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navik [9.2K]
3 years ago
9

Consider the following scenario

Chemistry
1 answer:
GarryVolchara [31]3 years ago
7 0

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)

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Ethylene produced by fermentation has a specific gravity of 0.787 at 25 degree Celsius. What is the volume of 125g of ethanol at
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<u>Answer:</u> The volume of given amount of ethanol at this temperature is 159.44 mL

<u>Explanation:</u>

Specific gravity is given by the formula:

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Putting values in above equation, we get:

0.787=\frac{\text{Density of a substance}}{0.997g/mL}\\\\\text{Density of a substance}=(0.787\times 0.997g/mL)=0.784g/mL

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Putting values in equation 1, we get:

\text{Volume of ethanol}=\frac{125g}{0.784g/mL}=159.44mL

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL

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