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astraxan [27]
2 years ago
6

What is the wavelength of a photon with an energy of 3.50 x 10^-19 J ?

Chemistry
1 answer:
Leokris [45]2 years ago
8 0

Answer:

λ = 5.68×10⁻⁷ m

Explanation:

Given data:

Energy of photon = 3.50 ×10⁻¹⁹ J

Wavelength of photon = ?

Solution:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = 3×10⁸ m/s

Now we will put the values in formula.

3.50 ×10⁻¹⁹ J =  6.63×10⁻³⁴ Js × 3×10⁸ m/s/ λ

λ =  6.63×10⁻³⁴ Js × 3×10⁸ m/s / 3.50 ×10⁻¹⁹ J

λ =  19.89×10⁻²⁶ J.m / 3.50 ×10⁻¹⁹ J

λ = 5.68×10⁻⁷ m

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Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s
Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g)  ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

8 0
3 years ago
Why does an exothermic reaction need activation energy?
andrey2020 [161]

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Activation energy is needed so reactants can move together, overcome forces of repulsion, and to begin breaking bonds.

Explanation:

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3 years ago
Which of the following actions does not require power
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Eating an apple

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The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38
adell [148]

Answer:

1.21x10^{-3} M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x10^{-3}

H = \frac{0.21}{1.38x10^{-3} }

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = 1.21x10^{-3} M

7 0
3 years ago
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