This is a basic stoichiometry question. Things you need to know:
1 mol = 6.022 x 10^23 atoms/particles/molecules/ions
Ratio of Mg in MgCl2 : 1mol Mg2+/1 mol MgCl2
Always start with what you are given:
3.00 mol MgCl2 X 1mol Mg2+/1 mol MgCl2 X 6.022 x 10^23 ions Mg2+/ 1mol Mg2+
=1.8066 x 10^24 ions Mg2+
There are 3 moles of sulfate ions (SO4^2-) present in 1 mole of Al2(SO4)3. Therefore, in 1.7 moles of Al2(SO4)3, there would be 3 * 1.7 = 5.1 moles of sulfate ions.
To find the number of moles of H ions in the solution, first calculate the moles of HNO3 using the given concentration and volume. Since each mole of HNO3 yields 1 mole of H ions in solution, the number of moles of H ions is the same as the moles of HNO3. Therefore, in this case, there are 0.4512 moles of H ions present in the solution.
The number of moles is 8,00944733981.10e23.
To find the number of moles of ions in the solution, we first calculate the total number of moles of KCl using the given concentration and volume. Multiply the number of moles of KCl by 2 (since one mole of KCl produces 2 moles of ions, one K+ and one Cl-) to find the total moles of ions present.
There are 5.2 moles of ions in 2.6 moles of K2SO4. This is because each formula unit of K2SO4 dissociates into three ions: 2K+ ions and 1 SO4^2- ion. Thus, for every mole of K2SO4, you have 3 moles of ions.
5,7 moles (SO4)3-.
There are 3 moles of sulfate ions (SO4^2-) present in 1 mole of Al2(SO4)3. Therefore, in 1.7 moles of Al2(SO4)3, there would be 3 * 1.7 = 5.1 moles of sulfate ions.
To find the number of moles of H ions in the solution, first calculate the moles of HNO3 using the given concentration and volume. Since each mole of HNO3 yields 1 mole of H ions in solution, the number of moles of H ions is the same as the moles of HNO3. Therefore, in this case, there are 0.4512 moles of H ions present in the solution.
The number of moles is 8,00944733981.10e23.
There are 1.35 moles of MgBr2 in 1 L of solution, which corresponds to 2 moles of bromide ions. Therefore, in 750.0 mL of 1.35 M MgBr2 solution, there will be 1.0125 moles of bromide ions.
I suppose that the answers are: - 0,9 moles aluminium ions - 2,7 moles chloride ions
There are 0.75 moles of sodium ions present in 2.50 L of 0.300 M Na3PO4. Each formula unit of Na3PO4 has 3 sodium ions. So, for every mole of Na3PO4, there are 3 moles of sodium ions. Calculating the moles of sodium ions: 2.5 L * 0.300 mol/L * 3 mol Na+ / 1 mol Na3PO4 = 0.75 moles of Na+.
When Na2CO3 dissociates, it produces 3 moles of ions: 2 moles of Na+ ions and 1 mole of CO3^2- ions. So, if you have 0.5 moles of Na2CO3, you would produce 1.5 moles of ions in total.
For every one mole of iron-III-chromate, there are three moles of chromate ions present. Therefore, in a sample containing 3.71 moles of iron-III-chromate, there would be 3.71 moles multiplied by 3, which is 11.13 moles of chromate ions.
Since the formula shows two sodium atoms in each formula unit of sodium sulfate and this compound normally completely ionizes in water solution, the number of sodium ions will be twice the number of moles of the salt; in this instance, 1.0 moles of sodium ions.
To find the number of moles of ions in the solution, we first calculate the total number of moles of KCl using the given concentration and volume. Multiply the number of moles of KCl by 2 (since one mole of KCl produces 2 moles of ions, one K+ and one Cl-) to find the total moles of ions present.
2 moles of Ca and 4 moles of OH