Which one will create a precipitate when mixed with a bromine ion?

question 3 options:

ag+1

cu+2

zn+2

The answer to your question is Ag⁺¹

Explanation:

To solve this problem we need to look at the table and find each of the cations given and the anion and observe if the cross of both lines is soluble or insoluble.

1.- For the first cation Ag⁺ and the anion Br⁻, we notice that the result is insoluble, so this compound will precipitate.

2.- For the second cation Cu⁺² and the anion Br⁻, we notice that the result is "this compound does not exist".

3.- For the third compound Zn⁺² and the anion Br⁻, we notice that the result is soluble so this compound will not precipitate.

Ag+

Pb2+ (cold water)

Explanation:

Step 1: What is a precipitate ?

When two aqueous solutions react, they sometimes form solids in the solution. The solid is called a precipitate. Precipitation reactions occur when the cations of one reactant and the anions of a second reactant found in aqueous solutions combine to form an insoluble ionic solid

Most bromide compounds are highly soluble in water.

We can see this on the table. When we combine a bromine ion (Br-) with Zn2+ or with Cu2+

there is a soluble compound formed.

An exception is when Br- ion reacts with Ag+. Here, an insoluble precipitate (AgBr or silverbromide).

Which one will create a precipitate when mixed with a bromine ion?

Question 3 options:

Pb2+

Li+

Na+

Since most bromide compounds are soluble in water, as we can see on the table, this is also true when Br- reacts with Li+ or Na+ . There will be formed a soluble compound.

When Br- reacts with Pb2+ there will be formed a compound that is (very) slightly soluble.

There will be formed PbBr2.

PbBr2 is very bad soluble in cold water ( will precipate).

In hot water (boiling temperature). The solubility of PbBr2 increases.

We can say that Br- forms a precipitate with Pb2+, but in cold water.

h let me think

increased stability in the completion of fluorine's outer shell.