how can there be so many different substances in the world if there are only a few elements that are common?

1 answer:

3 0

-- If there are only 10 elements in the universe that can make compound molecules, and a compound molecule can be formed by combining 1, 2, 3, or 4 different elements, then that's already the possibility of at least 400 different molecules.

-- There are many more than 10 elements that can combine to form compound molecules.

-- Every single "organic" molecule, of which there are thousands, is the combination of carbon with other elements.

-- Most all of the substances that can be distilled out of oil, including the paraffin waxes, the alcohols, gasoline, kerosene, butane, propane, octane, and natural gas, are made of just carbon, hydrogen, and oxygen, only with different numbers of each one.

-- Plastics, drugs, rubber, and DNA are examples of molecules that are made of hundreds of atoms.

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La luz roja visible tiene una longitud de onda de 680 nanómetros (6,8 x 10-7 m). La velocidad de la luz es de 3.0 x108 m / s. ¿C

Lina20 [59]

Answer:

Frequency, $f=4.41\times 10^{14}\ Hz$

Explanation:

Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?

It is given that,

Wavelength of a visible red light is, $\lambda=6.8\times 10^{-7}\ m$

Speed of light is, $c=3\times 10^8\ m/s$

We need to find the frequency of visible red light. It can be calculated using below relation.

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.8\times 10^{-7}}\\\\f=4.41\times 10^{14}\ Hz$

So, the frequency of visible red light is $4.41\times 10^{14}\ Hz$ .

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3 years ago

Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the

mash [69]

Answer:

The value is $} N = 66 \ blocks$

Explanation:

From the question we are told that

The weight of the block is $W_b = 100 \ N$

The dimension of the block is $d = 0.400 m \ X \ 0.250 \ m \ X \ 0.130 \ m$

Generally two atmosphere is equivalent to

$P_{2atm} = 2 * 1.013 *10^{5} = 202600 \ N/m^2$

Generally 1 atm = $1.013 *10^{5} N/m^2$

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

$A = 0.250 * 0.130$