Answer:
λ = 2.38 × 10^(-7) m
Explanation:
We are given the work function for palladium as 503.7 kJ/mol.
Now let's convert this to KJ/electron.
We know from avogadro's number that;
1 mole of electron = 6.022 × 10^(23) electrons
Thus,
503.7 kJ/mol = 503.7 × 1/(6.022 × 10^(23)) = 8.364 × 10^(-22) KJ/electron = 8.364 × 10^(-19) J/electron
Formula for energy of a photon is;
E = hv
Where;
h is Planck's constant = 6.626 × 10^(-34) J.s
v is velocity
Now, v = c/λ
Where;
c is speed of light = 3 × 10^(8) m/s
λ is wavelength of light.
Thus;
E = hc/λ
Making λ the subject, we have;
λ = hc/E
λ = (6.626 × 10^(-34) × 3 × 10^(8))/(8.364 × 10^(-19))
λ = 2.38 × 10^(-7) m
The reaction equation is:
CaF₂ + H₂SO₄ → 2HF + CaSO₄
The molar ratio between fluorite and hydrogen fluoride is 1 : 1.
The moles of fluorite supplied are:
Moles = 15.6 / 78.07
Moles = 0.200
The moles of hydrogen fluoride produced will be 0.2.
Now, we may use the ideal gas equation to determine the temperature:
PV = nRT
T = PV/nR
T = (899 * 7.4) / (0.2 * 62.36)
T = 533.40 K
The temperature will be 260.25 °C
The temperature is the same but the heat flow is the opposite.
Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.
Answer:
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Explanation:
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