Answer:
sp³
Explanation:
Number of hybrid orbitals = ( V + S - C + A ) / 2
Where
H is the number of hybrid orbitals
V is the valence electrons of the central atom = 5
S is the number of single valency atoms = 4
C is the number of cations = 1
A is the number of anions = 0
For PCl₄⁺
Applying the values, we get:
H = ( 5+4-1+0) / 2
= 4
<u>This corresponds to sp³ hybridization.</u>
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)
Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol
Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol
<em>Answer: -1367.5 kJ/mol</em>
The answer is low frequency and long wavelength
Answer: El carbono, que en estado sólido, puede adoptar muchas formas alotrópicas, siendo las más comunes el diamante (red tridimensional) y el grafito (láminas), aunque también puede formar nanoestructuras en forma de balón de fútbol (fullerenos) o tubos diminutos (nanotubos de carbono), entre otras posibilidades.
Explanation:
Answer:
168°C is the melting point of your impure sample.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= ?
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample = 0.275 mol/kg
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( non electrolyte)
168°C is the melting point of your impure sample.
