Answer:
Reducing or increasing the amount of H+ ions / hydronium (H3O+) ions
Explanation:
To reduce the pH (reducing the strength of the acid) can be done by adding a base (including a conjugate base such as bicarbonate ion) which will absorb the H+ ions either through adsorption or reaction.
Adding more H+ decreases the pH of the acid making it stronger. This can be done by adding HCL that will dissociate and increase the H+ ions.
Answer:
The reaction is a double displacement reaction
Explanation:
Let us consider the reaction equation of the reaction between ammonium oxalate and lithium acetate.
(NH4)2C2O4(aq) +2 CH3COOLi(aq) -------> 2NH4CH3CO2(aq) + Li2C2O4(s)
This is a displacement reaction. A double displacement reaction is a type of reaction in which two reactants exchange their ions to form two new compounds. Double displacement reactions usually lead to the formation of a solid product which is also called a precipitate.
The general form of a Double displacement reaction is of the format:
AB + CD → AD + CB
Where A,B,C and D represents different ions respectively.
A double displacement reaction can also be referred to as salt metathesis reaction, double replacement reaction, exchange reaction, or a double decomposition reaction, although the latter term is more strictly used when one or more of the reactants does not dissolve in the solvent.
Answer:
see explanation below
Explanation:
You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.
According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:
R: cyclopentane
R - C ≡ C (-)
Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:
R - C ≡ C - CH3
Finally in the last step, Na in NH3 are reactants to promvove the hydrogenation of alkines. In this case, it will undergo hydrogenation in the triple bond and will form an alkene:
R - CH = CH - CH3
In picture 2, you have the reaction and mechanism.