Answer:
To consume the 2.8 moles of CH4 we need 5.6 moles of O2 since the molar ratio is 1:2. We have only 3 moles of O2 ; therefore, O2 is the limiting reactant.
Explanation:
Answer:
Concentration AgBr at saturation = 7.07 x 10⁻⁷M
Explanation:
Given AgBr(s) => Ag⁺(aq) + Br⁻(aq) ; Ksp = 5 x 10⁻¹³ = [Ag⁺][Br⁻]
I --- 0 0
C --- +x +x
E --- x x
[Ag⁺][Br⁻] = (x)(x) = x² = 5 x 10⁻¹³ => x = SqrRt(5 x 10⁻¹³) = 7.07 x 10⁻⁷M
Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m
<span>Energy emitted
Ee = hc/ 1736^-9m </span>
Energy retained ..
∆E = Ea - Ee = hc(1/92.05<span>^-9 - 1/1736^-9) </span>
<span>∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J </span>
<span>Converting J to eV (1.60^-19 J/eV)
∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV </span>
<span>Ground state (n=1) energy for Hydrogen = - 13.60eV </span>
<span>New energy state = (-13.60 + 12.70)eV = -0.85 eV </span>
<span>Energy states for Hydrogen
En = - (13.60 / n²) </span>
n² = -13.60 / -0.85 = 16
n = 4
Answer:
Reactant : A combustion of hydrocarbon.
Explanation:
It is known that when hydrocarbons are involved in combustion, they produce carbon dioxide and water.
CxHy + (x+y/4)O2 ===> xCO2 + y/2H2O
Answer:
a) Aqueous LiBr = Hydrogen Gas
b) Aqueous AgBr = solid Ag
c) Molten LiBr = solid Li
c) Molten AgBr = Solid Ag
Explanation:
a) Aqueous LiBr
This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.
b) Aqueous AgBr
This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.
c) Molten LiBr
In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.
c) Molten AgBr
The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.