step one
calculate the % of oxygen
from avogadro constant
1moles = 6.02 x 10 ^23 atoms
what about 4.33 x10^22 atoms
= ( 4.33 x 10^ 22 x 1 mole ) / 6.02 10^23= 0.0719 moles
mass= 0.0719 x16= 1.1504 g
% composition is therefore= ( 1.1504/3.25) x100 = 35.40%
step two
calculate the % composition of chrorine
100- (25.42 + 35.40)=39.18%
step 3
calculate the moles of each element
that is
Na = 25.42 /23=1.1052 moles
Cl= 39.18 /35.5=1.1037moles
O= 35.40/16= 2.2125 moles
step 4
find the mole ratio by dividing each mole by 1.1037 moles
that is
Na = 1.1052/1.1037=1.001
Cl= 1.1037/1.1037= 1
0=2.2125 = 2
therefore the empirical formula= NaClO2
Density = mass/volume
Density (iron) = 117 g/15.0 cm³ = 7.80 g/cm³
Al is the reducing agent.That is answer B is the above answer
Al acts as a strong reducing agent. It reduces crO3 to form cr while Al is oxidized to Al2O3. Al is capable to reduce cr since Al is higher in reactivity series than cr.
Mass of sodium carbonate = 7.022 g
<h3>Further explanation</h3>
Reaction
Na₂CO₃ + 2HNO₃ ⇒ 2NaNO₃ + H₂O + CO₂
mol of HNO₃ (MW=63,01 g/mol):
mol Na₂CO₃
mass Na₂CO₃(MW = 105,9888 g/mol) :
Answer:
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