All categories

3 years ago

What is the energy of a photon with wavelength of 670.8?

1 answer:

8 0

<span>The velocity of a wave is calculated by v = λf. From this equation we can take the frequency f = v / λ and substitute in the energy equation to find a new equation for the calculation of the energy from the wavelength: E = hc / λ</span>

To calculate the photon energy, in eV (Electrovolt), when we have the value of the wavelength λ (necessarily given in Å - angstrom), we use an alternative equation given by:
$E = \frac{12400}{\lambda}$

Data: 670.8 Å

Solving:
$E = \frac{12400}{\lambda}$
$E = \frac{12400}{670.8}$
$\boxed{E \approx 18,4eV}$

You might be interested in

Please get me the right answer

prisoha [69]

Answer:

Explanation:

7 0

3 years ago

Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind

Leya [2.2K]

Answer:

The molecular formula is SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

Moles F = 37.23 grams / 19.00 g/mol

Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

3 0

3 years ago

Which of the atoms shown has an atomic number of 4?

spin [16.1K]

Answer:

B. Be

Explanation:

8 0

3 years ago

A pound is approximately 0.45 kilogram. A person weighs 87 kilograms. What is the person’s weight, in pounds, when expressed to

Sauron [17]

Answer:

87 kilograms =191.8 pounds

4 0

3 years ago

Read 2 more answers

The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M

anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0

3 years ago

Read 2 more answers