Explanation:
chemical analysis of a compound shows that it contains 0.99g lead ,0.154g sulfur and 0.306g oxygen. what is the empirical formula of the compound.
Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
Answer:
Left hand side:-
Carbon - 12
HYdrogen - 28
Oxygen - 38
Right hand side:-
Carbon - 12
Hydrogen - 28
Oxygen - 38
Since, the number of atoms each side are equal, the reaction is balanced.
Explanation:
The given reaction is:-
Left hand side:-
Carbon - 12
HYdrogen - 28
Oxygen - 38
Right hand side:-
Carbon - 12
Hydrogen - 28
Oxygen - 38
<u>Since, the number of atoms each side are equal, the reaction is balanced.</u>
Number 4 is
-Oxidation occurs at the anode, while reduction occurs at the cathode. Recharging a battery involves the conversion of electrical energy to chemical energy. During recharging, there is movement of electrons from an external power source to the anode, and on the other side electrons are removed from the cathode.
Answer: See below
Explanation:
1. a) 0.15 moles calcium carbonate (15g/100g/mole)
b) 0.15 moles CaO (molar ratio of CaO to CaCO3 is 1:1)
c) 8.4 grams CaO (0.15 moles)*(56 grams/mole)
2. a) 0.274 moles Na2O (17g/62 grams/mole)
b) 46.6 grams NaNO3 (2 moles NaNO3/1 mole Na2O)*(0.274 moles Na2O)*(85 g/mole NaNO3)
