Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.
There are five states of matter out of which we encounter three states of matter in our day today life
a) gas b) solid and c) liquid
the main difference between the three is of
a) Inter molecular forces of attraction
b) thermal energy
due to this
a) solid has high intermolecular forces and low thermal energy: thus they have fix shape and occupy fix volume
b) liquid has intermediate forces and medium themal energy. Thus they may have fixed volume and but no fix shape
c) gas has weak intermoelcular forces and high thermal energy. thus they have no fixed volume no fix shape
so in the given problem
the state of the substance D- Gas.
<span>answer is A : attracted to the negative terminal of the voltage source
I think, that the "hole"moves as it captures a free electron leaving another hole in a slightly different place. The electron leaving leaves a net + charge, which is attracted to the negative terminal. Because the "hole" behaves as a positive charge it is attracted towards the negative terminal.</span>
Answer:
11.0 L
Explanation:
The equation for this reaction is given as;
2H2 + O2 --> 2H2O
2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O
At STP;
1 mol = 22.4 L
This means;
44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O
In this reaction, the limiting reactant is H2 as O2 is in excess.
The relationship between H2 and H2O;
44.8 L = 44.8 L
11.0 L would produce x
Solving for x;
x = 11 * 44.8 / 44.8
x = 11.0 L
