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3 years ago

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 ?

Chemistry

1 answer:

Dima020 [189]3 years ago

5 0

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!

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Answer:

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Explanation:

At STP, when pressure is 1.000 atm the value of temperature is 273,15 K.

If the pressure at temperature T reads as 1.250 atm.

$P_1=1.000 atm, T_1=298.15K$

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Applying Gay Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$ (At constant volume)

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3 years ago

In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5

Rufina [12.5K]

Answer:

The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.

Explanation:

The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:

MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol

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3 years ago

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Steno's law are laws that applied to sedimentary rocks. These laws helps in understanding sedimentary sequences.

Sedimentary rocks are derived from the deposition of pre-existing rocks in basins. In order to understand some important relationships between these rock layers, Steno's law offer a good insight.

Steno's law are often applied when we want to do relative dating of rock layers. Some of the laws are:

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4 years ago

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<h3>Further explanation</h3>

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the following solutions

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3 years ago

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