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Basile [38]
3 years ago
11

Given that a 12.00 g milk chocolate bar contains 8.000 g of sugar, calculate the percentage of sugar present in 12.00 g of milk

chocolate bar keeping in mind that the answer should have four significant figures (two decimal places).
Chemistry
2 answers:
kakasveta [241]3 years ago
6 0

Answer:

\%of\;sugar\;present = 66.67\%

Explanation:

Given:

Amount of milk chocolate bar = 12.00 g

Mass of sugar present in milk chocolate bar = 8.00 g

\%of\;sugar\;present=\frac{Amount\;of\;sugar}{Amount\;of\;milk}\times 100

Substitute the values in the above formula,

\%of\;sugar\;present=\frac{8.00\;g}{12.00\;g}\times 100=66.67\%

neonofarm [45]3 years ago
5 0
So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!
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<u>Explanation</u>:

Step 1: Obtain the mass of each element present in grams

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Step 2: Determine the number of moles of each type of atom present

                m/atomic mass = Molar amount (M)

Molar amount of carbon = (83.884 1 mol ) / 12 g = 6.99

Molar amount of hydrogen = (10.486  1 mol) / 1 g = 10.49

Molar amount of oxygen = (18.64  1 mol) / 16 g = 1.17

Molar amount of nitrogen = (86.99  1 mol) / 14 g = 6.21

Step 3: Divide the number of moles of each element by the smallest number of moles

            M / least M value = Atomic Ratio (R)

Atomic radius of carbon = 6.99 / 1.17 = 5.9 = 6

Atomic radius of hydrogen = 10.49 / 1.17 = 8.9 = 9

Atomic radius of oxygen = 1.17 / 1.17 = 1

Atomic radius of nitrogen = 6.21 / 1.17 = 5

Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.

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The empirical formula of a given compound is C6H9ON5.

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Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

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Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

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[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

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