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Aliun [14]
2 years ago
8

Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.

Chemistry
1 answer:
zepelin [54]2 years ago
3 0

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

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Which compounds are lewis acids? select all that apply. (ch3ch2)2hc+ cl3c− ch3nh2 bbr3?
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The Lewis acids are electron pair acceptors due to the availability of vacant orbitals they accept electrons from the donor compounds.

  • (CH_3CH_2)_2CH^{+} is a Lewis acid due to the unfilled valence shell on carbon atom (possessing positive charge) it accepts electron readily.
  • CCl_{3}^{-} is not a Lewis acid due to the presence of negative charge which represents that there are electrons available for donation.
  • CH_3NH_2 is not a Lewis acid due to the presence of lone pair on nitrogen which is easily donated to the acceptors.
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Hence, (CH_3CH_2)_2CH^{+} and BBr_3 are Lewis acids.

3 0
3 years ago
How many parts per million of lead is found in 250 ml of water if there is 1.30g of lead in the water? 1ml=1g
SSSSS [86.1K]

Answer:

5200 ppm

Explanation:

As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.

Given in the question,

Water = 250 ml = 250 g

Lead = 1.30 g

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ppm of Lead = \frac{Lead}{Water} \times 10^6 = \frac{1.30}{250} \times 10^6 = 5200 ppm

So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.

6 0
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