Answer:
373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.
Explanation:
Mass of silver to be precipitated on ecah spoon = 0.500 g
Number of silver spoons = 250
Total mass of silver = 250 × 0.500 g = 125 g
Moles of AgCN = n = 
Volume of AgCN solution =V
Molarity of the AgCN = 2.50 M
(1 L = 1000 mL)
373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.
Answer:
they must have same atomic number and different atomic mass
Answer:
Esterification reaction
Explanation:
An esterification reaction is an organic reaction involving an organic acid and an alkanol to give an ester or an ethanoate and water
Like the name suggests, an ester is the product formed in an esterification reaction alongside water. It is like a neutralization reaction but this time it solely contains organic molecules. These molecules react with each other to give rise to another organic molecule which is a member of a different homologous series.
Practically, to form ethyl ethanoate, ethanoic acid react with ethanol in the presence of concentrated sulphuric acid which catalyses the reaction.
Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
Q1. TI (210/81Thallium)
Q2.
The answers are opposite from each other
