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4 years ago

How many grams of nh3 can be produced from 4.08 mol of n2 and excess h2. express your answer numerically in grams?

1 answer:

7 0

The reaction of N2 and H2 to generate NH3 is as follows:N2 + 3H2 ⇒2NH3

Since H2 is in excess amount in this reaction, the determine factor is N2. The ratio of N2 and product NH3 is 1:2. Therefore, 4.08 moles of N2 can generate 4.08*2 = 8.16 mole of NH3.

Grams of NH3 = moles of NH3 * molar mass of NH3 = 8.16 mole * 17 g/mole = 139 g.

139 grams of nh3 can be produced from 4.08 mol of n2 and excess h2.

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Wet snow is best for snow man building<br>

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Answer:

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Explanation:

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3 years ago

What happens to the composition of an atom to cause it to become positively or negatively charged?

cupoosta [38]

<span>The addition and subtraction of negatively charged electrons can easily change an atom’s charge, because they perpetually spin in valence shells outside the nucleus. It is easier for a neighboring atom to share or steal an electron rather than a positively charged proton, which is found in the nucleus. It requires a strong energy input to split a proton free from other protons and neutrons. thus, the atoms lose or gain electrons from neighboring ones and become what is known as "ions". Hope it helped!</span>

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4 years ago

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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3 years ago

The answer to (3.540)x(0.0065)x(401) should have ____.

miv72 [106K]

B. Two significant figures

Your final answer can only have as many significant figures as the number that has the least amount of significant figures in it, which is (0.0065)

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3 years ago

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

$MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}$

Thus, we can conclude that half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

3 0

3 years ago