Answer:
Mechanism for top reaction is combustion while the side product at bottom path oximes
Explanation:
Hydroxiamine is a reducing agent that requires high temperature to combust in it'd ageous state to produced oxime while water is eliminated. The reverse is the case because the production of water during combustion I does not yield the fight result as it can e very irritating.
Increased temperature stimulates the protons to become free to combust and react with 2 butene and other aldehyde during chemical reaction.
Find the mass of C in the 2.657 g CO2:
(2.657 g CO2) / (44.01 g/mol) = 0.06037 mol CO2
Since each mole of CO2 also has 1 mole of C, this is equivalent to 0.06037 mol C.
Find the mass of H in the 1.089 g H2O:
(1.089 g H2O) / (18.02 g/mol) = 0.06043 mol H2O
Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.
Taking the ratio of H to C: 0.1209 / 0.06037 = 2.002 ~ 2
Therefore, the empirical formula of isobutylene is CH2.
<span>The appropriate response is filtration. Filtration is any of different mechanical, physical or organic operations that different solids from liquids by including a medium through which just the liquid can pass. The liquid that goes through is known as the filtrate</span>
<u>Answer:</u> The volume of concentrated solution required is 9.95 mL
<u>Explanation:</u>
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
pH = 0.70
Putting values in above equation, we get:
![0.70=-\log[H^+]](https://tex.z-dn.net/?f=0.70%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-0.70}=0.199M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-0.70%7D%3D0.199M)
1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.
Molarity of nitric acid = 0.199 M
To calculate the volume of the concentrated solution, we use the equation:

where,
are the molarity and volume of the concentrated nitric acid solution
are the molarity and volume of diluted nitric acid solution
We are given:

Putting values in above equation, we get:

Hence, the volume of concentrated solution required is 9.95 mL