CO+2 H2=CH3OH
2.85 mol Co x (2mol H2/1 mol Co)=5.70 mol just concert to grams
5.70 mol H2 x (2 g H2/1 mol H2) =11.40 grams of H2
When you wake up early, and go brush your teeth the toothpaste is made of chemical compounds, the water that you use to wash your faceis treated chemically. To make the coffee sugar or sweetener has chemical compounds.
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Aluminium oxide is amphoteric. It is easy to see that it is a Bronsted-Lowry base through the following reaction:
Al2O3+6HCl →2AlCl3+3H2O
AlX2OX3+6HCl →2AlClX3+3HX2O
The Alumnium oxide splits and the oxygen accepts a proton, forming water.
But what about the reaction with a base? In my textbook, they say:
Al2O3+NaOH →2NaAlO2+H2O
AlX2OX3+NaOH →2NaAlOX2+HX2O
Now, the textbook claims that Aluminium oxide is an acid merely because it reacts with a base to form a salt and water, as is characteristic of a neutralization reaction.
But I'm not satisfied with this definition. I mean, acids aren't defined as 'things that neutralize bases', we have well-established definitions for them.
I tried to figure out for myself how this could be. Clearly, the Bronsted-Lowry theory cannot be applied here since the compound in question has no protons to donate. Therefore, the only alternative is the Lewis concept. I cannot see how that is applicable in this case.
The most basic definition of "acid" is that it is a proton donor (or one which accepts a lone pair)
All of this stuff is done in an aqueous medium, so we can assume that all aqueous ions and molecules are present. With this assumption (in this case, we are assuming that OH−OHX− is available to react), we get the following equation:
Al2O3+OH−⟶2AlO2−+H+
AlX2OX3+OHX−⟶2AlOX2X−+HX+
Similarly, we get:
Al2O3+6H+⟶2Al3++H2O
AlX2OX3+6HX+⟶2AlX3++HX2O
where it is acting like a proton acceptor (base).
Answer:
2.04 atm
Explanation:
The pressure of the gas can be found by using the equation of state for an ideal gas:
where
p is the gas pressure
V is the gas volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
For this gas we have:
n = 0.316 mol (moles)
V = 4.00 L (volume)
T = 315 K (absolute temperature)
(gas constant)
Therefore, the gas pressure is