386 J = 1.386 KJ

Thus, the heat absorbed by the sample of the sample of copper is 1.386 KJ.

5 0
2 years ago
What type of property is mass
BabaBlast [244]

Answer:

I'm pretty sure its solid

6 0
3 years ago
Read 2 more answers
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