Answer:
[NH₃] → 3.24 M
Explanation:
Our solute: Ammonia
Our solvent: Water
Solution's mass = Mass of solute + Mass of solvent
Solution's mass = 15 g + 250 g = 265g
We use density to determine, the volume.
D = mass /volume → Volume = m / D → 265 g /0.974 g/mL = 272.07 mL.
We convert the mL to L → 272.07 mL . 1L /1000mL = 0.27207 L
To determine molarity we need the moles of solute in 1 L of solution.
Moles of solute are: 15g / 17g/mol = 0.882 moles
[NH₃] = 0.882mol /0.27207 L → 3.24 M
Answer:
Explanation:
The formula of the reaction:
KClO₂ → KCl + O₂
To assign oxidation numbers, we have to obey some rules:
- Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
- The charge on simple ions signifies their oxidation number.
- The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.
The oxidation number of K in KClO₂:
K + (-1) + 2(-2) = 0
K-5 = 0
K = +5
The oxidation number of K in KCl:
K + (-1) = 0
K = +1
The oxidation number Cl in KClO₂ is -1
For Cl in KCl, the oxidation number is -1
For O in KClO₂, the oxidation number is (2 x -2) = -4
For O in O₂, the oxidation number is 0
K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.
O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.
The answer would be b, or 2
The question is improperly formatted.
What is the concentration of H+ ions in a 2.2 M solution of HNO3.
Answer:-
2.2 moles of H+ per litre
Explanation:-
M stands for molarity. 2.2 M means 2.2 moles of HNO3 is present per litre of the solution.
Now HNO3 has just 1 H in it's formula. HNO3 would give H+. So 2.2 moles of HNO3 would mean 2.2 moles of H+ per litre.
Oxidation reaction
In ---> In³⁺ + 3e ---1)
reduction reaction
Cd²⁺ + 2e ---> Cd ---2)
when balancing the reactions, electrons have to be balanced. to balance the electrons multiple 1st reaction by 2 and 2nd reaction by 3
1) x 2
2) x 3
2In ---> 2In³⁺ + 6e
3Cd²⁺ + 6e ---> 3Cd
add the 2 equations to obtain the overall reaction
2In + 3Cd²⁺ ---> 2In³⁺ + 3Cd
