M = 2 . 8 . 2
Valence Electron of M = 2
M ==> M⁺² + 2 e⁻
a. M⁺² + OH⁻ ==> M(OH)₂
b. M⁺² + PO₄⁻³ ==> M₃(PO₄)₂
Missing question: Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of ammonium carbonate and cobalt(II) bromide are combined.<span>Balanced chemical reaction:
(NH</span>₄)₂CO₃(aq) + CoBr₂(aq) → CoCO₃(s) + 2NH₄Br(aq).
Net ionic reaction:
2NH₄⁺(aq) + CO²⁻(aq) + Co²⁺(aq) + 2Br⁻(aq) → CoCO₃ + 2NH₄(aq)+ 2Br(aq).
or CO²⁻(aq) + Co²⁺(aq) → CoCO₃(s).
The balanced chemical reaction:
C3H8 + 5O2 = 3CO2 + 4H2O
We are given the amount of the carbon dioxide to be produced. This will be the starting point of our calculations.
<span>43.62 L CO2 ( 1 mol CO2 / 22.4 L CO2 ) (5 mol O2 / 3 mol CO2 ) (
22.4 L O2 / 1 mol O2) = 72.7 L O2</span>
Formula unit mass is defined as the sum of the mass of all the atoms each multiplied its atomic masses that are present in the empirical formula of a compound. It is expressed in amu.
Atomic mass of calcium = 40 amu
Atomic mass of chlorine = 35.5 amu
Formula mass of CaCl2 = (1 x 40) + (2 x 35.5) = 111amu.
Answer:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2
