Answer:
[COF₂] = 0.346M
Explanation:
For the reaction:
2COF₂(g) ⇌ CO₂(g) + CF₄(g)
Kc = 5.70 is defined as:
Kc = [CO₂] [CF₄] / [COF₂]² = 5.70 <em>(1)</em>
Equilibrium concentrations of each compound after addition of 2.00M COF₂ will be:
[COF₂] : 2.00M - 2x
[CO₂] : x
[CF₄] : x
Replacing in (1):
5.70 = [X] [X] / [2-2x]²
22.8 - 45.6x + 22.8x² = x²
0 = -21.8x² + 45.6x - 22.8
Solving for x:
X = 1.265 <em>-False answer, will produce negative concentrations-</em>
<em>X = 0.827.</em>
Replaing, molar concentration of COF₂ is:
[COF₂] : 2.00M - 2×0.827 = <em>0.346M</em>
I hope it helps!
Answer:
Explanation:
The fundamental units of a measurement is known as its base unit. The units of these substances serves as the base through which other quantities depends. Examples of such quantities are mass, length, time, electric current, temperature, amount of substance and luminous intensity.
Derived units are those that results from the combination of the fundamental or basic units. Examples of derived quantities are force, volume, density, pressure e.t.c.
Derived units of Density:
kgm⁻³
gcm⁻³
Derived unit of volume:
m³
cm³
mL
L
dm³
<span>for every 1 mole of Cu3(PO4)2, there are 2 moles of phosphorus. We also know that 1 mole of Cu3(PO4)2 has 6.022 x 10^23 atoms (Avogadro's number
</span>8.90 moles Cu3(PO4)22 moles P1 mol Cu3(PO4)21 mol Cu3(PO4)2<span>6.022 x 10^23 atoms
</span><span> (53.5958 x 10^23)/2 = 26.7979 x 10^23 atoms of P</span>
There are 12 hydrogen atoms are in a fructose.
The tendency of an element to loose an electron and gain positive charge is called metallic character. So an element which easily looses its electron/electrons are called metals. Hope it helps
