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1 year ago

How much energy does a 445 nm wave of light have? (The speed of light in a

Chemistry

1 answer:

Dahasolnce [82]1 year ago

8 0

Answer:

Explanation:

To calculate questions like this, you need to make use of the equation below:

E = hν

E is the energy, h is Planck's constant, and ν is the frequency.

Also, another important formula you need for this question is the following:

ν = c / λ

ν is the frequency, c is the speed of light, and λ is the wavelength.

Therefore, E = hν = hc/λ.

Also remember to convert nm wavelength to meters.

To answer the question, E = hc/λ = ((6.626*10^-34 J•s)(3*10^8 m/s))/(4.45*10-7 m) = 4.42 • 10^-19 J = A

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What is the change in internal energy for each of the following situations? 1. q-7.9 J out of the system and w 3.6 J done on the

adoni [48]

Answer:

A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.

Explanation:

A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.

1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J

ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J

B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂

= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.

8 0

2 years ago

Why is performing extraction with several small portions of a solvent more officient than a single extraction with the same tota

elena55 [62]

With various extractions the amount of material left in the trash will be lower, ergo the extraction will be more perfect. Various extractions with fewer amounts of solvent are more efficient than a single extraction with a huge amount of solvent.

<u>Explanation:</u>

Surely multiple extractions are better than the single large extraction. Because extraction is about maximizing outside field communication between the two solvents, and you easily get more surface area contact with fewer amounts.

You can merge two smaller portions quicker and more completely than with large portions.

6 0

2 years ago

How many electrons should be shown in the lewis symbol for hydrogen?

fredd [130]

999,956,000,000,000,000

6 0

2 years ago

A 104 m3 thoroughly mixed pond has a water inflow and outflow of 5 m3/h. The inflow water contains 0.01 mol/m3 of chemical. Chem

Naily [24]

Answer:

C ≈ 1.44 × 10⁻³ mol/m³

Explanation:

The given information are;

The liquid volume the pond can hold = 104 m³

The volume of inflow into the pond = 5 m³/h

The volume of outflow into the pond = 5 m³/h

The concentration of the chemical in the inflow water = 0.01 mol/m³

The concentration of the chemical discharged directly into the water = 0.1 mol/h

The concentration, $c_{(inflow)}$ , of chemical that enters the water through inflow per hour is given as follows;

$c_{(inflow)}$ = 0.01 mol/m³ × 5 m³/h = 0.05 mol/h

The concentration, $c_{(discharge)}$ , of chemical that enters the water through direct discharge per hour is given as follows;

$c_{(discharge)}$ = 0.1 mol/h

The total concentration that enters the pond per hour is given as follows;

$c_{(inflow)}$ + $c_{(discharge)}$ = 0.1 mol/h + 0.05 mol/h = 0.15 mol/h

Whereby the water in the pond properly mixes with the pond, we have;

The concentration of chemicals (C) in the outflow water = 0.15 mol/(104 m³) ≈ 0.00144 mol/m³

C ≈ 1.44 × 10⁻³ mol/m³.

4 0

2 years ago

Determine the velocity of a 55-kg skier whose kinetic energy is 8900 J

goldfiish [28.3K]

Answer:

Kinetic energy = 1/2mv^2

8900 j =1/2*55*v^2

v^2=8900*2/55

v^2=323.6

v=17.98

Explanation:

4 0

2 years ago