C) can give up electrons more readily than others.
The number of chirps increases in relation to temperature.
Answer:
1. I shifts toward products and II shifts toward reactants.
Explanation:
Hello,
In his case, since the reaction I) is endothermic (positive ∆H°) and the reaction II) is exothermic (negative ∆H°):
- In the first case, the energy is understood as a reactant, so if the temperature increases, heat is added so the reaction will shift rightwards (towards products).
- In the second case, the energy is understood as a product, so if the temperature increases, heat is added so the reaction will shift leftwards (towards reactants).
Therefore, the answer is:
1. I shifts toward products and II shifts toward reactants.
Best regards.
The correct answer is reactants 2Na and 2H₂O and products 2NaOH and 1H₂.
The given unbalanced chemical equation is:
Na (s) + H2O (l) → NaOH (aq) + H₂ (g)
From the equation, it can be seen that there are three atoms of hydrogen on the products side, however, only two on the reactant's side. So, in order to balance the equation, one needs to multiply the sodium hydroxide by 2 to get a total of 4 atoms of hydrogen on the product's side.
This will enable one to readily double the number of molecules of water to get 4 atoms of hydrogen on the reactants side, and then balance the atoms of sodium by multiplying the sodium metal by 2. The balanced equation obtained is:
2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂ (g)
