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Andrew [12]
2 years ago
5

Calculate the average rate of decomposition of NH4NO2 by the reaction 2NH4O5(aq) --> N2(g) H2O(g) at the following time inter

val. At time 250 s, the concentration of NH4NO2 is 4.12 M. At time 330 s, the concentration of NH4NO2 is 1.48 M
Chemistry
1 answer:
igor_vitrenko [27]2 years ago
6 0

Answer:

r=-0.033\frac{M}{s}

Explanation:

Hello!

In this case, since the average rate of reaction is computed as a change given by:

r=\frac{\Delta [NH_4NO_2 ]}{\Delta t}

In such a way, given the concentrations at the specified times, we plug them in to obtain:

r=\frac{(1.48M-4.12M)}{(330s-250s)}\\\\r=-0.033\frac{M}{s}

Whose negative sign means the concentration decreased due to the decomposition.

Best regards!

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ohaa [14]

Answer:

It should b KNO3

Explanation:

one Potassium (K) and three Nitrite (NO3)

8 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
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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>
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Materials:
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Answer:

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Explanation:

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142 g ethanol produced
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3 years ago
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