From other sources, the given mass of the solute that is being dissolved here is 7.15 g Na2CO3 - 10H2O. We use this amount to convert it to moles of Na2CO3 by converting it to moles using the molar mass then relating the ratio of the unhydrated salt with the number of water molecules. And by the dissociation of the unhydrated salt in the solution, we can calculate the moles of Na+ ions that are present in the solution.
Na2CO3 = 2Na+ + CO3^2-
7.15 g Na2CO3 - 10H2O (1 mol / 402.9319 g) (1 mol Na2CO3 / 1 mol Na2CO3 - 10H2O) ( 1 mol Na2CO3 / 1 mol Na2CO3-10H2O ) ( 2 mol Na+ / 1 mol Na2CO3) = 0.04 mol Na+ ions present
The empirical formula of a compound is the simplest ratio of components making up the compound.
In 100 g of compound,there's <span>66.6 g of C, 11.2 g of H and 22.2 g of O
lets calculate for 100 g of compound
C H O
mass </span> 66.6 g 11.2 g 22.2 g
number of moles 66.6/12 g/mol 11.2/1 g/mol 22.2/ 16 g/mol
= 5.55 mol =11.2 mol =1.3875 mol
divide by the least number of moles
5.55/1.3875 11.2/1.3875 1.3875/1.3875
= 4 = 8.08 = 1
round them off to the nearest whole number
C - 4
H - 8
O - 1
Therefore empirical formula of compound is C₄H₈O
The element is Am and since you lose e- there must be a postive charge. Am+6 is the symbol
Answer:
Average atomic mass = 63.553 amu.
Explanation:
Given data:
Abundance of Y-63 = 69.17%
Abundance of Y-65 = 100 - 69.17 = 30.83%
Atomic mass of Y-63 = 62.940 amu
Atomic mass of Y-65 = 64.928 amu
Atomic mass of Y = ?
Solution:
Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass= (62.940×69.17)+(64.928×30.83) /100
Average atomic mass = 4353.560 + 2001.730 / 100
Average atomic mass = 6355.29 / 100
Average atomic mass = 63.553 amu.
