Explanation:
b. What useful functions do oxidation numbers serve?
It is used to show oxidation and reduction (loss and gain of electrons)
b. How many molecules are in 1 mole of molecules?
1 mole = 6.022 * 10^23 molecules
c. What is the name given to the number of molecules in 1 mole?
Avogadro's Number of molecules
21. a. What is the molar mass of an element?
This is the mass of an element divided by the number of moles.
Molar mass = Mass / Number of moles
b. Write the molar mass rounded to two decimal places of carbon, neon, iron and uranium.
amu = Atomic Mass Unit
Carbon = 12.01 amu
Neon = 20.18 amu
Iron = 55.85 amu
Uranium = 238.03 amu
Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
![0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)](https://tex.z-dn.net/?f=0.80%3D%5Cfrac%7B1%7D%7B95%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.86%29%7D%5Cright%29)
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M
At 50 degrees Celsius and standard pressure inter-molecular forces of attraction are strongest in a sample of ethanoic acid.
Ethanoic acid has hydrogen atom bonded with a more electronegative atom; Oxygen. As a result, the molecule possesses strong intermolecular Hydrogen Bonds. Therefore; ethanoic acid, and all other carboxyllic acids have the tendency to form dimers.
If the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.
<h3>What is Charles's law?</h3>
Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.
It is expressed as;
V₁/T₁ = V₂/T₂
Given the data in the question;
- Initial temperature of gas T₁ = 100K
- Initial volume of gas V₁ = 300mL
- Final temperature T₂ = 200K
V₁/T₁ = V₂/T₂
V₂ = V₁T₂ / T₁
V₂ = ( 300mL × 200K ) / 100K
V₂ = 60000mLK / 100K
V₂ = 600mL
Therefore, if the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.
Learn more about Charles's law here: brainly.com/question/12835309
#SPJ1