Balance the equation and state the limiting reagent in the following reaction:

What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 11.630 g to 4.000 g ?

Dima020 [189]

The amount of sugar is 2621 mg

Why?

The complete question is:

A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar.

Part B. What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 12.630 g to 4.000 g ?

To find the answer we have to determine first the amount of sugar in milligrams per gram of chocolate bar. We can find that by applying the following conversion factor:

$\frac{8.315 gSugar}{12.630 g Chocolate}*\frac{1000mg}{1g}=658.35mgSugar/gChocolate$

Now, we have to determine the amount of sugar in milligrams if we had a chocolate bar with 4.000 g:

$4.000gChocolate*\frac{658.35mgSugar}{1gChocolate}=2621mgSugar$

Have a nice day!

5 0

3 years ago

Compound A serves as a prodrug for the analgesic benzocaine. (A prodrug is a pharmacologically inactive compound that is convert

NemiM [27]

Answer:

The hydrolysis in aqueous HCl of compound A can lead to the formation of a carboxylic acid and an alcohol.

Explanation:

The picture shows the structures of compound A, benzoncaine and the possible products of the proposed reaction.

The acidic hydrolysis is the inverse of the esterification reaction. Therefore, the ester group of compund A will react to form the equivalent carboxylic acid and alcohol.

In order to form benzocaine, the hydrolysis happens in with the nitrile group.

3 0

3 years ago

The bomb calorimeter in Exercise 102 is filled with 987g water. The initial temperature of the calorimeter contents is 23.32. A

Sholpan [36]

Answer:

25.907°C

Explanation:

In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C

The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.

Thus:

$-q_{combust} = q_{water} + q_{calori}$

$q_{combust}$ = heat of combustion of benzoic acid

$q_{water}$ = heat energy released to water

$q_{calori}$ = heat energy released to the calorimeter

Therefore,

$-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})$

1.056*26.42 = [0.987*4.18 + 6.66]( $T_{f}$ - 23.32)

27.8995 = [4.12566+6.660]( $T_{f}$ - 23.32)

( $T_{f}$ - 23.32) = 27.8995/10.7857 = 2.587

$T_{f}$ = 23.32 + 2.587 = 25.907°C

4 0

3 years ago

A mixture of propane and butane is burned with pure oxygen. The combustion products contain 46.7 mole% H2O. After all the water

wlad13 [49]

Answer:

28%

Explanation:

Basically, all o did was write the equations, balance it and solve for them. Also, at the place I stared, I used simultaneous equation to solve it. Multiplying by 8 and also 3.

It's a pretty straightforward question.

At the final step that's missing, I Did

(y)C3H8 = 2.8 / ( 2.8 + 7.1)

(y)C3H8 = 0.28