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Nimfa-mama [501]
3 years ago
12

A) A 100-watt lightbulb radiates energy at a rate of 100 J/s. (The watt, a unit of power, or energy over time, is defined as 1 J

/s.) If all of the light emitted has a wavelength of 520 nm, how many photons are emitted per second? (Assume three significant figures in this calculation.)
Express your answer using three significant figures.

B) How much energy (in \({\rm J}\)) is contained in 1.00 mole of 490\({\rm nm}\) photons?
Express your answer to three significant figures and include the appropriate units.

C) What is the de Broglie wavelength of an electron traveling at 1.48

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

The answers are as given in the attachment

Explanation:

The application of the de brogile equation was used and appropriate substitution were made as shown in the attachment

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3 years ago
What is the value for AG at 100 Kif AH = 27 kJ/mol and AS = 0.09 kJ/(mol-K)?
Elina [12.6K]

Answer:

ΔG =   18KJ/mol

Explanation:

Given data:

ΔS = 0.09 Kj/mol.K

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Temperature  = 100 K

ΔG = ?

Solution:

Formula:

ΔG = ΔH - TΔS

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7 0
2 years ago
An ideal gas in a 1.25-gallon container is at a temperature of 125 degrees Celsius and pressure of 2.5 atmospheres. If the gas i
Tresset [83]
You will need the equation PV = nRT

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V = Volume in L
n = moles
R = 8.314 (constant)
T = Temperature in Kelvin 

First convert 2.5 atm into kPa:

2.5 X 101.3 = 253.25 kPa

Convert 125 Celsius into Kelvin:

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Plug your values into the equation to solve for n:

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8 0
3 years ago
Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).
muminat

Hello!

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

We have the following data:

m (mass) = ? 

n (number of moles) = 5.20 moles

MM (Molar mass of C6H12) ≈ 84.2 g/mol

Now, let's find the mass, knowing that:

n = \dfrac{m}{MM}

5.20\:\:\diagup\!\!\!\!\!\!\!mol = \dfrac{m}{84.2\:g/\diagup\!\!\!\!\!\!\!mol}

m = 5.20*84.2

\boxed{\boxed{m = 437.84\:g}}\end{array}}\qquad\checkmark

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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