What is the empirical formula for a compound that is 94.1% oxygen and 5.90 % hydrogen?

Chemistry

1 answer:

RSB [31]3 years ago

6 0

Answer:

The empirical formula would be N₂Os * Page 2 Calculate the empirical formulaof a compound that is 94.1% oxygen, 5.9% hydrogen.

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What is the frequency and energy per quantum (in Joules) of :

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(a) f = 5.00 × 10²⁰ Hz, E = 3.32 × 10⁻¹³ J;

(b) f = 1.20 × 10¹⁰ Hz, E = 7.96 × 10⁻²⁴J.

<h3>Explanation</h3>

What's the similarity between a gamma ray and a microwave?

Both gamma rays and microwave rays are electromagnetic radiations. Both travel at the speed of light at $3.00 \times 10^{8}\;\text{m}\cdot\text{s}^{-1}$ in vacuum.

$f = \dfrac{c}{\lambda}$

where

f is the frequency of the electromagnetic radiation,
c is the speed of light, and
$\lambda$ is the wavelength of the radiation.

(a)

Convert all units to standard ones.

$\lambda = 0.600\;\text{pm} = 0.600 \times 10^{-12} \;\text{m}$ .

The unit of $f$ shall also be standard.

$f = \dfrac{c}{\lambda} = \dfrac{3.00\times 10^{8}\;\text{m}\cdot\text{s}^{-1}}{0.600\times 10^{12}\;\text{m}} = 5.00 \times 10^{20}\;\text{s}^{-1}= 5.00\times 10^{20}\;\text{Hz}$ .

For each particle,

$E = h\cdot f$ ,

where

$E$ is the energy of the particle,
$h$ is the planck's constant where $h = 6.63\times 10^{-34}\;\text{J}\cdot\text{s}^{-1}$ , and
$f$ is the frequency of the particle.