Answer:
- <em>The net charge of the ionic compound calcium fluoride is </em><u><em>zero (0).</em></u>
<em>Explanation:</em>
<em>Ionic compounds,</em> such as covalent ones, have zero net charge; this is, they are neutral.
Substances with net positive charge are cations and substances with net negative charge are anions.
The charges in the <em>ionic compound calcium flouride</em> are distributed in this way:
- Calcium charge: Ca²⁺: this is, each calcium ion has a 2 positive charge
- Fluoride charge: F⁻: each fluoride ion has a 1 negative charge.
- Then, the <em>net charge</em> is: 1 × (2+) + 2 × (1-) = +2 - 2 = 0.
So, a two positve charge, from one calcium ion, is equal to two negative charges, from two fluoride tions, yielding a <u>zero net charge</u>.
To Tell how much of each reactant will be used in a reaction, we need to find which reactant is the Limiting Reagent.
All the reactants will be consumed in equal amount as that of L.R.
First, since l = n-1,
5,4,-5,1/2 and 2,1,0,1/2 are the only answer choices left.
Next, since ml = -l to l,
2,1,0,1/2
is the answer because in 5,4,-5,1/2, the ml value of -5 is not in the range of -4 to 4, as notes by the value 4 for l.
Given:
<span>CS2 + 3O2 → CO2 + 2SO2
</span><span>114 grams of CS2 are burned in an excess of O2
</span>
moles CS2 = 114 g/76.143 g/mol → 114g * mol/76.143 g = 1.497 mol
<span>the ratio between CS2 and SO2 is 1 : 2 </span>
moles SO2 formed = 1.497 x 2 = 2.994 moles → 2nd option