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galina1969 [7]
3 years ago
7

A photon detector captures a photon with an energy of 5.45 ✕ 10^−19 J. What is the wavelength, in nanometers, of the photon?

Chemistry
1 answer:
GarryVolchara [31]3 years ago
5 0

Answer:

364 nm

Explanation:

The formula for the energy (E) of a photon in terms of its wavelength λ is

E = hc/λ

We can rearrange this formula to give

λ = hc/E

If E = 5.45 × 10⁻¹⁹ J, then  

\begin{array}{rcl}\lambda& = & \dfrac{6.626 \times 10^{-34} \text{ J$\cdot$s} \times 2.998 \times 10^{8}\text{ m$\cdot$ s}^{-1}} {5.45 \times 10^{19} \text{ J}}\\\\& = & 3.64 \times 10^{-7} \text{ m}\\& = & \textbf{364 nm}\\\end{array}\\

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9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
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Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

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