Answer:
pocomoag I hope it helsps
Explanation:
I don't know but I'll solve it xd
answer= 0.912 L or 912 mL
M(KClO3) = 122.55 g/mol
3.00 g KClO3 * 1 mol/122.55 g = 3.00/122.55 mol =0.02449 mol
2KCIO3(s)=2KCI(s) + 3O2(g)
from reaction 2 mol 3 mol
given 0.02449 mol x
x = 0.02449*3/2 =0.03673 mol O2
T = 24 + 273.15 = 297.15 K
PV = nRT
V= nRT/P = (0.03673 mol*0.082057 L*atm/K*mol*297.15 K)/0.982 atm =
= 0.912 L or 912 mL
Answer:
The standard cell potential increases with increasing temperature.
Explanation:
Equatio 1: ΔG° =ΔH° -TΔS°
Equation 2: ΔG° = -nFE°
Isolating E° in equation 2:
E° = - ΔG° / nF (Equation 3)
Substituting equation 1 in equation 3:
E° = (- ΔH° +TΔS°)/ nf
We can rearrange the equation:
E° = (ΔS°/nF) T + (ΔH°/nF)
Now it is clear that the higher the temperature, the higher the standard cell potential.
Answer: -
6.26 grams
Explanation: -
Atomic Mass of sodium Na = 23 gram
Mass of Na = 18 g
Atomic mass of oxygen O = 16 gram
Molar mass of O2 = 16 x 2 = 32 gram
The balanced chemical equation for the reaction is
4 Na + O2 -- > 2 Na 2 O
From the balanced chemical equation, we see
4 Na requires 1 O2
4 x 23 gram of Na requires 32 gram of O2
18 g of Na gives

= 6.26 grams of O2
For the reaction of formation of Na2O, for 18.0 g of Na, 6.26 grams of O2 are needed for reaction.
Answer: 75.7 % yield
Explanation:
The balanced chemical equation is:
According to stoichiometry :
68.2 g of
will require = 64 g of 
Thus 7.5 g of
will require =
of 
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 68.2 g of give = 96 g of
Thus 7.5 g of
give =
of 
% yield=
Thus 75.7 % yield is there.