Answer:
K = 8.6x10⁻⁶
Explanation:
<em>a chemist finds that a 5.2L reaction vessel...</em>
To solve this question we need first to find the equation of the equilibrium constant using the chemical eqiation:
TiCl₄(l) ⇄ Ti(s) + 2Cl₂(g)
The equilibrium constant expression is:
K = [Cl₂]²
<em>Because equilibrium constant is defined as the ratio berween concentrationa of products over reactant powered to its reaction coefficient. But pure liquids as TiCl₄(l) and pure solids as Ti(s) are not taken into account</em>
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Now, we need to find the molar concentration of Cl₂, [Cl₂]:
<em>Moles Cl₂ -Molar mass: 70.9g/mol-:</em>
1.08g * (1mol / 70.9g) = 0.0152 moles / 5.2L =
2.93x10⁻³M = [Cl₂]
K = (2.93x10⁻³)²
<h3>K = 8.6x10⁻⁶</h3>
The difference between <span>"formula unit" and "molecule" lies in the fact that a</span> molecule is made up of two or more elements that are bonded covalently while a formula nit represents the lowest whole number ratio of ions in an ionic compound which makes it applicable only to ionic substances.
It's the other name of Water and it's formula would be H2O
Answer:
The correct answer is 0.09 percent.
Explanation:
Based on the given information, the atomic radius is 2 × 10⁻¹⁰ m. Now the atomic volume can be determined by using the formula, 4/3πr³. by putting the values we get,
v = 4/3πr³ = 4/3π(2×10⁻¹⁰ m)³
v = 3.35 × 10⁻²⁹ m³
The no. of atoms present in one mole of xenon is 6.02 × 10²³ atoms. Therefore, the total atomic volume will be,
Va = 6.02 × 10²³ × 3.35 × 10⁻²⁹
Va = 2.02 × 10⁻⁵ m³
The %age of total volume occupied can be determined by dividing the total atomic volume with the volume under STP. Under STP, the volume is 22.4 m³ or 2.24 × 10⁻² m³.
Va/V = 2.02 × 10⁻⁵ m³ / 2.24 × 10⁻² m³
= 9.02 × 10⁻⁴ or 0.09 %
Answer:
Hey!I will try to help with the ones I know.
solubility is a means of comparing the extent to which different solutes can dissolve in a particular solvent at a definite temperature.
A saturated solution of a solute at a particular temperature is one which contains as much solute as it can dissolve at that temperature in the presence of undissolved solute particles....
This is what I know I hope it helps you