A result of the common ion effect is that the selective precipitation of a metal ion, such as Ag+, is promoted by the addition of an appropriate counter ion (X-) that produces a compound (AgX) with a very low solubility.

3 0

3 years ago

How many electrons does N 3-have?

Allushta [10]

Answer: 10 electrons

Explanation:

N represents Nitrogen. Nitrogen has an atomic number of 7, this means in ground state it has 7 electrons also.

But N-3, means Nitrogen has gained 3 more electrons. So, we have 10 electrons

8 0

3 years ago

Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ...

KatRina [158]

Answer:

The oxidation state of silver in $\rm Ag_2O$ is $+1$ .

The oxidation state of sulfur in $\rm SO_2$ is $+4$ .

Explanation:

The oxidation states of atoms in a compound should add up to zero.

There are two silver $\rm Ag$ atoms and one oxygen $\rm O$ atom in one formula unit of $\rm Ag_2O$ . Therefore:

$\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}$ .

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is $-2$ . Silver oxide $\rm Ag_2O$ isn't an exception. Therefore:

$\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}$ .

Solve this equation for the (average) oxidation state of $\rm Ag$ :

$\text{Oxidation state of $\rm Ag$} = 1$ .

Similarly, because there are one sulfur $\rm S$ atom and two oxygen $\rm O$ atoms in each $\rm SO_2$ molecules:

$\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}$ .

The oxidation state of $\rm O$ in $\rm SO_2$ is also $-2$ , not an exception, either.

Therefore:

$\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}$ .

Solve this equation for the oxidation state of $\rm S$ here:

$\text{Oxidation state of $\rm S$} = 4$ .

8 0

4 years ago

In the UNBALANCED chemical reaction for the combustion of acetylene (used in welding torches), determine at standard temperature

kiruha [24]

Answer:

2 C2H2 + 5 O2 = 4 CO2 + 2 H2O

I've checked this multiple times this should be it

4 0

3 years ago