Answer:2.27 g H2O
Explanation:
Step 1. Get the limiting reagent
1.24 atm of H2 would react completely with 1.24 x (1/2) = 0.62 atm of O2, but there is more O2 present than that (0.98). O2 is in excess and H2 is the limiting reactant.
If the reaction goes to completion:
(1.24 atm H2) x (2 mol H2O / 2 mol H2) = 1.24 atm H2O vapor
n = PV / RT = (1.24 atm) x (2.5 L) / ((0.08205746 L atm/K mol) x (28 + 273)K) = 0.126 mol H2O
(0.126 mol H2O) x (18.01532 g H2O/mol) = 2.27 g H2O
Answer:
Using the Iodoform test, we can differentiate both compounds.
Explanation:
Benzaldehyde (C6H5CHO -an aldehyde) and Acetophenone (C6H5COCH3 - a methyl ketone) can be differentiated by reacting both compounds with iodine in a basic (NaOH) solutions.
The methyl ketone (acetophenone) gives a pale yellow precipitate of triiodomethane (iodoform) while the aldehyde (benzaldehyde) would not react.
This is known as the IODOFORM test and is indicative for methyl ketones
Answer:
(b) matter is lost and energy is released
Explanation:
When atoms are being formed from its constituent components it weighs less this is called mass defect so the answer would be (b) matter is lost and energy is released.
Answer:
If temperature stays the same, the amount of carbon dioxide dissolved in the ocean would increase.
Explanation:
The carbon dioxide in the air (
) and the carbon dioxide dissolved in the ocean (
) are in a solution equilibrium:
.
Assume that the concentration of carbon dioxide in the air () increased while temperature stayed the same.
By Le Ch
atelier's Principle, the solution equilibrium would shift to offset this increase in concentration. Specifically, this equilibrium would reduce the amont of 
in the system by converting more atmospheric 
to 
dissolved in the ocean. Therefore, the concentration of 
dissolved in the ocean would increase.
