Answer:
D. 76%
Explanation:
Percent yield=actual/theoretical * 100
38/50 * 100 = 76%
Answer:
major enantiomer = 90.5 %
minor enantiomer = 9.5 %
Explanation:
Assuming that the major isomer is in excess;
From, ee/2 + 50
ee = enantiomeric excess = 81%
% major
=81/2 + 50 = 90.5 %
% minor
= 100% - 90.5 %
% minor
= 9.5 %
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
To find moles : moles= Mass (C₄H₂O₄) / RFM (C₄H₂O₄)
so moles = 147.7 / 114 = <span>1.2956mol
hope that helps </span>
Answer:
Na
Explanation:
<u>Reactant side:</u>
1 Na
1 Mg
2 F
<u>Product side:</u>
2 Na
1 Mg
2 F
There is one more Na atom on the product side than the reactant side. To balance this, multiply the Na on the reactant side by 2.