Answer:
Ka = 0.1815
Explanation:
Chromic acid
pH = ?
Concentration = 0.078 M
Ka = ?
HCl
conc. = 0.059M
pH = -log(H+)
pH = -log(0.059) = 1.23
pH of chromic acid = 1.23
Step 1 - Set up Initial, Change, Equilibrium table;
H2CrO4 ⇄ H+ + HCrO4−
Initial - 0.078M 0 0
Change : -x +x +x
Equilibrium : 0.078-x x x
Step 2- Write Ka as Ratio of Conjugate Base to Acid
The dissociation constant Ka is [H+] [HCrO4−] / [H2CrO4].
Step 3 - Plug in Values from the Table
Ka = x * x / 0.078-x
Step 4 - Note that x is Related to pH and Calculate Ka
[H+] = 10^-pH.
Since x = [H+] and you know the pH of the solution,
you can write x = 10^-1.23.
It is now possible to find a numerical value for Ka.
Ka = (10^-1.23))^2 / (0.078 - 10^-1.23) = 0.00347 / 0.0191156
Ka = 0.1815
Answer:t pans
Explanation:it’s the cool thing
Answer:
1040%
Explanation:
To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:
Percent yield = Actual yield (5.40g) / Theoretical yield * 100
<em>Moles Fe -Molar mass: 55.845g/mol-:</em>
10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.
For a complete reaction of these moles there are necessaries:
0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.
As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>
The moles of H2 produced are:
0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2
The mass is:
0.277 moles H2 * (2.016g/mol) = 0.558g H2
Percent yield is:
5.40g / 0.558g * 100 = 1040%
It is possible the experiment wasn't performed correctly
Answer:
well plus da cuh cause opp homie freak jit.on 390/it = 30
Answer:
I think the answer is 6.022*10^23
