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Rudiy27
3 years ago
10

Nitric oxide, NO·, is a radical thought to cause ozone destruction by a mechanism similar to that of hydrofluorocarbons. One sou

rce of NO· in the stratosphere is supersonic aircraft whose jet engines convert small amounts of N2 and O2 to NO·. Write the propagation steps for the reaction of O3 with NO·.
Chemistry
1 answer:
nika2105 [10]3 years ago
7 0

Answer:

O3 + NO. = O2. + NO2

2O. +NO2 = NO. + 2O2

Explanation:

NO. = Radical

O. & O2. = Radical

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harina [27]

11 LIMITING react this is the answer

7 0
3 years ago
Assume that concentrated aqueous NH3 has a density of 0.252 g/mL (0.252 g of NH3 per mL of liquid). Calculate the volume of NH3
Kobotan [32]

The question is incomplete, here is the complete question:

Assume that concentrated aqueous NH₃ has a density of 0.252 g/mL (0.252 g of NH₃ per mL of liquid). Calculate the volume of NH₃ required to contain 0.442 mol?

<u>Answer:</u> The volume of ammonia required is 29.82 mL

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 0.442 moles

Putting values in above equation, we get:

0.442mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(0.442mol\times 17g/mol)=7.514g

To calculate the volume of ammonia, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ammonia = 0.252 g/mL

Mass of ammonia = 7.514 g

Putting values in above equation, we get:

0.252g/mL=\frac{7.514g}{\text{Volume of ammonia}}\\\\\text{Volume of ammonia}=\frac{7.514g}{0.252g/mL}=29.82mL

Hence, the volume of ammonia required is 29.82 mL

4 0
3 years ago
A 4.0 g sample of iron was heated from 0°C to 20.°C. It absorbed 35.2 J of energy as heat. What is the specific heat of this pie
Crazy boy [7]

Answer:

Explanation:i joule is equal to 0.238902957619 calories so 1251 joules is equal to 298.87 calories divided by 25.0 degrees centigrade is equal to 11.95 calories divided by the 35.2 gram sample weight to get the calories per gram per degree centigrade would come to 0.3396 calories/gram degree centigrade. Presumably this, if correct, could be used to obtain the metal in question by consulting a chart or table with specific heats of various metals because they should always be the same specific heat for each metal.

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3 years ago
What is the difference between bose- einstein condensates and plasma
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Plasma's are gases of electrically charged particles, with equal amounts of both positive and negative charge. Only certain types of particles, called bosons, can form Bose-Einstein condensates. The other type of particle- Fermions- can only have at most one particle in each state.

4 0
3 years ago
Plllllllllzzzzz helppppzzzz mezzzzz
ValentinkaMS [17]

Answer:

i think that the answer would be 7 because 8 or less is for 1 shell so I think that it makes most sence

Explanation:

hope this helps

7 0
3 years ago
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