<h3>
Answer:</h3>
56.11 g/mol
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Compound] KOH
<u>Step 2: Identify</u>
[PT] Molar Mass of K - 39.10 g/mol
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mass of H - 1.01 g/mol
<u>Step 3: Find</u>
39.10 + 16.00 + 1.01 = 56.11 g/mol
ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
Learn more about ΔG° here:
brainly.com/question/17214066
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Answer:
a. The second run will be faster.
d. The second run has twice the surface area.
Explanation:
The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.
The area of the 10.0 cm³-sphere is:
The area of each 1.25 cm³-sphere is:
The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²
The ratio of 8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00
Since the surface area is doubled, the second run will be faster.
Psolution = X · PH_20
= 0.966 · 31.8 torr
= 30.7 torr
