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PtichkaEL [24]
3 years ago
6

T he thermochemical equation for the reaction between hydrazine, N2H4, and dinitrogen tetroxide is given as: 2N2H4(2) N204 (1) 3

N2 (g) + 4H20 (g),AHO-1049 kJ What is the corresponding thermochemical equation for this reaction when 6 mol of nitrogen are formed?
Chemistry
1 answer:
SpyIntel [72]3 years ago
6 0

Answer:

4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ

Explanation:

2N2H4(2) + N204 (1) --> 3N2 (g) + 4H20 (g),AHO-1049 kJ

when 6 mol of nitrogen are formed?

The ratio of Nitrogen in the previous therrmochemical equation to when 6 mol f nitrogen are formed is 2; 6/3 = 2.

So to get the new thermoochemical equation, multiply all parameters in the given equation by 2.

We have;

4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ

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mass = density x volume
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What is the lower concentration limit (vol%) at which a mixture of ethanol in air can explode?
liubo4ka [24]

Answer:

Lower explosive limit (LEL) of ethanol = 3.3%

Explanation:

In the case of alcohol, ethanol presents certain fire hazards. Its momentary flash point is 55ºF (12.9ºC), while the momentary flash point of gasoline is -45ºF (-42.8ºC), and the E85 mixture ranges between -20ºF and -4ºF (between -28 , 9ºC and -20ºC), and has a wider range of flammability limits than gasoline. For emergency response teams, this implies that during a release of the typical ethanol / gasoline mixture, the fuel can be expected to behave like gasoline: It is heavier than air - as we mentioned earlier - and can produce vapors and form flammable mixtures in the air, under most environmental conditions.

General properties and comparison with other inflambles products:

Flash point momentary Gasoline = -45 ° F

<u>Ethanol</u> = 55 ° F

E 85 = between -20º and -4º F

<u>Flammability limits </u>

Lower explosive limit (LEL) of ethanol = 3.3%

Upper Explosive Limit (UEL) = 19%

Lower explosive limit (LEL) of the mixture E 85 = 1.4%

Upper Explosive Limit (UEL) 85 = 19%

Lower explosive limit (LEL) of gasoline = 1.4%

Upper Explosive Limit (UEL) = 7.6%

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4 0
3 years ago
The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis
Dimas [21]
The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

        borneol  →  isoborneol
I         0.0972           0.0486
C         -x                     +x
E     0.0972 - x        0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832 

Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
6 0
3 years ago
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