Answer:
See explaination
Explanation:
Going by the clues that it is between Silver Flouride (AgF) and Sodium Fluoride (NaF) and since it is an aqueous solution , the 1 liter bottle is likely to be Sodium Chloride( NaCl). Going by the reaction,
AgF + NaCl= AgCl + NaF
Here, the color of AgCl is white, hence the solution cannot be AgCl.
Determination of NaCl
Determination of NaCl can be done by Mohr's Method or Volhard's method. But results in Volhard's method are more accurate . Its uses the method of back titration with Potassium Thiocynate which forms a AgCl precipitate . Prior to titration,excess AgNO3 ( The problem also has a clue that excess reagents are present in the lab ) is added to the NaCl solution so that all the Cl- ions react with Ag+. Fe3+ is then added as an indicator and the solution is titrated with KSCN to form a silver thiocyannite precipitate (AgSCN). Once all the silver has reacted, a slight excess of SCN- reacts with Fe3+ to form Fe(SCN)3 dark red complex. The concentration of Cl- is determined by subtracting the titer findings of Ag+ ions that reacted to form AgSCN from the Ag NO3 moles added to the solution. This is used because pH of the solution is acidic. If the pH of solution is basic, Mohr's method is used.
Reactions
Ag+ (aq)+ Cl-(aq) = AgCl(aq)
Ag+(aq) + SCN-(aq) = AgSCN(aq)
Fe3+(aq) + SCN-(aq) = [FeSCN]2- (aq)
Explanation:
I hope you interested about the chemical what they add in tea
Answer:
C
Explanation:
This is because matter is anything that has mass and occupies space.Therefore the space occupied by matter is volume
Answer:
Berryllium
Explanation:
its the most reactive (mark brainliest plz)
Answer:
The percent yield of this reaction is 42.1 % (option B)
Explanation:
Step 1: Data given
Theoretical yield of beryllium chloride = 10.7 grams
actual yield of beryllium chloride = 4.5 grams
Step 2: Calculate the percent yield for this reaction
Percent yield = (actual yield / theoretical yield) * 100%
Percent yield = (4.5 grams / 10.7 grams ) * 100 %
Percent yield = 42.1 %
The percent yield of this reaction is 42.1 % (option B)
