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Maru [420]
3 years ago
14

What is the energy of a microwave photon that has a frequency of 9.86 x 1012 Hz?

Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
5 0

Answer:

6.533 × 10^-21J

Explanation:

The energy of the microwave photon can be calculated using:

E = hf

Where;

E = energy of photon (J)

h = Planck's constant (6.626 × 10^-34 J/s)

f = frequency (9.86 x 10^12 Hz)

Hence, E = hf

E = 6.626 × 10^-34 × 9.86 x 10^12

E = 65.33 × 10^(-34 + 12)

E = 65.33 × 10^(-22)

E = 6.533 × 10^-21J

The energy of the microwave photon is

6.533 × 10^-21J

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A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach
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Answer:

% = 5.69%

Explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ +  I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

Finally the % of this, in the bleach sample would be:

% = 0.036 / 0.634 * 100

<h2>% = 5.69%</h2>
6 0
3 years ago
How do you find mole ratios?
Luba_88 [7]

In order to determine the mole ratio, you need to begin with a balanced chemical equation.

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Explanation:

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Elements present in the same period will have same atomic orbitals.

For example, electronic distribution of Na is 2, 8, 1 and it is a third period element.

Similarly, electronic distribution of Cl is 2, 8, 7 and it is also a third period element.

Hence, both Na and Cl will have K, L, M shells, that is, they have three atomic orbitals.

Thus, we can conclude that a pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.

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