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Gwar [14]
3 years ago
10

The work function of an element is the energy required to remove an electron from the surface of the solid. The work function fo

r palladium is 503.7 kJ/mol (that is, it takes 503.7 kJ of energy to remove 1 mole of electrons from 1 mole of Pd atoms on the surface of Pd metal). What is the maximum wavelength of light that can remove an electron from an atom in palladium metal
Chemistry
1 answer:
uranmaximum [27]3 years ago
4 0

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

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Convert .0150 km to meters and express the correct significant figures
weeeeeb [17]
<h3>Answer:</h3>

15.0 meters

<h3>Explanation:</h3>
  • Kilometers are units used to measure length.
  • Length is also measured using other units which include meters, centimeters, decimeter, millimeter, etc
  • To convert one unit to another we use conversion factors.

In this case to know inter-conversion between units of measuring length, we consider the following;

Kilometer(km)

10

Hectometer(Hm)

10

Decameter(dam)

10

Meter (M)

10

Decimeter (dm)

10

Centimeter (cm)

10

Millimeter (mm)

Therefore, to convert km to meter we use the conversion factor (10×10×10) 1000 m/km.

Thus;

0.0150 km is equivalent to;

= 0.0150 km × 1000 m/Km

= 15.0 meters

Thus, 0.0150 km is equivalent to 15.0 meters.

4 0
3 years ago
The following sequence of reactions occurs in the commercial production of aqueous nitric acid: 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=
castortr0y [4]

Answer:

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

Explanation:

<u>Step 1</u>: Data given

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=−907kJ 2NO(g)+O2(g)⟶2NO2(g)ΔH=−113kJ 3NO2+H2O(l)⟶2HNO3(aq)+NO(g)ΔH=−139kJ

<u>Step 2:</u> Multiply equations

Multiply the first equation by 3:

12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(l) ΔH = −2721 kJ

Multiply the second equation by 6:

12 NO(g) + 6 O2(g) → 12 NO2(g) ΔH = −678 kJ

Multiply the third equation by 4:

12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g) ΔH = −556 kJ

<u>Step 3:</u> Get the equations together

12 NH3(g) + 15 O2(g) + 12 NO(g) + 6 O2(g) + 12 NO2(g) + 4 H2O(l) →

12 NO(g) + 18 H2O(l) + 12 NO2(g) + 8 HNO3(aq) + 4 NO(g)

ΔH = −2721 kJ − 678 kJ − 556 kJ

We can simplify as followed:

12 NH3(g) + 21 O2(g) → 14 H2O(l) + 8 HNO3(aq) + 4 NO(g) ΔH = −3955 kJ

<u> Step 4:</u> Determine the total energy change for the production of one mole of aqueous nitric acid by this process:

−3955 kJ/8 moles HNO3= −494 kJ

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

3 0
3 years ago
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