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mel-nik [20]
3 years ago
14

Which group or groups of atoms are the only atoms with f orbitals?

Chemistry
1 answer:
garri49 [273]3 years ago
5 0

Answer:

Lanthanides and Actinides

Explanation:

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The radius of an atom of gold (Au) is about 1.35 Å.
Eduardwww [97]
Two radius of an atom is equal to the diameter. Adding up all the diameter of  the atoms, it should be equal to 9.5 mm. Therefore, we simply convert the units to the same units then divide 1.35 A to 9.5 mm. We calculate as follows:

no. of atoms = 0.0095 m / 1.35x10^-9 m = 7037037 atoms 

Hope this answers the question. Have a nice day.
6 0
3 years ago
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Feliz [49]
I think the correct answer would be the third option. The reason I2 has a higher melting point than F2 is because I2 possesses a more polarizable electron cloud. I2 contains more electrons than F2 which would result to a stronger intermolecular forces. Having stronger intermoleculer forces would mean more energy is needed to break the bonds so a higher melting point would be observed.
4 0
4 years ago
Conducting a monohybrid cross of F1 generation plants, the recessive trait reappears in F2 plants in a ratio of _____.
lisabon 2012 [21]

the answer is 3:1

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8 0
3 years ago
Look at the above table.
MariettaO [177]

Answer:

<em> 1</em>. A. 0

<em>2</em>. B. 7

<em>3. </em>C<em>.</em><em> </em>4

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7 0
3 years ago
Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr
RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

The net reaction is  :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

4 0
3 years ago
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