Answer:
50 g of K₂CO₃ are needed
Explanation:
How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?
We analyse data:
500 g is the mass of the solution we want
10% m/m is a sort of concentration, in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution
Therefore we can solve this, by a rule of three:
In 100 g of solution we have 10 g of K₂CO₃
In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃
Format Method - Writing the symbol of the cation and then the anion. Add whatever subscripts in order to balance the charges.
Crisscross Method - The numerical value of the charge of each ion is crossed over and becomes the subscripts for the other ion.
The Avogadro number represents the number of units in one mole of a chemical substance.
So to find the mole number of a chemical element, you divide its atom number of the Avogadro number which Na = 6.02*10^23 approx.
So n=N/Na (n=mole number, N=number of atoms, Na=Avogadro number)
n=1.0*10^15/6.02*10^23
n=1.6 * 10^-9
So 1.0*10^15 atoms of Sodium represent 1.6*10^-9 mol.
Hope this Helps! :)
Answer : There are mainly three isotopes of magnesium found in nature; namely Mg-24, Mg-25 and Mg-26. Out of which Mg-24 has 12 neutrons, Mg-25 has 13 neutrons and Mg-26 has 14 neutrons in their atoms. The number of protons and their atomic masses remains the same for the atom. The relative abundance in nature differs for all the three isotopes. Mg-24 has abundance nearly 80% in nature, Mg-25 has abundance as 10% and Mg-26 has 11.01% abundance.
