The missing question is:
<em>What is the percent efficiency of the laser in converting electrical power to light?</em>
The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.
A particular laser consumes 130.0 Watt (P) of electrical power. The energy input (Ei) in 1 second (t) is:
The laser produced photons with a wavelength (λ) of 1017 nm. We can calculate the energy (E) of each photon using the Planck-Einstein's relation.
where,
The energy of 1 photon is 6.52 × 10⁻²⁰ J. The energy of 2.67 × 10¹⁹ photons (Energy output = Eo) is:
The percent efficiency of the laser is the ratio of the energy output to the energy input, times 100.
The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.
You can learn more about lasers here: brainly.com/question/4869798
It should be a chemical change
Hey there!:
Given the mass of PbCl(OH) :
0.135 Kg = 0.135 Kg*(1000g / 1Kg) = 135 g
Molecular mass of PbCl(OH) = 207+35.5+16+1 = 259.5 g / mol
Atomic mass of Pb = 207 g/mol
Hence mass of Pb in 135 g PbCl(OH) :
(207 g Pb / 259.5 g PbClOH) * 135g PbClOH =
0.79768 * 135 => 107.68 g of Pb
For Pb2Cl2CO3 :
Given the mass of Pb2Cl2CO3 :
0.135 Kg = 0.135 Kgx(1000g / 1Kg) = 135 g
Molecular mass of Pb2Cl2CO3 = 2*207+2*35.5+12+3*16 = 545 g / mol
Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3 = 2*207 = 414 g
Hence mass of Pb in 135 g Pb2Cl2CO3:
(414 g Pb / 545 g PbClOH) * 135g PbClOH =
0.75963 * 135 => 102.55 g of Pb2Cl2CO3
Hope that helps!
Reactivity trends of halogen:
1) Melting point and boiling points increased down the
group
2) Colour becomes darker.
E.g. Fluorine (pale yellow)
Chlorine (yellowish-green)
Bromine (reddish-brown)
Iodine (purplish-black)
Astatine (black)
3) The reactivity decreases down the group.
Reactivity:
F > Cl > Br > I > At
Answer:
- 602 mg of CO₂ and 94.8 mg of H₂O
Explanation:
The<em> yield</em> is measured by the amount of each product produced by the reaction.
The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.
The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:
To calculate the yield follow these steps:
<u>1. Mole ratio</u>
<u />
<u>2. Convert 175mg of fluorene to number of moles</u>
- Number of moles = mass in grams / molar mass
<u>3. Set a proportion for each product of the reaction</u>
a) <u>For CO₂</u>
i) number of moles
ii) mass in grams
The molar mass of CO₂ is 44.01g/mol
- mass = number of moles × molar mass
- mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg
b) <u>For H₂O</u>
i) number of moles
ii) mass in grams
The molar mass of H₂O is 18.015g/mol
- mass = number of moles × molar mass
- mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
