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Sphinxa [80]
3 years ago
6

List the symbols for all the elements with electron configurations that end as follows. Each n represents an energy level.

Chemistry
1 answer:
Gwar [14]3 years ago
6 0

List the symbols for all the elements with electron configurations that end as follows. Each n represents an energy level.

a. ns1

b. ns2np4

Answer: a) H, Li, Na, K, Rb, Cs, and Fr

b) O, S, Se, Te, Po

Explanation:

a. Determining the elements correlating to the electron configuration of "ns1" , you should understand that this suggests that the elements are in Group 1A.

This is because that there is one electron in an s-orbital. The orbital is located in the first column of the periodic table.

Thus, the elements that have this electron configuration are H, Li, Na, K, Rb, Cs, and Fr.

b. Determining the elements correlating to the electron configuration of ns^2np^4 characterizes group 6A of the periodic table.

This is because that there is have six valence electrons in their highest-energy orbitals (ns2np4). The orbital is located in the first column of the periodic table.

Thus, the elements that have this electron configuration are the nonmetals oxygen (O), sulfur (S), and selenium (Se), the metalloid tellurium (Te), and the metal polonium (Po).

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A mixture of calcium carbonate, CaCO3, and barium carbonate, BaCO3, weighing 5.40 g reacts fully with hydrochloric acid, HCl(aq)
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CaCO₃ = 85.18%

BaCO₃ = 14.82%

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CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O

So, the total amount of CaCO₃ BaCO₃ will form the CO₂.

Using the ideal gas law to calculate the number of moles of CO₂:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature (50ºC + 273 = 323 K).

0.904*1.39 = n*0.082*323

26.486n = 1.25656

n = 0.05 mol

So, the number of moles of the mixture is 0.05 mol.

The molar masses of the components are:

CaCO₃ = 40 g/mol of Ca + 12 g/mol of C + 3*16 g/mol of O = 100 g/mol

BaCO₃ = 137.3 g/mol of Ba + 12 g/mol of C + 3*16 g/mol of O = 197.3 g/mol

Let's call x the number of moles of CaCO₃ and y the number of moles of BaCO₃, so:

100x + 197.3y = 5.4

x + y = 0.05 mol

y = 0.05 - x

100x + 197.3*(0.05 - x) = 5.4

100x - 197.3x = 5.4 - 9.865

97.3x = 4.465

x = 0.046 mol of CaCO₃

y = 0.004 mol of BaCO₃

So, the masses are:

CaCO₃ = 100* 0.046 = 4.60 g

BaCO₃ = 137.3*0.004 = 0.80 g

The percentages in the mixture are:

CaCO₃ = (4.60/5.40)*100% = 85.18%

BaCO₃ = (0.80/5.40)*100% = 14.82%

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