Answer:
Elements form compounds to satisfy the octet rule. Noble gasses never form compounds because they already satisfy the octet rule.
Explanation:
The octet Rule is the theory that an element will attempt to gain a valence of 8 by binding with another element in it's vicinity. This can happen in a variety of ways, but the main thing to remember is that they will take the "shortest path" to 8(I.e an element will sometimes lose an electron or 2 if it has a valence 1 or 2 to loop back around to 8, while an element with a valence of 6 or 7 will attempt to gain 2 or 1 electrons).
Valence of elements can be counted by group in the image attached.
Group 1 has a valence of 1, Group 2 has a valence of 2, then we move to group 13 which has a valence of 3, group 14 has a valence of 4, group 15 has a valence of 5, group 16 has 6, group 17 has 7, and group 18 is the noble gasses which have 8.
We can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = 303.15 x 300 / 333.15
<span>V2 = 272.99 cm³</span>
The density of the sample is 0.96 g/mL.
Density is found by dividing the mass of an object by its volume.
D=m÷v
D=35.4 g÷36.82 mL
D=0.96 g/mL
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
CH₄ + 2O₂ → CO₂ + 2H₂O
From the equation, we know that methane and carbon dioxide have the same number of moles.
no. of moles of CO₂ produced = no. of moles of methane
= 4.5 × 10⁻³ ÷ (12 + 1×4)
= 2.8125 × 10⁻⁴
∴ mass of CO₂ = 2.8125 × 10⁻⁴ × (12 + 16×2)
= 12.375 × 10⁻³ g
