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ehidna [41]
3 years ago
13

An 8 oz. bottle of energy drink contains 6.0 g of protein, 2.0 g of fat, and 16.3 g of carbohydrate. The fuel value of this ener

gy drink bottle is ________ kJ. The fuel values for protein, fat, and carbohydrate are 17, 38, and 17 kJ/g, respectively.
a. 280
b. 720
c. 72
d. 460
e. 520
Chemistry
1 answer:
valina [46]3 years ago
4 0

<u>Answer:</u> The correct option is d) 460 kJ

<u>Explanation:</u>

We are given:

Content of fat in energy drink = 2.0 g

Content of protein in energy drink = 6.0 g

Content of carbohydrate in energy drink = 16.3 g

Also,

The fuel value of fat = 38 kJ/g

The fuel value of protein = 17 kJ/g

The fuel value of carbohydrate = 17 kJ/g

So, the fuel value of the energy drink will be:

Total fuel value = (2.0g\times 38 kJ/g)+(6.0g\times 17 kJ/g)+(16.3g\times 17 kJ/g)

Total fuel value = [76+102+277]=460kJ

Hence, the correct option is d) 460 kJ

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What type of bonding is shown in the diagram
Nutka1998 [239]

Answer:

Covalent Bond

Explanation:

In the diagram Carbon and each of the 4 Hydrogens are sharing electrons. They are also both non metals. Both of these are characteristics of Covalent Bonds.

4 0
3 years ago
A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d
Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

8 0
3 years ago
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO
Irina-Kira [14]

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

5 0
3 years ago
In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0
3 years ago
Some nitrogen for use in synthesizing ammonia is heated slowly, maintaining the external pressure close to the internal pressure
swat32

Answer: The work done for the given process is -2188.7 J

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

W=-P\Delta V=-P(V_2-V_1)

W = amount of work done = ?

P = pressure = 50.0 atm

V_1 = initial volume = 542 L

V_2 = final volume = 974 L

Putting values in above equation, we get:

W=-50.0atm\times (974-542)L=-21600L.atm

To convert this into joules, we use the conversion factor:

1L.atm=101.33J

So, -21600L.atm=-21600\times 101.33=-2188728J=-2188.7kJ

The negative sign indicates the system is doing work.

Hence, the work done for the given process is -2188.7 J

7 0
3 years ago
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