Answer:
0.21mol Ar (g)
Explanation:
To convert from litres to moles at STP we must divide the amount of litres by 22.4.
4.7 / 22.4 = 0.21mol Ar (g)
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻
1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions
the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol
NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol
total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
Answer : The molar mass of unknown compound is 152.38 g/mole
Explanation :
Depression in freezing point = 
Mass of unknown compound = 16.5 g
Mass of water = 106.0 g
Formula used :
where,
= depression in freezing point
i = Van't Hoff factor = 1 (for non-electrolyte)
= freezing point constant for water = 
m = molality
Now put all the given values in this formula, we get
Therefore, the molar mass of unknown compound is 152.38 g/mole
There can be three definitions of a base: Arrhenius base, Lewis base or Bronsted-Lowry base. Isoquinoline acts specifically as a Bronsted-Lowry base, which is a proton acceptor. Also, it acts as an Arrhenius base because it produces OH- ions after the reaction. The net ionic equation is:
<span><em>C</em></span><em>₉</em><span><em>H</em></span><em>₇</em><span><em>N (aq) + H</em></span><em>₂</em><span><em>O ---> C</em></span><em>₉</em><span><em>H</em></span><em>₇</em><span><em>NH</em></span><em>⁺</em><span><em> (aq) + OH</em></span><em>⁻</em><span><em> (aq)</em></span>
