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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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The Moon doesn't make its own light, it just reflects the Sun's light as all the planets do. The Sun always illuminates one half of the Moon. Since the Moon is tidally locked, we always see the same side from Earth, but there's no permanent "dark side of the Moon." The Sun lights up different sides of the Moon as it orbits around Earth – it's the fraction of the Moon from which we see reflected sunlight that determines the lunar phase.
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Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a
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Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

  • Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT
  • Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g
  • Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT
  • Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g
  • Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) <u>THIS IS THE RIGHT ANSWER</u> (the more negative= the more released)

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Answer:

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Explanation:

When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:

NH4Br(aq) + AgNO3(aq) —>

In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:

NH4Br(aq) —> NH4+(aq) + Br-(aq)

AgNO3(aq) —> Ag+(aq) + NO3-(aq)

The double displacement reaction will occur as follow:

NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)

NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)

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4 years ago
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