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Sedbober [7]
3 years ago
11

Write a balanced half-reaction for the oxidation of gaseous nitric oxide NO to nitrate ion NO−3 in acidic aqueous solution. Be s

ure to add physical state symbols where appropriate.
Chemistry
1 answer:
crimeas [40]3 years ago
3 0

Answer:

2H_2O(l)+NO(g)\rightarrow NO_3^-(aq) +4H^+(aq)+3e^-

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the initial version of the half-reaction as follows:

NO\rightarrow NO_3^-

Now, we assign the oxidation numbers to both nitrogen atoms:

N^{2+}O\rightarrow N^{5+}O_3^-

It means three electrons are carried, two water molecules are needed on the left side in order to balance the oxygen atoms and consequently four hydronium ions to balance hydrogen atoms in acidic media:

2H_2O(l)+NO(g)\rightarrow NO_3^-(aq) +4H^+(aq)+3e^-

Regards!

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According to Reference Table H, what is the vapor pressure of propanone at 45°C?
makkiz [27]
If we work it out with direct proportions, the vapor pressure of propanone is 56 degrees Celsius. The atmospheric pressure is about 101 kPa. If the vapor pressure is equal to the atmospheric pressure, the substance boils. Therefore, if we hold a linear proportion, 45/56 = x/101. 56x = 101*45, x = 101*45/56 = 81. The closest choice here is (3) 70 kPa.

Next time, please provide us with Reference Table H.
8 0
3 years ago
What is the mole fraction of acetic acid in a solution containing 2 moles of vinegar and 3 moles of water? A. 2 B. 3 C. 0.4 D. 0
9966 [12]

Answer:

C. 0.4.

Explanation:

<em>∵ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) = (no. of moles acetic acid)/(no. of moles of acetic acid + no. of moles of water).</em>

<em></em>

- no. of moles of acetic acid = 2, no. of moles of water = 3.

- Total no. of moles = no. of moles of acetic acid + no. of moles of water = 2 + 3 = 5.

<em>∴ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) =</em> (2)/(5)<em> = 0.4.</em>

3 0
3 years ago
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If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start wit
Svetllana [295]

Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

4Al + 3O_{2} ⇒ 2Al_{2}O_{3}

We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

16 gAl_{2}O_{3} * \frac{1 mol}{101.96 g} =0.157 mol Al_{2}O_{3}

10 g Al * \frac{1mol}{29.98 g}= 0.333 mol Al

19 g O_{2}*\frac{1 mol}{31.99 g} =0.594 mol O_{2}

Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}

0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

Finally, we have to use the yield formula to calculate the percent yield of the reaction:

Percent of yield = \frac{actual yield}{theoretical yield} * 100 = \frac{0.157 mol Al_{2}O_{3}}{0.1665 mol Al_{2}O_{3}} * 100 = 94.25

Therefore, the percent of yield is 50.03%.

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3 years ago
PLZ HELP IT BEING TIMED!!!!!! PLZ ANSWER!!!!!
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Answer:

YIKES. a bit late. Answers include 1, 2, 3

Explanation:

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Select the person responsible for the following: cotton gin
Stells [14]
The inventor of the cotton gin in Eli Whitney
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