Answer:
52.45g
Explanation:
The computation of the mass of pure acetic acid in 125mL of this solution is shown below:
The percentage of mass would be equivalent to the g of solute in each 100g of water
As we know that
density = mass ÷ volume
So,
Volume = mass ÷ density
V = 100g / 1.049 (g / ml)
V = 95.328 mL
Now In every 95,328 ml of C_2H_4O_2 there are 40g of C_2H_4O_2
i.e.
each 125ml of C_2H_4O_2 there are 52.45g
SO,
x = 40g. 125ml ÷ 95.328
x = 52.45g
KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol
HNO3= 1 mol KOH (keep in mind this because it will be used later).
We also know that 0.100 M KOH aqueous
solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the
definition of molarity).
First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/
1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.
Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.
The final answer is </span>(2) 20.0 mL.<span>
Also, this problem can also be done by using
dimensional analysis.
Hope this would help~
</span>
Phenolphthalein<span> is often used as an indicator in acid–base titrations. For this application, it </span>turns<span> colorless in acidic solutions and </span>pink<span> in basic solutions. </span>Phenolphthalein<span> is slightly soluble in water and usually is dissolved in alcohols for use in experiments.
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I’m sorry no one helped you with those questions. But could you please be a doll and give me the answer for the multiple-choice questions 1-15. Pleaseee I’m so desperate.?
