Na₂SO₄ → 2Na⁺ + SO₄²⁻
c₁(Na⁺)=2c(Na₂SO₄)
n₁(Na⁺)=c₁(Na⁺)v₁=2c(Na₂SO₄)v₁
NaCl → Na⁺ + Cl⁻
c₂(Na⁺)=c(NaCl)
n₂(Na⁺)=c₂(Na⁺)v₂=c(NaCl)v₂
n(Na⁺)=n₁(Na⁺)+n₂(Na⁺)=2c(Na₂SO₄)v₁+c(NaCl)v₂
v=v₁+v₂
c(Na⁺)=n(Na⁺)/v={2c(Na₂SO₄)v₁+c(NaCl)v₂}/(v₁+v₂)
c(Na⁺)={2*0.400*0.1+0.600*0.2}/(0.1+0.2)=0.667 mol/L
c(Na⁺)=0.667M
There are two big advantages of using molarity to express concentration. The first advantage is that it's easy and convenient to use because the solute may be measured in grams, converted into moles, and mixed with a volume.
The second advantage is that the sum of the molar concentrations is the total molar concentration. This permits calculations of density and ionic strength
Metallic I’m pretty sure. :)
<u>Answer:</u> The equation is written below.
<u>Explanation:</u>
For the formation of nitrogen dioxide, the reactants used are nitrogen monoxide and oxygen gas.
The balanced chemical equation for the above reaction is given as:
By Stoichiometry of the reaction:
2 moles of nitrogen monoxide gas reacts with 1 mole of oxygen gas to produce 2 moles of nitrogen dioxide gas.
