Complete ionic:
Cu(aq) + 2Cl(aq) + 8O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s) + 2Na(aq) + Cl(aq) + 4O(aq)
Net ionic:
Cu(aq) + Cl(aq) + 4O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s)
So write everything out as IF it will dissociate in water. So everything that is aq splits but solid just floats to the bottom of the mixture. Cancel what you can (in this case the two from the ClO4 on the left of the equation cancels with the ClO4 from the right) and the 2Na cancels. Then, write out the whole solution and you are done!
Answer is: A. 1.1 3 1023 NiCl2 formula units.
m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.
M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.
n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).
n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.
n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.
Na = 6.022·10²³ 1/mol; Avogadro constant.
N(NiCl₂) = n(NiCl₂) · Na.
N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.
N(NiCl₂) = 1.13·10²³; number of formula units.
Answer:
43. 6 sig figs
44.6 sig figs
45. 2 sig figs
Explanation:
I used the Atlantic Pacific rule
P if decimal is present
A if decimal is absent
Pacific Ocean (number) Atlantic Ocean
D. Electrons
Atoms can share their electrons in order to create bonds with other atoms
Answer:
19.264×
atoms are present in 3.2 moles of carbon.
Explanation:
It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.
So, as 1 mole of any element = Avagadro's number of atoms = 6.02× atoms
It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.
As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.
1 mole of carbon = 6.02 × atoms
Then, 3.20 moles of carbon = 3.20 × 6.02 × atoms
Thus, 19.264× atoms are present in 3.2 moles of carbon.
