A compound has to be chemically bonded, however, air is not chemically bonded.
This can be proven by freezing air. By freezing air, it yields different liquids at different temperature. Liquid nitrogen has a different boiling point than liquid oxygen.
If air was a compound, they would all have a single boiling point and a single freezing point.
Hope this helps :)
Answer:
See explanation
Explanation:
Let us recall that the basic rule in writing balanced chemical reaction equations is that the number of atoms of each element on the right hand side of the reaction equation is the same of the number of atoms of the same element on the left hand side of the reaction equation.
The reaction of red hot iron and steam is written as follows;
3Fe + 4H2O → Fe3O4 + 4H2.
The decomposition reaction of ammonium dichromate is written as;
(NH4)2Cr2O7 → N2 + Cr2O3 + 4H2O
Reaction of aluminium, sodium hydroxide and water is as follows,
2Al + 2NaOH + 2H2O ----> 2NaAlO2 + 3H2
Reaction of potassium bicarbonate with sulphuric acid;
2KHCO3 + H2SO4 -------> K2SO4 + 2H2O + 2CO2
Reaction of zinc and sodium hydroxide is as follows;
Zn + 2NaOH→Na2ZnO2 + H2
You would need to utilize Mole ratios found in the adjusted condition;
for each mole of hydrogen utilized, 2 moles of HCl are delivered.
Thusly:
10 mol H2 x 2 mol HCl/1 mol H2 = 20 mol HCL.
For the second question:
you would need to change over 2.0x10^23 particles of Oxygen to moles of oxygen, utilizing Avogadro's number:
2.0x10^23 particles oxygen x 1 mol oxygen/6.022x10^23 atoms oxygen = 0.33 mol Oxygen
utilizing mole proportions once more:
0.66 mol H2O = 2 mol H2O/1 mol Oxygen x 0.33 mol Oxygen
45.0 mol H2O = 2 mol H2O/1 mol Oxygen x 22.5 mol Oxygen
fundamentally to answer stoichiometry, you should take a gander at the adjusted condition to make sense of the mole proportions between components/mixes, and utilizing mole proportions you can change over from moles of one component/compound to moles of another component/compound
Answer:
Really wished I could help
Right around the face/nose area
