What conversion factor is needed to calculate the number of atoms in 8.6 moles of Aluminum?

Chemistry

1 answer:

Rudiy272 years ago

7 0

Answer:hey did you ever get an answer? I need it for this exam

Explanation:

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It is necessary to make 225 mL of 0.222 M solution of nitric acid. Looking on the shelf, you see only 16 M nitric acid. How much

nalin [4]

Explanation:

The required concentration of $HNO_3$ M1 =0.222 M.

The required volume of $HNO_3$ is V1 =225 mL.

The standard solution of $HNO_3$ is M2 =16 M.

The volume of standard solution required can be calculated as shown below:

Since the number of moles of solute does not change on dilution.

The number of moles $n=molarity * volume$

$M_1.V_1=M_2.V_2$

$V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL$

Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.

6 0

2 years ago

An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of calcium nitrate.

Salsk061 [2.6K]

<u>Answer:</u> The net ionic equation contains $Ca^{2+}(aq.)$ ions

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium sulfate and calcium nitrate is given as:

$(NH_4)_2SO_4(aq.)+Ca(NO_3)_2(aq.)\rightarrow CaSO_4(s)2NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ca^{2+}(aq.)+2NO_3^{-}(aq.)\rightarrow CaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonia and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)$