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damaskus [11]
3 years ago
6

dinitrogen pentoxide decomposes to form nitrogen dioxide oxygen, following the equation 2N2O5 -> 4NO2. at a certain time poin

t, N2O5 is being consumed at the rate of 0.1 M/s. what’s the rate of production of NO2 and O2 at the same time point?

Chemistry
2 answers:
luda_lava [24]3 years ago
6 0

Answer:

NO2 is produced at 0.2 M/s, and O2 is produced at 0.05 M/s.

Explanation:

Temka [501]3 years ago
5 0

Rate of production of NO₂ = 0.2 M/s

Rate of production of O₂ = 0.05 M/s

<h3>Further explanation</h3>

Given

The rate of N₂O₅ : 0.1 M/s

Reaction

2N₂O₅⇒4NO₂+O₂

Required

The rate of production of NO₂ and O₂ at the same time point

Solution

Rate of disappearance of N₂O₅ = reaction rate x coefficient of N₂O₅

0.1 M/s=reaction rate x 2

<em>reaction rate = 0.05 M/s</em>

The rate of production of NO₂=

Rate of production of NO₂ = reaction rate x coefficient of NO₂

Rate of production of NO₂ = 0.05 M/s x 4

Rate of production of NO₂ = 0.2 M/s

The rate of production of O₂=

Rate of production of O₂ = reaction rate x coefficient of O₂

Rate of production of O₂ = 0.05 M/s x 1

Rate of production of O₂ = 0.05 M/s

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[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con
telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = 284 \ grams

solid soil volume =205 \ cc

saturated mass soil = 361 \ g

The weight of the soil after drainage is =295 \ g

Water weight for soil saturation = (361-284) = 77 \ g

Water volume required for soil saturation =\frac{77}{1} = 77 \ cc

Sample volume of water: = \frac{\text{water density}}{\text{water density input}}

= 361- 295 \\\\ = 66 \ cc

Soil water retained volume = (draining field weight - dry soil weight)

                                             = 295 - 284 \\\\ = 11 \ cc.

\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}

                    = \frac{77}{(205 + 77)}  \\\\= \frac{77}{(282)}  \\\\ = 27.30 \%

(Its saturated water volume is equal to the volume of voids)

\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}

                              = \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23

\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}

                            = \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04

6 0
3 years ago
A mixture of oxygen, hydrogen, and nitrogen exerts a total pressure of 378 kPa. If the partial pressures of oxygen and hydrogen
sertanlavr [38]
What's the relationship between total and partial pressure? The total pressure is the sum of the parcial pressures!


So for us, it would be:

378= 212+101+x

where x is the parcial pressure of nitrogen.

Now we count:
378= 212+101+x
378=313+x
378-313=x
65=x

So the parcial pressure exerted by nitrogen is 65!

8 0
3 years ago
Help meee??
Luden [163]
Answer: Monoprotic Acid
6 0
3 years ago
Which of the following happens to a molecule of an object when the object is heated? (1 point)
vfiekz [6]

Answer:

They get more energy, so they vibrate!

Explanation:

7 0
3 years ago
Write a nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90.
rusak2 [61]

Answer:

235/92U+10n→144/54Xe+90/38Sr+2/10n

Explanation:

  • The nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90 is represented by;

235/92U+10n→144/54Xe+90/38Sr+2/10n

  • In nuclear fission reactions a heavy nuclide is split into two light nuclides and is coupled by the release of energy.
3 0
3 years ago
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