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Rina8888 [55]
3 years ago
11

A 2-g sample of sucrose is 6.50% hydrogen. What is the mass percentage of hydrogen in 300 g of sucrose? Show all work to support

your answer.
Chemistry
1 answer:
grin007 [14]3 years ago
3 0

The mass percentage of hydrogen in 300 g of sucrose is 9.75%

From the question given above, we were told that:

2 g of sucrose contains 6.50% (ie 0.065) hydrogen

Thus, we can obtain the percentage of hydrogen in 300 g of sucrose as follow:

2 g of sucrose contains 0.065 hydrogen.

Therefore,

300 g of sucrose will contain = \frac{300 * 0.065}{2} = 9.75% of Hydrogen.

Thus, 300 g of sucrose contain 9.75% of Hydrogen.

Learn more: brainly.com/question/16559402

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What is the mass percent of oxygen (0) in SO2?
vazorg [7]

Answer:

D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

Explanation:

Step 1: Detemine the mass of O in SO₂

There are 2 atoms of O in 1 molecule of SO₂. Then,

m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g

Step 2: Determine the mass of SO₂

m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g

Step 3: Detemine the mass percent of oxygen in SO₂

We will use the following expression.

m(O)/m(SO₂) × 100%

(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

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2 years ago
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3 years ago
Select True or False: A "gas" is a substance in which the molecules are separated on the average by distances that are large com
marta [7]

Answer:

True

Explanation:

The gaseous state is characterized in that the cohesion forces are usually null, in which the particles have their maximum mobility. The particles tend to occupy all the available volume, so their shape and volume are variable. The gaseous state is a dispersed state of matter, which means that the molecules are separated by distances much larger than the diameter of the gas molecules.

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The Law of Superposition states that
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<span>C. older material is located below younger material 
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g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

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