Answer:
Mass of Ag produced = 64.6 g
Note: the question is, how many grams of Ag is produced from 19.0 g of Cu and 125 g of AgNO3
Explanation:
Equation of the reaction:
Cu + 2AgNO3 ---> 2Ag + Cu(NO3)2
From the equation above, 1 mole of Cu reacts with 2 moles of AgNO3 to produce 2 moles of Ag and 1 mole of Cu(NO3)2.
Molar mass of the reactants and products are; Cu = 63.5 g/mol, Ag = 108 g/mol, AgNO3 = 170 g/mol, Cu(NO3)2 = 187.5 g/mol
To determine, the limiting reactant;
63.5 g of Cu reacts with 170 * 2 g of AgNO3,
19 g of Cu will react with (340 * 19)/63.5 g of AgNO3 =101.7 g of AgNO3.
Since there are 125 g of AgNO3 available for reaction, it is in excess and Cu is the limiting reactant.
63.5 g of Cu reacts to produce 108 * 2 g of Ag,
19 g of Cu will react to produce (216 * 19)/63.5 g of Ag = 64.6 g of Ag.
Therefore mass of Ag produced = 64.6g
Answer:
Destructive interference will occur, causing the new wave to have less energy than Wave A or Wave B.
Explanation:
Destructive interference has lesser intensity.
Answer:
72.22 g
Explanation:
975 mL Mercury× 13.5 g/mL = 72.22 g
The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.
<u>Explanation:</u>
An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.
So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as 
ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to 
ions. This process of acceptance of electrons is termed as reduction.
