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eduard
3 years ago
8

If the visible light spectrum is from 400 to 700 nm, would light with an energy of 2.79 x 10^-19 J be visible with the naked eye

? What is the wavelength of this light?
Chemistry
1 answer:
sladkih [1.3K]3 years ago
5 0

Answer:

713 nm. It is not visible with the naked eye.

Explanation:

Step 1: Given data

  • Energy of light (E): 2.79 × 10⁻¹⁹ J
  • Planck's constant (h): 6.63 × 10⁻³⁴ J.s
  • Speed of light (c): 3.00 × 10⁸ m/s
  • Wavelength (λ): ?

Step 2: Calculate the wavelength of the light

We will use the Planck-Einstein equation.

E = h × c / λ

λ = h × c / E

λ = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s / 2.79 × 10⁻¹⁹ J

λ = 7.13 × 10⁻⁷ m

Step 3: Convert "λ" to nm

We will use the relationship 1 m = 10⁹ nm.

7.13 × 10⁻⁷ m × (10⁹ nm/1 m) = 713 nm

This light is not in the 400-700 nm interval so it is not visible with the naked eye.

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160.32 grams of Ca or 160 if rounded

Explanation:

Multiply moles of Ca by the conversion factor (molar mass of calcium) 40.08 g Ca/ 1 mol Ca, which then allows the cancelation of moles, leaving grams of Ca.

4 mol*40.08g/mol = 160.32 grams of Ca

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Answer: short answer short 0.65 / 5.37

15% NaOH by mass is 15g of NaOH in (100-15)g of solution.

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Explanation:

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3 years ago
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What volume of 0.140 HCl is needed to neutralize 2.58 of Mg(OH)2<br> ? ...?
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8 0
3 years ago
Ammonia (NH3) reacts with oxygen to form nitric oxide (NO) and water vapor: 4NH3 + 502 4NO + 6H2O b) When 20.0 g NH3 and 50.0 g
Solnce55 [7]

Answer: a) . Ammonia is the limiting reagent

b. Oxygen is left over and 0.1375 g of oxygen is left over.

c. The theoretical yield of NO is 35.29 g.

d. The theoretical yield of H_2O is 31.74 g.

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

For NH_3

Given mass of ammonia = 20.0 g

Molar mass of ammonia = 17.031 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{20.0g}{17.031g/mol}=1.17mol

For O_2

Given mass of oxygen gas = 50.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{50.0g}{32g/mol}=1.6mol

The chemical equation for the reaction is

4NH_3+5O_2\rightarrow 4NO+6H_2O

By Stoichiometry of the reaction:

4 moles of ammonia reacts with = 5 moles of oxygen

So 1.17 moles of ammonia will react with = \frac{5}{4}\times 1.17=1.4625mol of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.

Thus ammonia is considered as a limiting reagent because it limits the formation of product.

1. By Stoichiometry of the reaction:

4 moles of ammonia produces = 4 moles of NO

1.17 moles of ammonia will produce = \frac{4}{4}\times 1.17=1.17moles of NO

Mass of NO=moles\times {\text{Molar Mass}}=1.17\times 30=35.29g

Thus Theoretical yield of NO is 35.29 grams.

2. By Stoichiometry of the reaction:

4 moles of ammonia produces = 6 moles of H_2O

1.2 moles of ammonia will produce = \frac{6}{4}\times 1.2=1.8moles of H_2O

Mass of H_2O=moles\times {\text{Molar Mass}}=1.8\times 18.015=31.74g  H_2O

Thus Theoretical yield of H_2O is 31.74 grams.

8 0
3 years ago
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