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Flura [38]
3 years ago
5

One photon of a certain type of light has an energy of 7.32x10^-19 J. a.) What is the frequency of this type of light? b.) What

is the wavelength of this type of light?
Chemistry
1 answer:
iren [92.7K]3 years ago
7 0

Answer:

f = 1.1041 × 10¹⁵ s⁻¹

λ = 2.72  × 10⁻⁷ m

Explanation:

Given data:

Energy of photon = 7.32 × 10⁻¹⁹ J

Wavelength = ?

Frequency = ?

Solution:

Formula

E = h. f

h = planck's constant = 6.63 × 10⁻³⁴ Kg.m²/s

Now we will put the values in equation

f = E/h

Kg.m²/s² = j

f = 7.32 × 10⁻¹⁹ Kg.m²/s² / 6.63 × 10⁻³⁴ Kg.m²/s

f = 1.1041 × 10¹⁵ s⁻¹

Wavelength of photon.

E = h.c /λ

λ =  h. c / E

λ = (6.63 × 10⁻³⁴ Kg.m²/s × 3 × 10⁸ m/s) / 7.32 × 10⁻¹⁹ Kg.m²/s²

λ = 19.89 × 10⁻²⁶ / 7.32 × 10⁻¹⁹ m

λ = 2.72  × 10⁻⁷ m

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The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
Qu 4.
Anna11 [10]

Answer:

It's the second one down.

Explanation:

Gold

Mass 197

Number of Protons: 79

Number of Neutrons: 197 - 79 = 118

Number of electrons: = number of protons = 97

3 0
3 years ago
Will alka-seltzer fizz when added to apple juice?
Naddik [55]
Yes but only if you're talking about treated apple juice with a naturally small amount of water mixed in.
4 0
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The amount of matter in a object is referred as:
mestny [16]
Volume is the thing
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4 0
3 years ago
3. Given the following equation:
miskamm [114]

Answer:

4.767 grams of KCl are produced from 2.50 g of K and excess Cl2

Explanation:

The balanced equation is

2 K+ Cl2 --->2 KCI

Here the limiting agent is K. Hence, the amount of KCl will be calculated as per the mass of 2.50 gram of K

Mass of one atom/mole of potassium is 39.098 grams

Number of moles is 2.5 grams = \frac{2.5}{39.098} = 0.064

So, 2 moles of K produces 2 moles of KCL

0.064 moles of K will produces 0.064 moles of KCl

Mass of one molecule of KCl is 74.5513 g/mol

Mass of 0.064 moles of KCl is 4.767 grams

4 0
3 years ago
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