Answer:
HCN < HOCl < HF
Explanation:
The larger the Kₐ value, the stronger the acid.
6.2 × 10⁻¹⁰ < 4.0 × 10⁻⁸ < 6.3 × 10⁻⁴
HCN < HOCl < HF
weakest stronger strongest
Product of mixing acids and bases describes salt is a physical property.
Product of mixing acids and bases
<u>Explanation:</u>
When an acid and a base are put together, they respond to kill the corrosive and base properties, creating a salt which portrays the physical property. The physical properties of table salt will be: Salt is a white cubic gem. At the point when the salt is unadulterated it clear.
It likewise shows up in white, dim or caramel shading relying on immaculateness. It is unscented yet has a solid salty taste. Fundamental salts contain the conjugate base of a feeble corrosive, so when they break down in the water, they respond with water to yield an answer with a pH more than 7.0.
- E(Bonds broken) = 1371 kJ/mol reaction
- E(Bonds formed) = 1852 kJ/mol reaction
- ΔH = -481 kJ/mol.
- The reaction is exothermic.
<h3>Explanation</h3>
2 H-H + O=O → 2 H-O-H
There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb
- E(Bonds broken) = 2 × 436 + 499 = 1371 kJ/mol reaction.
- ΔH(Breaking bonds) = +1371 kJ/mol
Each mole of the reaction will form two moles of water molecules. Each mole of H₂O molecules have two moles O-H bonds. Two moles of the molecule will have four moles of O-H bonds. Forming all those bond will release
- E(Bonds formed) = 2 × 2 × 463 = 1852 kJ/mol reaction.
- ΔH(Forming bonds) = - 1852 kJ/mol
Heat of the reaction:
is negative. As a result, the reaction is exothermic.
To calculate for the final temperature, we need to remember that the heat rejected should be equal to the absorbed by the other system. We calculate as follows:
Q1 = Q2
(mCΔT)1 = (mCΔT)2
We can cancel m assuming the two systems are equal in mass. Also, we cancel C since they are the same system. This leaves us,
(ΔT)1 = (ΔT)2
(T - 80) = (0 - T)
T = 40°C
If I remember correctly, you would have to heat the reaction beaker over a burner..
I apologize if I'm wrong
