Write the complete symbol for a positron

Sodium is much more apt to exist as a cation than chlorine because:

8090 [49]

Answer:

(D) chlorine has a greater ionization energy than sodium

Explanation:

Sodium is the element of the group 1 and period 3 which means that the valence electronic configuration is $[Ne]3s^1$ .

Chlorine is the element of the group 17 and period 3 which means that the valence electronic configuration is $[Ne]3s^23p^5$ .

Ionization energy is the minimum amount of energy which is required to knock out the loosely bound valence electron from the isolated gaseous atom.

Thus, removal of one electron in sodium is easy as it will gain noble gas configuration and become stable. But this case does not exist in chlorine and hence, chlorine has a greater ionization energy than sodium.

So,

Sodium is much more apt to exist as a cation than chlorine because chlorine has a greater ionization energy than sodium.

7 0

3 years ago

Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration

gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

$K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$9.84 \times 10^{-21} = s \times s$

s = $9.92 \times 10^{-11}$

Also, $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

$2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

$K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

$[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

= $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

$9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

$[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.

7 0

3 years ago

[06.04]On a potential energy diagram, the activation energy is greatest under which of the following conditions?

Nata [24]

The first one is the correct answer: The potential energy of the products is greater than that of the reactants and the change in enthalpy is positive.

5 0

3 years ago

Read 2 more answers

2H2S + 3O2 2SO2 + 2H2O Which option gives the correct mole ratios? H2S:SO2 = 2:2 and O2:H2O = 3:2 H2S:SO2 = 2:3 and O2:H2O = 3:2

UkoKoshka [18]

Correct Answer: Option (A)

H₂S : SO₂ = 2:2 and,

O₂: H₂O = 3:2

Explanation:

The mole ratio is the ratio of coefficients of the compounds involved in the reaction. Reactants, as on the left hand side of the chemical reaction which reach to form new products.

In the given reaction,

2H₂S + 3O₂ ⇒ 2SO₂ + 2H₂O

The reactants are : H₂S (hydrogen sulphide) & O₂ (oxygen)

The products are : SO₂ (sulphur dioxide) & H₂O (water)

2 moles of hydrogen sulphide react with 3 moles of oxygen
2 moles of sulphur dioxide & 2 moles of water are produced

Therefore correct option is

H₂S : SO₂ = 2:2 and,

O₂: H₂O = 3:2

7 0

3 years ago

The Battle of Gettysburg proved to be the turning point of the war. true or false

xxTIMURxx [149]

The Battle of Gettysburg proved to be the turning point of the war.

TRUE

6 0

3 years ago