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2 years ago

Which metalloid has three valence electrons? boron arsenic silicon lithium HElp PLEase

Chemistry

2 answers:

qaws [65]2 years ago

7 0

Answer:

the answer is boron

Explanation:

got it right on edge

Sever21 [200]2 years ago

3 0

Answer:

Boron

Explanation:

First, locate the metalloids on the periodic table. (First picture) The metalloids are in the yellow section. Next, remember that the group your element is in determines its valence electrons. For example, group 17/7A has 7 valence electrons. Look at the group numbers above each column on the periodic table to find out which metalloid is the answer. In this case, we need to look at 13/3A since we know that this metalloid has three valence electrons. Not all periodic tables will be labeled with the 3A part, so remember that groups 13 and over have the amount of valence electrons after you take out the one in front. For example, group 16 has 6 valence electrons because I removed the one in front of it. Now your options should be limited down to boron because we know that it is in the yellow section in group 13.

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MrRa [10]

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3 years ago

What is the molarity of H3PO4?

masya89 [10]

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2 years ago

Determine the ground-state electron configuration for each of the following elements: A. chlorine ______________________________

ki77a [65]

In ground-state electron configuration, the principal quantum number (n), the orbital (s, p, d, or f), and the total number of electrons are used to represent electron configurations.

Case 1 Ground-state electron configuration of elements.

A) Chlorine: 1s²2s²2 $p^{6}$ 3s²3 $p^{5}$

B) Cesium: 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s¹

C) Vanadium: 1s²2s²2p⁶3s²3p⁶3d³4s²

D) Rhenium: 1s² 2s² 2p⁶ 3s² 3p⁶3d¹⁰ 4s²4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d⁵ 6s²

Case 2 : Number of valence electron in atoms

A) Sn -Tin - Four valence electrons

B) La - Lanthanum - Three valence electrons

C) Mn - Manganese - Seven valence electrons

D) Zn- Zinc - Two valence electrons

Case 3 Ground-state electronic configuration of ions

A) Co⁺³ : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹3d⁵

B) Mo⁺² : 1s² 2s² 2p⁶3s²3p⁶ 4s² 3d¹⁰ 4p⁶5s² 4d²

C) Ra⁺² : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6p⁶

D) I⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶

Learn more about Quantum numbers here brainly.com/question/2292596

#SPJ4

8 0

1 year ago

BALANCE the equation

Anna11 [10]

Answer:

1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

2. 6 moles of Cl2

Explanation:

1. The balanced equation for the reaction. This is illustrated below:

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.

This is illustrated below:

From the balanced equation above,

4 moles of FeCl3 reacted with 3 moles of O2.

Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.

Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:

Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.

From the balanced equation above,

4 moles of FeCl3 will react to produced 6 moles of Cl2.

8 0

2 years ago

A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following

marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}$

V₂ = 4,00L

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}$

V₂ = 16,00L

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

$\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}$

V₂ = 12,31L

I hope it helps!

6 0

3 years ago