Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
Answer:
Project scope statement?
Explanation:
it's good one just do research on it
Answer:
b) Delta S < 0
Explanation:
The change in the entropy (ΔS) is related to the change in the number of gaseous moles of the reaction: Δn(g) = n(g, products) - n(g, reactants).
- If Δn(g) > 0, the entropy increases (ΔS > 0).
- If Δn(g) < 0, the entropy decreases (ΔS < 0).
- If Δn(g) = 0, there is little or no change in the entropy
Let's consider the following equation.
2 H₂S(g) + 3 O₂(g) → 2 H₂O(g)
Δn(g) = 2 - 5 = - 3. Since Δn(g) < 0, the entropy decreases and ΔS < 0.
Answer:(I didn’t really know where to Wrigh-)
Explanation: Gas bubbles appear after a chemical reaction has occurred and the mixture becomes saturated with gas. The chemical change that creates the gas is completed after the gas bubbles leave the mixture.