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Tpy6a [65]
3 years ago
15

Consider the electron configuration. mc011-1.jpg mc011-2.jpg mc011-3.jpgmc011-4.jpgmc011-5.jpg mc011-6.jpg mc011-7.jpgmc011-8.jp

gmc011-9.jpg 1s 2s 2p 3s 3p Which element has this electron configuration? astatine bromine chlorine fluorine
Chemistry
2 answers:
vfiekz [6]3 years ago
8 0

Answer:

The element which has the electronic configuration is CHLORINE.

The atomic number of chlorine is 17 and it has 7 valence electrons in its outermost shell. Because it needs only one more electrons to have a stable octet, it usually react with metals from group one of the periodic table who are normally willing to donate the single electrons in their outermost shells. The ground state electronic configuration of chlorine atom is 1S^2 2S^2 2P^6 3S^2 3P^5.

Explanation:

zepelin [54]3 years ago
5 0
The element which has the electronic configuration is CHLORINE.
The atomic number of chlorine is 17 and it has 7 valence electrons in its outermost shell. Because it needs only one more electrons to have a stable octet, it usually react with metals from group one of the periodic table who are normally willing to donate the single electrons in their outermost shells. The ground state electronic configuration of chlorine atom is 1S^2 2S^2 2P^6 3S^2 3P^5.
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AleksAgata [21]

for 39g water solute dissolved at 20C = solubility ( g/ 100 g H2O ) × mass of water = ( 11g / 100g H2O ) × 39g H2O = 4.29 g

amount of solute dissolved at 30 C =

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Amount of extra solute dissolved = 8.97 - 4.29 = 4.7 g

3 0
4 years ago
How many mols of a 3M NaOH solution are in 1L of the solution?
Alex

To calculate this, we need the Molarity formula. This formula tell us that Molarity, which is a concentration unit, is equal to the number of moles divided by the volume. In this question we already have the Molarity and the Volume, so let's build our equation:

C = n/V (You can see Molarity with the letter "C" because it means concentration)

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2 years ago
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3 years ago
onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀
Sveta_85 [38]

Answer:

9.34x10^-4

Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

Mass of PbCl2 = 0.2393 g

Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

Molarity of PbCl2 = 0.01722 M

Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

5 0
3 years ago
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lakkis [162]

Answer:

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Explanation:

7 0
3 years ago
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