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Marrrta [24]
3 years ago
14

Consider the following reaction.

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
7 0

Answer:

NO2= 1.764 atm

N2O4= 0.1556 atm

Explanation:

This values of pressure has doubled, i.e multiplied by 2, because as the volume is halved, the pressure doubles, this is in accordance with Boyle's law

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Which statement is true regarding the circuit schematic below?
LenKa [72]
Unless you're able to provide a diagram representing the problem, there's not much I can do so solve this problem.
4 0
3 years ago
A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0
3 years ago
How strong are bonds between subatomic particles?
Aleks [24]

Answer:

they are indeed very strong

8 0
3 years ago
Naphthalene combustion may be used to calibrate the heat capacity of a bomb calorimeter. The heat of combustion of naphthalene i
tino4ka555 [31]

Answer:

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Explanation:

5 0
3 years ago
Asap Help!!! I just wanted to know if my answer is correct i am not sure if its not please correct me
Elenna [48]

Answer:

Precipitate

Explanation:

A precipitate is a solid formed from a chemical solution

7 0
2 years ago
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