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olga2289 [7]
3 years ago
7

Barium (Ba) has two valence electrons, and these electrons are located in the 6s subshell. Without using the periodic table, in

which group and period is barium located?
Group 2A, Period 7

Group 1A, Period 6

Group 1A, Period 7

Group 2A, Period 6
Chemistry
2 answers:
Katena32 [7]3 years ago
7 0
Barium is located in Group 2A, Period 6
artcher [175]3 years ago
5 0

Answer:

Group 2A, Period 6

Explanation:

To know where an atom is located, we need to see its electronic distribution. The major number in it will be the period (not necessary is the number of the valence, because subshell 4s is less energetic than 3d, so it will occur first!).

To know the group, we need to check the subshell more energetic, if is subshell <em>s</em>, it will be in group 1A or 2A, if it has only 1 valence electron, then its group 1A, with two valence electrons, is group 2A.

If the subshell more energetic is the subshell <em>p</em>, so it will be from 3A to 8A, we just add the number of electrons of subshell <em>p</em> with the numbers of electrons of subshell <em>s</em> of the same layer, and we'll have the number of the group.

If the subshell more energetic is the subshell <em>d</em>, it will be in group B, and the number of electrons will be the number of the group.

If the subshell more energetic is subshell <em>f</em>, then it is on group 3B, because will be in latanids or actinides.

So, Barium has the subshell 6s with 2 valence electrons, then it is in group 2A, Period 6.

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You are given an unknown gaseous binary compound (that is, a compound consisting of two different elements). when 10.0 g of the
Luda [366]

Answer is: a possible identity for the unknown compound is C₃H₈.

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m(H₂O) = 16.3 g; mass of water.

M(O₂) = 32 g/mol; molar mass of oxygen.

M(binary compound) = 1.38 · 32 g/mol.

M(binary compound) = 44.16 g/mol.

n(binary compound) = 10 g ÷ 44.16 g/mol.

n(binary compound) = 0.225 mol; amount of substance.

n(H₂O) = 16.3 g ÷ 18 g/mol.

n(H₂O) = 0.9 mol; amount of water.

m(H₂O) : n(binary compound) = 0.9 mol ÷ 0.225 mol.

m(H₂O) : n(binary compound) = 4 : 1.

Unknown cpmpound has 4 times more hydrogen than water, it has 8 hydrogen atoms.

Second element in compound is carbon:

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8 0
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Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is starte
Katarina [22]

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

3 0
3 years ago
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