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anzhelika [568]
3 years ago
11

Will give medal! What is the reason the group 13 metals have a typical charge of 3+?

Chemistry
1 answer:
Art [367]3 years ago
8 0
My answer:

13 group of the periodic table represented by boron, aluminum and gallium subgroup. It includes gallium, indium, thallium. Typical steper oxidation in the subset gallium 3 is explained by the presence of (n-1)d^10 E-configuration.
Aluminium oxidation degree has +3 an electronic configuration of noble gases S^2P^6

Hope this helps yah!!!
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Question 5
pantera1 [17]

Answer: The coefficient for the diatomic oxygen (O2) is 3.

Explanation:

To know the coefficient for the diatomic Oxygen, we need to balance the equation.

Fe + O2 ------->   Fe2O3

LHS of the equation; Fe =  1    , O2 = 1

RHS of the equation; Fe = 2 ,  O = 3

∴ Multiply 'Fe' on the LHS of the equation by 4 and O2 by 3

   Doing that will give the balance equation which is;

   4 Fe +  3 O2  --------> 2 Fe2O3

The coefficient for the diatomic oxygen (O2) as seen from the equation is 3.

7 0
3 years ago
How can atoms have more than 32 electons?
guajiro [1.7K]
86 atoms have more electrons than Germanium
3 0
3 years ago
¿Cuál es la cantidad de electrones (e-) de Níquel si tiene una masa atómica de 58.6 y un número
daser333 [38]

Answer:

Hi do we translate a this

Explanation:

4 0
3 years ago
Plz help I’m pretty confused on this question
sergeinik [125]

Answer:

Bohr model A

Explanation:

It has more valance electrons therefore has more interaction between the atoms and has more electronegativity.

7 0
3 years ago
If the starting volume of a hot air balloon is 55,500 m3and the initial temperature is 21 °C, what is the temperature inside the
Dmitry_Shevchenko [17]

Answer:

T₂ = 392 K

Explanation:

Given that,

Initial volume of the hot air balloon, V₁ = 55500 m³

Initial temperature, T₁ = 21°C = 294 K

Final volume, V₂ = 74000 m³

We need to find the final temperature inside the balloon. The relation between the temperature and volume is given by charles law i.e.

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Where

T₂ is the final temperature

So,

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{294\times 74000 }{55500 }\\\\T_2=392\ K

So, the new temperature is 392 K.

8 0
3 years ago
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