First, let us calculate the moles of solute or sodium
bicarbonate is in the 1 ml solution.
<span>moles = 1 mL * (1 g
/ 9 mL) = 0.11 moles</span>
The molar mass of sodium bicarbonate is 84 g/mol,
therefore the mass is:
mass = 0.11 moles * 84 g/mol
<span>mass = 9.33 g</span>
The reaction of an Arrhenius acid with an Arrhenius base produces water and <span>A) a salt</span>
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~
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