Answer:
B and D could be true
Explanation:
A volume of sodium hydroxide less than expected could occurs for two reasons:
The real concentration of sodium hydroxide was higher than expected or the amount of vinegar added was less than expected:
A. The sodium hydroxide solution had been allowed to stand exposed to the air for a long time prior to the titration. FALSE. A long expose to the air decreases concentration of the NaOH.
B. The volumetric flask used to prepare the diluted vinegar solution was rinsed with water prior to use. TRUE. You add a less amount of vinegar doing you require less amount of NaOH than expected.
C. The burette used to deliver the sodium hydroxide solution was rinsed with water prior to use. FALSE. Thus, you add a less amount of NaOH than expected. To explain the matter, you add more NaOH than expected.
D. The pipette used to deliver the vinegar solution was rinsed with water prior to use. TRUE. Again, you are adding a less amount of Vinegar than expected doing the necessary NaOH during titration less than expected
Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
Answer:
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
Explanation:
2 NO (g) + O₂ (g) ⇄ 2NO₂ (g)
Let's apply the thermodynamic formula to calculate the ΔG
ΔG = ΔG° + R .T . lnQ
We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q
How can we know Q? By the partial pressures (Qp)
P NO = 0.450atm
PO₂ = 0.1 atm
PNO₂ = 0.650 atm
Qp = [NO₂]² / [NO]² . [O₂]
Qp = 0.650² / 0.450² . 0.1 = 20.86
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
I'm sorry but could you tell me IE2
I couldn't asks for details
Answer:
2NaCl+H2SO4-->Na2SO4+2HCl
Explanation:
There are two Na on the right, so put a 2 in front of NaCl on the left. This makes 3 Cl also, so put a 2 in front of HCl on the right. There are already 2 H on the left, so the equation is balanced.
