Explanation:
The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.
This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated
The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.
Answer:
48 molecules of CO₂
Explanation:
I think you made a mistake in your question. The formula for propane is C₃H₈, not C₃H₃. But, I will give you the answer for both cases.
For C₃H₃:
First you have to balance the equation.
4 C₃H₃ + 15 O₂ ⇒ 12 CO₂ + 6 H₂O
Next, you need to use the mole ratios between C₃H₃ and CO₂ to find the amount of molecules of CO₂ you will produce with the given amount of C₃H₃.
(16 mol's C₃H₃) × (12 mol's CO₂/4 mol's C₃H₃) = 48 mol's CO₂
You will get 48 molecules of CO₂.
For C₃H₈:
Balance the equation.
C₃H₈ + 5 O₂ ⇒ 3 CO₂ + 4 H₂O
Use the mole ratios between C₃H₈ and CO₂.
(16 mol's C₃H₈) × (3 mol's CO₂/1 mol's C₃H₈) = 48 mol's CO₂
You will get 48 molecules of CO₂ for this equation as well.
The correct choices are: (2) the particles in the Warner cup has more kinetic energy and (3) if there are two different amounts of the same liquid the temp is a good way to compare thermal energies
Answer:
C
Explanation:
84.6 kJ of energy released when 2.50 moles of nitrogen dioxide is decomposed.
Answer: option A, J, K, N, O
Explanation: