Answer:
Ld₂Mz
General Formulas and Concepts:
<u>Chemistry - Compounds</u>
- Valence shells
- Writing chemical compounds
- Balancing charges
Explanation:
<u>Step 1: Define</u>
"Lindenium" Ld w/ oxidation charge of +1
"Mendezine" Mz w/ oxidation charge of -2
<u>Step 2: Write Compound</u>
To balance out the charges to be neutral (0), we must counter balance the amount of Ld's with Mz's. We need 2 Ld's with 1 Mz to create a compound with a neutral charge:
2Ld¹⁺ + Mz²⁻
2(+1) + (-2) = (+2) + (-2) = 0 (neutral charge)
Therefore, our compound must be Ld₂Mz.
Answer:
Molecular Formula = C₆H₁₂O₆
Solution:
Molecular formula is calculated by using following formula,
Molecular Formula = n × Empirical Formula ---- (1)
Also, n is given as,
n = Molecular Weight / Empirical Formula Weight
Molecular Weight = 180.2 g.mol⁻¹
Empirical Formula Weight = 12 (C) + 2 (H₂) + 16 (O) = 30 g.mol⁻¹
Son
n = 180.2 g.mol⁻¹ ÷ 30 g.mol⁻¹
n = 6
Putting Empirical Formula and value of "n" in equation 1,
Molecular Formula = 6 × CH₂O
Molecular Formula = C₆H₁₂O₆
Lol. It is now a solution and not really inseparable, so your only logical option would be to throw it away and try again.
Answer:- Ca = +2, S = +6 and O = -2
Solution:- There are certain rules for oxidation numbers. As per the rule, oxidation number of alkaline earth metals in their compounds is +2.
Oxidation number of oxygen in it's compounds is -2(except peroxides) and the sum of oxidation numbers of all the elements of a neutral compound is zero.
Since, Ca is +2 and O is -2, the oxidation number of S could easily be calculated for the given compound as:
Let's say the oxidation number of S in 
is 
. Let's make the algebraic equation and solve it.
Hence. the oxidation number of Ca is +2, O is -2 and S is +6.
