F. None of the above (Four electrons are transferred).
2Mg → 2Mg^(2+) + <em>4e^(-)</em>
O_2 + 2H_2O + <em>4e^(-)</em> → 4OH^(-)
2Mg + O_2 + 2H_2O + <u>4e</u>^(-) → 2Mg^(2+) + 4OH^(-) + <u>4e</u>^(-)
2Mg + O_2 + 2H_2O → 2Mg^(2+) + 4OH^(-)
This is a <em>redox reaction</em> in which two atoms of Mg transfer <em>four electrons</em> to one molecule of O_2.
Just took the test
It's B = original battery, thinner wire
Answer:
The hydrolysis in aqueous HCl of compound A can lead to the formation of a carboxylic acid and an alcohol.
Explanation:
The picture shows the structures of compound A, benzoncaine and the possible products of the proposed reaction.
The acidic hydrolysis is the inverse of the esterification reaction. Therefore, the ester group of compund A will react to form the equivalent carboxylic acid and alcohol.
In order to form benzocaine, the hydrolysis happens in with the nitrile group.
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7.4x10^23 = molecules of silver nitrate sample
6.022x10^23 number of molecules per mole (Avogadro's number)
Divide molecules of AgNO3 by # of molecules per mol
7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)
(I leave off the x10^23 because they both will divide out)
Use your periodic table to find the molar weight of silver nitrate.
107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3
Now multiply your moles of AgNO3 with your molar weight of AgNO3
1.229mol x 169.868g/mol = 208.767g AgNO3
