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>
Answer:
23
Explanation:
we do not care about electrons, so 11 + 12 = 23
Its b Fe(s) <span> Fe</span>2+(aq) + 2e– <span><span> </span>E</span><span> = </span><span>+0.44 V</span>
The equation for calculating a mass is as follows:
m=n×M
Molar mass (M) we can determine from Ar that can read in a periodical table, and a number of moles we can calculate from the available date for N:
n(H2SO4)=N/NA
n(H2SO4)= 1.7×10²³ / 6 × 10²³
n(H2SO4)= 0.3 mole
Now we can calculate a mass of H2SO4:
m(H2SO4) = n×M = 0.3 × 98 = 27.8 g
9.0532e-5 should be correct! hope this helps
