A formula unit is the same as the empirical formula of a compound or an ionic molecule. It is the lowest ratio of the atoms in the compound or ion. Zinc acetate ions dissociates into zinc ions and acetate ions. The dissociation reaction is expressed as follows:
Zn(O2CCH3)2 = Zn2+ + 2(O2CCH3)1-
We determine the amount of acetate ions produced as follows:
Moles Zn(O2CCH3)2 = (3 formula units Zn(O2CCH3)2) ( 1 mol / 6.022x10^23 formula units) = 4.98x10^-24 mol Zn(O2CCH3)2
moles (O2CCH3)1- = 4.98x10^-24 mol Zn(O2CCH3)2 ( 2 mol (O2CCH3)1- / 1 mol Zn(O2CCH3)2 ) = 9.96x10^-24 mol (O2CCH3)1-
# of acetate ions = 9.96x10^-24 mol (O2CCH3)1- ( 6.022x10^23 ions / 1 mol (O2CCH3)1-) = 6 acetate ions
Answer:
Explanation:
Electrons are the negatively charged particles of an atom that are involved in chemical reactions. They are located outside the nucleus of an atom
Answer:
2Na + Br2 = 2NaBr
Explanation:
In order to balance a chemical equation you make the make sure both sides have the same number of atoms on each side, you do this by multiplying on both sides as if it was a algebraic equation.
Na+ Br2 = NaBr
Na × 2 = Na2
Na × 2 = Na2
Br × 2 = Br2
2Na + Br2 = 2NaBr
Hope this helps.
solution:
You need to find the frequency, and they have already given you the wavelength. And since you already know the speed of light, you can use formula (2) to answer this problem. Remember to convert the nano meters to meters because the speed of light is in meters. 
The three sub-atomic particles that made up an atom are electron, proton and neutron. The proton and neutron are present in the nucleus of an atom whereas electrons revolve around the nucleus in the orbits. The atoms are electrically neutral that is they have no charge due to the presence of equal number of protons (which is positively charged) and electrons (which is negatively charged).
All the atoms have negatively charged electrons which are counter balanced by the presence of equal number of protons in the nucleus of an atom and thus, the element is neutral in nature and possess no charge.
