Determine the volume of Fe(NO3)3 stock solution needed to prepare a dilute solution, then decide what volume of dilute solution should be prepared.

Volume stock solution: 0.25 mL.

Volume of resulting diluted solution: 50 mL.

Concentration of the diluted solution: 0.001 M.

Explanation:

Hello!

In this case, according to the given scheme, we need to consider that 10-mL pipets are not able to measure out more than 10 mL, it means that we need to take 0.25 mL from the stock solution in order to prepare the diluted solution.

Next, since the given volumetric flask at the right side, probably has a volume of 50 mL, rather than 0.5 mL, we infer that 50 mL of the diluted solution should be prepared.

Finally, considering that the concentration and volume of the stock solutions are respectively 0.2 M and 0.25 mL and the volume of the diluted solution is 50 mL, we can compute the concentration of the diluted solution as follows

Best regards!

i just learnt this recently but i need to look through my notes again

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.