Answer:
number of ions = 12.04 x 10^²³
Explanation:
n = number of ions/Avogadro's constant
2 = number of ions/6.02 x 10^²³
number of ions= 2 x 6.02 x 10^²³
number of ions = 12.04 x 10^²³
A combustion reaction of an will generally produce CO2 and H20 -- carbon dioxide and water and/or an oxide
looking at the combustion material C2H2, you know that the end products will be CO2 and H20, so the question is how much of each will you get
well, look at the total amount of carbon atoms, 2 C2, which means a total of 4 carbon atoms in this reaction, since only CO2 has carbon atoms, that means there must be 4 CO2 as an end product and 4 CO2 will use up 4 of 5 O2 molecule leaving only 1 O2 molecule for the H2 reaction.
now O2 has a total of 2 oxygen molecules whereas H20 has only a single oxygen molecule, hence the end product must have 2 H20
check that the H atoms balance out on both sides
The correct answer is B.
1 mol O2 x 15.999 O2/ 1 mol O2 = 15.999 O2
16 O2 when rounded.
I think the correct answer would be D. The reaction that involves an acid and a covalent base would be the reaction of sulfuric acid and water or H2SO4 + 2H2O → 2H3O+ + SO42– . The acid would be H2SO4 and the covalent base would be H2O since it is being held by covalent bonds and when in solution it will have equal amounts of OH- and H+ ions.
<u>Answer:</u> The net ionic equation is written below.
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of ammonium carbonate and lead (II) nitrate is given as:
Ionic form of the above equation follows:
As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
Hence, the net ionic equation is written above.
