Answer:
We need 41.2 L of propane
Explanation:
Step 1: Data given
volume of H2O = 165 L
Step 2: The balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Step 3: Calculate moles of H2O
1 mol = 22.4 L
165 L = 7.37 moles
Step 4: Calculate moles of propane
For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 7.37 moles H2O we need 7.37/4 = 1.84 moles propane
Step 5: Calculate volume of propane
1 mol = 22.4 L
1.84 moles = 41.2 L
We need 41.2 L of propane
Answer:
[O₂(g)] = 0.0037M
Explanation:
2SO₂(g) + O₂(g) => 2SO₃(g)
Conc: [SO₂(g)] [O₂(g)] [SO₃(g)] and [SO₂(g)] = [SO₃(g)]
Kc = [SO₃(g)]²/[O₂(g)][SO₂(g)]² => Kc = 1/[O₂(g)] = 270 if [SO₂(g)] = [SO₃(g)]
∴ [O₂(g)] = (1/270)M = 0.0037M
Weathering is the process of breaking large rocks and boulders into much smaller ones. Weathering can be brought about by wind and water mostly. Sometimes even biological forces account for some types of weathering.
The pKa of formic acid is 3.75. At pH of 5.00, b) [formate] > [formic acid].
Formic acid is a weak acid. Thus, together with its conjugate base (formate) they form a buffer system. We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.
![pH = pKa + log \frac{[formate]}{[formic\ acid]} \\\\5.00 = 3.75 + log \frac{[formate]}{[formic\ acid]}\\\\\frac{[formate]}{[formic\ acid]} = 17.8](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%20%5C%5C%5C%5C5.00%20%3D%203.75%20%2B%20log%20%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%5C%5C%5C%5C%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%20%20%3D%2017.8)
As expected, at a pH above the pKa, the concentration of formate is higher than that of the formic acid.
The pKa of formic acid is 3.75. At pH of 5.00, b) [formate] > [formic acid].
Learn more: brainly.com/question/22821585
Answer: The answer is there is none
Explanation:
