Answer:
Lord Rayleigh's observation that the density of nitrogen extracted from the air was always greater than nitrogen released from various chemical compounds.
Explanation:
Answer: only the VSEPR mode shows the geometric shape of a formula.
P1V1=P2V2
(1.8)(1.5)=(1.0)V2
2.7=1.0 V2
2.7/1.0= 1.0/1.0V2
2.7=V2
Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a)
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ
2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ
2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ
2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ
Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction
Answer:
It is 20. g HF
Explanation:
H2 + F2 ==> 2HF ... balanced equation
Since the question is asking us to find the mass of product formed, we will want to first convert the molecules of H2 into moles of H2 (we could do this at the end of the calculations, but it's just as easy to do it now).
moles of H2 present (using Avogadro's number):
3.0x1023 molecules H2 x 1 mole H2/6.02x1023 molecules = 0.498 moles H2
From the balanced equation, we see that 1 mole H2 produces 2 moles HF. Therefore, we can now find the theoretical mass of HF produced from 0.498 moles H2:
0.498 moles H2 x 2 moles HF/1 mol H2 = 0.996 moles HF formed.
The molar mass of HF = 20.01 g/mole, thus...
0.996 moles HF x 20.01 g/mole = 19.93 g HF = 20. g HF formed (to 2 significant figures)
