You can use the formula M1 x V1 = M2 x V2 where M1 is the molarity of the first substance and V1 is the volume of the substance. M2 is the molarity of the 2nd substance and V2 is the volume of the substance
if substance 1 is HCl and 2 is KOH we can set up the following equation
x moles / liter (unknown) x .02 liters = .5 moles/ liter x .032 liters
x moles / liter x .02 liters = .016 moles
x moles / liter = .016/.02 liters
x moles / liter = .8 moles/ liter
You have . 8M HCL
Answer:
both the substances will evaporate
Explanation:
The oxidation number of elements in equation below are,
4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂
O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1
Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).
Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).
Result:
<span>N is oxidized and Cl is reduced.</span>
