First, we write the balanced equation for this reaction:
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:
Moles = volume (in L) * molarity
We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:
M₁V₁ = 2M₂V₂
V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L
The volume required is 173.6 mL
In oil and gas industry:
When crude oil get extracted from well, salt water and some other stuff needs to be removed before oil can be sued in the car
Answer:
enantiomers
Explanation:
L and D stand for levorotatory and dextrorotatory respectively. A levorotatory molecule will rotate the plane of plane polarised light left and a dextrorotatory molecule will rotate the plane of plane polarised light right. L and D molecules are non superimposable mirror image of each other. Therefore they are also known as enantiomers.
So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
2NaOH + H₂SO₄ → Na₂SO₄<span>
+ 2H₂O</span>
So now it's time to identify what reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)
If 1000 ml of H₂SO₄ contain 0.750 mol [0.750 M is the amount of moles in
1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol [15 ml is the amount of the acid that took part in the reaction]
⇒
x =
= 0.01125 molMole ratio of NaOH to H₂SO₄ can be obtained from the balanced equation
0
2NaOH +
1H₂SO₄ → Na₂SO₄ + 2H₂O
mole ratio of NaOH to H₂SO₄ is 2 : 1∴ if mole of of H₂SO₄ = 0.01125 mol then moles of NaOH = (0.01125 mol) × 2 = 0.0225 molIf 17.5 ml of NaOH contain 0.0225 mol [this was given in the question]
then let 1000 ml of NaOH contain x⇒ x =
= 1.286 mol∴ concentration of NaOH is 1.286 mol/L