Answer:
a mole have have wrong becouse i understand
<u>Answer:</u>
<u>For 1:</u> The amount of potassium iodate that were titrated is 
moles
<u>For 2:</u> The amount of sodium thiosulfate required is 
moles
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of 
solution = 0.0100 M
Volume of solution = 25 mL
Putting values in above equation, we get:
Hence, the amount of potassium iodate that were titrated is moles
The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:
By Stoichiometry of the reaction:
2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate
So, 0.00025 moles of potassium iodate will react with = 
of sodium thiosulfate
Hence, the amount of sodium thiosulfate required is moles
The mass of chlorine that react with 9.00 g of Al to form AlCl3 is 35.465 grams
Explanation
write the equation for reaction
that is
2 Al + 3 Cl2 = 2 Al CL3
find the moles of Al reacted
moles = mass/molar mass
9 g/ 27 g/mol = 0.333 moles of Al
by use of mole ratio between Al to Cl2 which is 2:3 find the moles of Cl2
mole of cl2 = 0.333 x3/2 = 0.4995 moles
mass of Cl2 is therefore = moles x molar mass
= 0.4995 x71 = 35.465 moles
2.0 L
The key to any dilution calculation is the dilution factor
The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.
In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to
DF=18.5M1.5M=12.333
So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.
The volume of the stock solution needed for this dilution will be
DF=VdilutedVstock⇒Vstock=VdilutedDF
Plug in your values to find
Vstock=25.0 L12.333=2.0 L−−−−−
The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.
So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.
IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!
In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.
Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!
So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.
Always remember
Water to concentrated acid →.NO!
Concentrated acid to water →.YES!
Answer: 24.1%, under below assumptions.
Justification:
The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:
- sodium bicarbonate, and
- ammonium bicarbonate
.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.
Given that later, the question is <span>what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.
With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.
In formulas:
100% = %m/m sodium bicarbonate + %m/m ammonium bicarbonate = 100%
=> % m/m sodium bicarbonate = 100% - % m/s ammonium bicarbonate
=> % sodium bicarbonate = 100% - 75.9% = 24.1%
Answer: 24.1%
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