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Debora [2.8K]
2 years ago
7

What is the number of electrons that can be held in the first orbit closest to the nucleus

Chemistry
1 answer:
Juliette [100K]2 years ago
5 0
The first shell can hold up to 2 electrons, the second shell can hold up to 8 (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on.
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How many moles of oxygen are in 5.6 moles of al(oh)3?
Mashcka [7]
5.6 Al(OH)3
5.6 Al, 16.8 O, 16.8 H

16.8 mols of oxegyn in 5.6 mols of Al(OH)3
7 0
3 years ago
Mixtures can be separated by filtration and distillation. Both methods take advantage of ...
bagirrra123 [75]
<span>differences in the physical properties of the mixture's components</span>
5 0
3 years ago
Calcula la masa atómica del Hierro y las partículas subatómicas de cada uno de sus isótopos. Fe-54 (5.82%), Fe-56 (91.66%), Fe-5
tino4ka555 [31]

Answer:

La masa atómica del hierro es 55.847 gramos por mol.

Explanation:

Las masas molares de Fe_{54}, Fe_{56}, Fe_{57} y Fe_{58} son 53.940 gramos por mol, 55.935 gramos por mol, 56.935 gramos por mol y 57.933 gramos por mol, respectivamente. La masa atómica del hierro se determina mediante el siguiente promedio ponderado:

M_{Fe} = \frac{5.82}{100}\times \left(53.940\,\frac{g}{mol} \right)+\frac{91.66}{100}\times (55.935\,\frac{g}{mol})+\frac{2.19}{100}\times \left(56.935\,\frac{g}{mol} \right)+\frac{0.33}{100}\times \left(57.933\,\frac{g}{mol} \right)

M_{Fe} = 55.847\,\frac{g}{mol}

La masa atómica del hierro es 55.847 gramos por mol.

8 0
2 years ago
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

6 0
3 years ago
What’s the oxidation state of Se in H2SeO3
Amiraneli [1.4K]

Answer:

The oxidation states are as follows:

+1 for each H, -2 for each O,

+4 for Se

Explanation: I hope this attachment helps!!

6 0
3 years ago
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