Answer:
the correct answer is four decimal places
Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
Kc= (nh4)
--------
(nh3) + (h2o)
Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
ACIDIC BEHAVIOR OF SOLUTION
