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Leona [35]
4 years ago
6

A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but additio

n of (SO4)2- ion resulted in a precipitate. Which cation is present
Chemistry
1 answer:
zhannawk [14.2K]4 years ago
7 0

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate SO_4~^-^2 and the chloride Cl^-:

<u>Sulfate</u>

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.(Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2), which are NOT soluble.

<u>Chloride</u>

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). (Pb^+^2~,Ag^+~,Hg_2~^+^2), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

<u>"Calcium, strontium, or barium".</u>

I hope it helps!

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If the volume is decreased from 5.15L to
hichkok12 [17]

Answer:

Increase

Explanation:

Boyle's law states volume and pressure have an inverse relationship.

- Hope that helped! Let me know if you need further explanation.

3 0
3 years ago
What is the reaction that corresponds to the first ionization energy of rubidium, rb?
dedylja [7]

Ionization is a process in which an electron present in the outermost orbital of an atom gets knocked out by absorbing an appropriate amount of energy. Release of electron from a neutral atom results in the formation of a cation- a positively charged ion.

Rubidium (Rb) is an alkali metal with a valence electron configuration of 5s¹. Ionization results in the loss of this '5s' electron forming a Rb+ ion. The reaction can be expressed as:

Rb (5s¹) → Rb⁺ (5s⁰) + e⁻



6 0
3 years ago
How many moles of oxygen are produce dif 11.0 mol of al are produced
ycow [4]

Answer:

8.25 moles

Explanation:

Given parameters:

Number of moles of Al produced = 11 moles

Unknown:

Number of moles of oxygen produced = ?

Solution:

To solve this problem, we need to understand the problem.

The decomposition of an aluminium oxide must has produced oxygen and aluminium,

              2Al₂O₃   →  4Al  + 3O₂

now since the known is the oxygen gas; we can find the unknown aluminium:

                3 mole of O₂ was produced with 4 mole of Al

               x mole of O₂ will be produced with 11 moles of Al

 x  = \frac{3 x 11}{4}  = 8.25 moles

7 0
3 years ago
COPY out the following sentences and fill in the gaps.
malfutka [58]

Answer:

1. Hydrogen ions; acidic

2. Alkali; hydroxide ions; alkaline

3a. Sulfuric acid --> 2 Hydrogen ions + sulfate ion

H₂SO₄ --> 2H+ + SO₄²-

3b. Sodium hydroxide --> Sodium ion + Hydroxide ion

NaOH --> Na+ + OH-

Explanation:

1. Sulfuric acid releases hydrogen ions in solution. This makes the solution acidic.

Acids produce hydrogen ions when dissolved in aqueous solutions.

2. Sodium hydroxide is an alkali. It releases hydroxide ions in solution. This makes the solution alkaline.

Alkalis are soluble bases that produce hydroxide ions in solution.

3a. Sulfuric acid --> 2 Hydrogen ions + sulfate ions

H₂SO₄ --> 2H+ + SO₄²-

The equation above is for the ionization of sulfuric acid

b. Sodium hydroxide --> Sodium ion + Hydroxide ion

NaOH --> Na+ + OH-

The equation above is for the ionization of sodium hydroxide

4 0
3 years ago
The photo shows wires made of pure copper, an element,
Strike441 [17]
I can’t see the picture but a atom I think
3 0
3 years ago
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