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:
The reaction is not spontaneous in the forward direction, but in the reverse direction.
Explanation:
<u>Step 1: </u>Data given
H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol
Temperature = 25°C = 25+273 = 298 Kelvin
The initial pressures are:
pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
<u>Step 2</u>: Calculate ΔG
ΔG = ΔG° + RTln Q
with ΔG° = 2.60 kJ/mol
with R = 8.3145 J/K*mol
with T = 298 Kelvin
Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]
with pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
Q = (3.10²)/(1.5*1.75)
Q = 3.661
ΔG = ΔG° + RTln Q
ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)
ΔG =5815.43 J/mol = 5.815 kJ/mol
To be spontaneous, ΔG should be <0.
ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.
Hey there!
Exothermic reactions release heat, causing its temperature to fall. If the reaction is lessening the temperature of the object while releasing all the heat, then you know that the reaction is indeed exothermic. Hope this helps!
I think your answer is “true”, sorry if I’m wrong and hope this helps
