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3 years ago

Given that a particular photon has a frequency of 2.2 x 10'7 Hz, calculate the photon's energy. Submit your answer in

Chemistry

1 answer:

BlackZzzverrR [31]3 years ago

5 0

Answer:

E = 15×10⁻²⁹ J

Explanation:

Given data:

Frequency of photon = 2.2× 10⁷ Hz

Energy of photon = ?

Solution:

Formula:

E = h.f

h = 6.63×10⁻³⁴ Js

by putting values,

E = 6.63×10⁻³⁴ Js × 2.2× 10⁷ s⁻¹

E = 14.586 ×10⁻²⁹ J

E = 15×10⁻²⁹ J

The energy of photon is 15×10⁻²⁹ J.

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In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of t

Radda [10]

Answer: The empirical formula for the given compound is $CuCl_3$

Explanation:

We are given:

Mass of copper chloride in 1 L or 1000 mL of solution = 42.62 grams

Taking Trial A:

Volume of solution = 49.6 mL

Applying unitary method:

In 1000 mL of solution, the mass of copper chloride present is 42.62 grams

So, in 49.6 mL of solution, the mass of copper chloride will be = $\frac{42.62}{1000}\times 49.6=2.114g$

We are given:

Mass of filter paper = 0.908 g

Mass of filter paper + copper = 1.694 g

Mass of copper = [1.694 - 0.908] g = 0.786 g

Mass of chlorine in the sample = [2.114 - 0.786]g = 1.328 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Copper = $\frac{\text{Given mass of Copper}}{\text{Molar mass of Copper}}=\frac{0.786g}{63.5g/mole}=0.0124moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{1.328g}{35.5g/mole}=0.0374moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0124 moles.

For Copper = $\frac{0.0124}{0.0124}=1$

For Chlorine = $\frac{0.0374}{0.0124}=3.02\approx 3$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Cu : Cl = 1 : 3

Hence, the empirical formula for the given compound is $CuCl_3$

3 0

3 years ago

A 29.2 mL sample of a 0.458 M aqueous hydrofluoric acid solution is titrated with a 0.337 M aqueous sodium hydroxide solution. W

Svetach [21]

Answer:

3.51

Explanation:

Before any sodium hydroxide has been added, the pH is that of the aqueous hydrofluoric acid solution.

HF is a weak acid that dissociates according to the following equation.

HF(aq) ⇄ H⁺(aq) + F⁻(aq) Ka = 6.76 × 10⁻⁴

We can find [H⁺] using the following expression.

[H⁺] = √(Ca × Ka)

where

Ca: concentration of the acid

Ka: acid dissociation constant

[H⁺] = √(Ca × Ka)

[H⁺] = √(0.458 × 6.76 × 10⁻⁴)

[H⁺] = 3.10 × 10⁻⁴ M

The pH is:

pH = -log [H⁺]

pH = -log 3.10 × 10⁻⁴ = 3.51

4 0

3 years ago

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zloy xaker [14]

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The following: Jupiter, Saturn, Uranus, or Neptune.

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2 years ago

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Which is the formula for diarsenic pentoxide? Ar2O5 As5O2 As2O5 2As5O

yaroslaw [1]

As205

Hope this helps:)

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3 years ago

Read 2 more answers

Objects that can bew seen only under magnification

timofeeve [1]

The answer is (D) microscopic. You can remember this, because the name is very close to "microscope," an instrument used to greatly magnify and observe tiny organisms and objects.

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3 years ago