In sodium there are 11 electrons
In chlorine there are 17 electrons
There are 17 protons in chlorine
There are 11 protons in sodium
I’m not sure what an octet is I study in a British system if there’s anything else you’re not sure of search the Internet it has all the answers
Answer:
The atomic mass of the boron atom would be <em>10.135</em>
Explanation:
This is generally known as relative atomic mass.
Relative atomic mass or atomic weight is a physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass of 1/12 of the mass of a carbon-12 atom. Since both quantities in the ratio are masses, the resulting value is dimensionless; hence the value is said to be relative and does not have a unit.
<em>Note that the relative atomic mass of atoms is not always a whole number because of it being isotopic in nature.</em>
- <em>Divide each abundance by 100 then multiply by atomic mass</em>
- <em>Do that for each isotope, then add the two result. Thus</em>
Relative atomic mass of Boron = (18.5/100 x 11) + (81/100 x 10)
= 2.035 + 8.1
= 10.135
The answer would be B the transfer of electrons from one atom to another
Answer:
Explanation:
If the choices are:
A. Molarity is defined as moles of solute per liter of solvent.
B. % by mass is defined as grams of solute per 100 g of solvent.
C. % by volume is defined as grams of solute per 100 L of solution.
D. Molarity is defined as moles of solute per liter of solution.
E. All of the above.
then the ans is E
Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
