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nekit [7.7K]
3 years ago
7

Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen monoxide and oxygen according to the followi

ng chemical equation. 2 NO2(g) → 2 NO(g) + O2(g). A sample of NO2(g) is initially placed in a 2.50-L reaction vessel at 300°C. If the half-life and the rate constant at 300°C are 22 seconds and 0.54 M-1 s-1, respectively, how many moles of NO2 were in the original sample?
Chemistry
1 answer:
givi [52]3 years ago
7 0
T(1/2)=1/(k[NO2])
[NO2]=1/(kt(1/2))
[NO2]=1/(0.54M-1 s-1*22s) seconds cancel
[NO2]=1/11.88=0.0842
inital mole= 0.0842*volume(2.5)=0.210 mols
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Answer:

3750 cm.

Explanation:

You multiply the three side measurements to find the volume.

25cm·10cm·15cm

375cm·10cm

3750 cm.

<em><u>Hope this helps!</u></em>

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The mole is defined as the amount of a substance containing the same number of particles as exactly 12 g of C-12. The amu is def
Pavlova-9 [17]

Answer:

Because it wouldn't make any sense

Explanation:

First of all, I think it's important to highlight the definition of isotope.

Isotope (Wikipedia): Variants of a particular chemical element <u>which differ in neutron number</u>...

This means that two isotopes are the same element but have different net mass per atom, due to the different number of neutrons.

Therefore, it's important to make the definition on the same isotope so that the proportion is equal. If the definition would be made on different isotopes, the proportion wouldn't have any sense. Let me be clear with this example:

mass in grams of a C-12 atom = 1.9944235 × 10 ^ -23 g  --> this is the mass of a single C atom.

By definition --> 1 mol of anything = 6.02 x 10 ^ 23 particles of anything

Therefore, we know how much a single C atom weights. How many grams do you think that 6.02 x 10 ^ 23 atoms of C (i.e a mol of C) could weight??

1 single C atom ----------------------------- 1.9944235 × 10 ^ -23 g

6.02 x 10 ^ 23 atoms of C   ------------ <u>12.006 g  !!</u>

These 12 g is the same quantity than above! Therefore, 1 mol of C weights 12 g. If the definition were made with 13 g of C-13 (the other C isotope), these numbers will not be the same --> There would be a contradiction.

Regarding the second question, we need to search Ne-20 atomic mass in grams -->  3,3509177 × 10 ^ -23 g

Hence, if we follow the same rule, the amu would be 1/12 of Ne-20.

[ 3,3509177 × 10 ^ -23 g ] / 12 = 2.79 ^ -24 g

3 0
3 years ago
B. 10% is the correct answer
GarryVolchara [31]
No the right answer is c
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​29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.
arsen [322]

The answer for the following mention bellow.

  • <u><em>Therefore the final temperature of the gas is 260 k</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 150.0 kPa

Final pressure (P_{2}) = 210.0 kPa

Initial volume (V_{1}) = 1.75 L

Final volume (V_{2}) = 1.30 L

Initial temperature (T_{1}) = -23°C = 250 k

To find:

Final temperature (T_{2})

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas  constant

T represents the temperature of the gas

We know;

\frac{P*V}{T} = constant

\frac{P_{1} }{P_{2} } × \frac{V_{1} }{V_{2} } = \frac{T_{1} }{T_{2} }

Where;

(P_{1}) represents the initial pressure of the gas

(P_{2}) represents the final pressure of the gas

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(T_{1}) represents the initial temperature of the gas

(T_{2}) represents the final temperature of the gas

So;

\frac{150 * 1.75}{210 * 1.30} = \frac{260}{T_{2} }

(T_{2}) =260 k

<u><em>Therefore the final temperature of the gas is 260 k</em></u>

<u><em></em></u>

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Answer:

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