Physical. chemical only applies when it undergoes a reaction
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:
1. acid
2. neutral
3. acid
4. base
5. acid
6. base
7. neutral
8. acid
9. base
10. base
Explanation:
I'm not 100 percent positive about number three but the rest I believe are correct
Answer:
412 g Cl₂
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 3.50 × 10²⁴ molecules Cl₂
[Solve] grams Cl₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
412.072 g Cl₂ ≈ 412 g Cl₂
Answer:
electron, proton, nucleus, atom, molecule teacher