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NISA [10]
3 years ago
9

According to the periodic table, which of these elements will form an ion with a –2 charge? calcium

Chemistry
1 answer:
Serga [27]3 years ago
6 0

Answer:

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HELP ME PLEASE!!!
krok68 [10]

Answer:

Br (Bromine)

H (Hydrogen)

Explanation:

7 0
3 years ago
Are the following combinations allowed? If not, show two ways to correct them:
mafiozo [28]

The following combination of n=3 ; l=1 ; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

<h3>Is the following combination n=3; l=1; ml=-2 allowed or not.? If not, suggest two ways through which it can be corrected.</h3>

The following combination of n=3 ; l=1 ; ml=-2 is not allowed.

There are several rules that need to be followed for assigning electron quantum numbers. They are:

1. Principal quantum number should be 1 ≤ n

2. Azimuthal quantum number, 0 ≤ l ≤ n − 1

3. Magnetic quantum number, -l ≤ ml ≤ l

4. Spin quantum number as either -1/2 or +1/2

For n = 3,

l should be n - 1 or n - 2 or n - 3 = 2, 1, 0 respectively.

If we choose l = 1 then ml should be -1, 0 and +1

Therefore, one way to correct the combination would be to change the magnetic quantum number to -1

If we choose l = 2 then ml would be -2, -1, 0, +1, +2

Thus, another way to correct the combination is to choose the azimuthal quantum number as 2.

Thus, the following combination of n=3; l=1; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

To learn more about quantum numbers refer:

brainly.com/question/5927165

#SPJ4

5 0
2 years ago
A second cross was done using a different pair of mice.One parent was a mouse with white fur (bb).The was other parent was a mou
eimsori [14]
Bb, or mostly dominant with a little recessive
3 0
3 years ago
A solution contains the ions Ag , Pb2 , and Ni2 . Dilute solutions of NaCl, Na2SO4, and Na2S are available to separate the posit
Delicious77 [7]

Answer:

The solutions should be added in this order NaCl > Na2SO4 > Na2S

Explanation:

Silver is insoluble as a chloride, so the silver ions get precipitated on addition of chloride ion as silver chloride.  This means Ag+ would be removed the first.

So we will add NaCl in the first step.

The following reaction will occur.

Ag+ + Cl- → AgCl(s)

Both, Pb2 and Ni are soluble as chlorides. (lead chloride is soluble as a hot solution but will ppt when colder).

When we add Na2SO4, Pb2+ will get precipitated (because it's insoluble) as PbSO4 and Ni will remain soluble as NiSO4 is soluble in water.

The reaction that will occur is:

Pb^2+ + SO4^2- → PbSO4(s)

Nickel is insoluble as a sulfide. So when we will add Na2S, nickel will be precipitated as sulfide and be able to separate and be collected.

The solutions should be added in this order NaCl > Na2SO4 > Na2S

6 0
3 years ago
Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc
wolverine [178]

Answer:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)

∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

∴ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

∴ Kc = Kc = 1 / PO2∧6

Explanation:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)

∴ O /  Al: 0 → +2 ≡ 2e-

         Na: +1 → +2

∴ R /  H: +1 → 0

     2 - Al - 2

     2 - Na - 1

     8 - O - 8

     14 - H - 14

⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

           1 - S - 1

           4 - O - 4

           2 - H - 2

⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

       8 - P - 8

       12 - O - 12

⇒ Kc = 1 / PO2∧6

6 0
3 years ago
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