Answer-Figure P38.63a
is a three-dimensional sketch of a birefringent crystal. The dotted lines illustrate how a thin, parallefaced slab of material could be cut from the larger specimen with the crystal’s optic axis parallel to the faces of the plate.”
Sorry if it’s it correct
Answer:
Explanation:
Bromine >Tellurium > Phosphorus > Helium > Sodium
Electron affinity of Bromine , Tellurium , Phosphorus are positive , of helium is zero and of sodium is negative .
Answer:
54.99% yield
Explanation:
percent yield is just the amount you obtained over the amount expected times 100%.
(experimental value/theoretical value) x 100%
= (107.9 g/196.2 g) x 100%
=54.99% yield
Answer:
a) K = [ CO2(g) ]
⇒ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
Explanation:
a) CaCO3(s) ↔ CaO(s) + CO2(g)
⇒ K = [ CO2(g) ]
∴ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) H2(g) + F2(g) ↔ 2 HF(g)
⇒ K = [ HF(g) ] ² / [ F2(g) ] * [ H2(g) ]
⇒ Kp = PHF² / PF2 * PH2
for ideal gas:
PV = RTn
⇒ P = n/V RT = [ ] RT
⇒ Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same.
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
<u>Answer: </u>The molar mass of solute is 115 g/mol.
<u>Explanation:</u>
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:
OR
......(1)
where,
Boiling point of pure solvent (benzene) = 
Boiling point of solution = 
i = Vant Hoff factor = 1 (for non-electrolytes)
= Boiling point elevation constant = 
= Given mass of solute = 10 g
= Molar mass of solute = ? g/mol
= Mass of solvent = 200 g
Putting values in equation 1, we get:
Hence, the molar mass of solute is 115 g/mol.
