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IgorLugansk [536]
2 years ago
10

Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution

Chemistry
1 answer:
Zigmanuir [339]2 years ago
7 0

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

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The answer is that the pilot was tired of life and committed suicide with hundreds of passengers.
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3 years ago
What is the mass of 2.071x10^14 molecules of SF2
kolezko [41]

Answer:

Name of molecule Sulfur Difluoride ( SF2)

No of Valence Electrons in the molecule 20

Hybridization of SF2 sp3 hybridization

Bond Angles 98 degrees

Molecular Geometry of SF2 Bent

Explanation:

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3 years ago
The complete combustion of a sample of propane produced 2.641 grams of carbon dioxide and 1.442 grams of water as the only produ
ohaa [14]
A general equation for a combustion reaction would be expressed as follows:

CxHy + (x+y/2)O2 = xCO2 + y/2H2O

Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.

moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C

moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H

Then, we divide the smallest amount to the each mole of the atoms. We do as follows:

C = 0.06 / 0.06 = 1
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Then we multiply a number in order to obtain a whole number ratio between the atoms.

1        CH2.67
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8 0
3 years ago
The Hall process for the production of aluminum involves the reaction of aluminum oxide with elemental carbon to give aluminum m
topjm [15]

Answer:

  • <u><em>1.7 × 10³ kg of ore.</em></u>

Explanation:

Call X the amount of aluminum ore mined to produce 1.0 × 10³ kg the aluminum metal.

Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:

  •  X         ×          71%                ×       82%     =     1.0 × 10³ kg

         ↑                      ↑                              ↑                     ↑

(mass of ore)    (% of Al in the ore)        (yield)        ( Al metal to obtain)

You must just simplify, solve and compute:

  • 0.71 × 0.82 × X = 1,000
  • X = 1,000 / (0.71 × 0.82) = 1,000 / 0.5822 = 1,717.6 Kg

Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.

6 0
3 years ago
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The answer is C, hydrogen gas. This is because in single replacement reactions, the single element (here Magnesium) replaces whichever element in the compound it corresponds to. Because Mg loses electrons since it’s a metal, it will replace the element which also loses electrons, which is Hydrogen here. So when they switch places, MgCl2 and H2 are made— and H2 is the hydrogen gas.
6 0
3 years ago
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