I'm going to have to say A. My reason to that is because when you leave a cup of hot cocoa (assuming its hot), it gives off the heat (exothermic) the
It'd be a physical change. This is because it's a change in the state of matter and not altering the chemical structure of water
Answer:
1) <em>The correct answer is A. Collision</em>
2) A hot solvent helps a solid dissolve faster because an increase in <u><em>kinetic energy</em></u> that also increases the rate of collisions
Explanation:
When a solute is added into a solvent and stirred, the solute particles get distributed to all parts of the solvent as a result of stirring.
More collisions occur between the solute and the solvent due to stirring. This increases the rate of dissolving.
<em>When a solvent is heated, then the kinetic energy would increase and the atoms will collide with a much greater force. As a result, ore solute will be able to dissolve in the solvent. </em>
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Answer:
2Mg + O2 → 2MgO
Explanation:
In all conbustion you should know, that reactans are an specific compound and O2, so the products must be CO2 and H2O, or in this case, the corresponding oxide.
