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- if the compound is made of just two elements, if one is a metal (ie belongs to any of groups 1, 2 or 3) and the other element a non metal, (ie belongs to group 5, 6 or 7) then the compound is most likely to be an ionic compound. For example NaCl, MgO
- If the compound is made of identical non metalic elements as in O2, Cl2 then the compound is covalent
- If the compoud is made of just two elements that are both non metals such as in SO2, CO, NO, CCl4, the compound is covalent
- If the compound is made up of more than two elements, such as in HNO3, Na2CO3, CuSO4.5H2O, you may need to break the compound into dissociating parts. You will see that, the compounds are ionic.
- Hydrocarbons, compounds containing only hydrogen and carbon of varying molecular size are all covalent. Examples are C2H6, C2H4, C2H2
Note that there could be some little exceptions to the examples given. Mostly with first members of every group because of their small size which make them show substantial deviations from group behavior. For example HCl is covalent not ionic.
2Al + 3CuCl2 ➡ 2AlCl3 + 3Cu.
Moles of Al = 1.2/27 = 0.045 moles
2 moles of Al reacts with 3 moles of CuCl2.
Therefore, 1 mole of CuCl2 reacts with =2/3 x moles of Al
= 2/3 x 0.045
= 0.03 moles.
Answer:
Percentage yield is 41.21%
Explanation:
Equation of reaction,
N₂ + 3H₂ → 2NH₃
Actual NH3 = 6.83g
Mass of N2 = 5.77g
Theoretical yield = ?
5.77g of N2 = 6.83g of NH3
14g of N2 = xg
X = (14 × 6.83) / 5.77
X = 95.62 / 5.77
X = 16.57g of NH3
Theoretical yield of NH3 is 16.57g
Percentage yield = (actual yield / theoretical yield) × 100
% yield = (6.83 / 16.57) × 100
% yield = 0.4121 × 100
% yield = 41.21%
The percentage yield of NH3 is 41.21%
First, let's determine the number of moles of carbon atoms by using molar mass. Then, using Avogadro's number, we can find the number of C atoms:
*1 mole of C3H8O= (12.0x3)+(1.0x8)+(16.0x1) = 60.0g/mol
25.0 grams C3H8O x (1 mole C3H8O/60.0 grams) = 0.417 mol
0.417mol C3H8O has (3 x 0.417 moles) C atoms = 1.251 moles C atoms
1.251 moles C atoms x(6.022x10^23 atoms/mol) = 9.42x10^23 C atoms.
The answer is in 3 significant figures, as that's what we have in the given, and we matched it with our rounding of the atomic masses from the periodic table.
You can do this all in one equation written left to right, just exclude the intermediate answers. Just easier to show it this way on the computer screen.
