Answer:
Explanation:
Principal quantum no "n" = 3
Azimuthal quantum no "l"= 1
Magnetic quantum no "m"= +1/2
Over all is 3pz
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
Answer:
39.7 %
Explanation:
magnesium + oxygen ⟶ magnesium oxide
10.57 g 6.96 g 17.53 g
According to the <em>Law of Conservation of Mass</em>, the mass of the product must equal the total mass of the reactants.
Mass of MgO = 10.57 + 6.96
Mass of MgO = 17.53 g
The formula for mass percent is
% by mass = Mass of component/Total mass × 100 %
In this case,
% O = mass of O/mass of MgO × 100 %
Mass of O = 6.96 g
Mass of MgO = 17.53 g
% O = 6.96/17.53 × 100
% O = 0.3970 × 100
% O = 39.7 %
The number of half -lives that has passed after 105 hours for krypton-79 that has half-life of 35 hours is calculated as below
if 1 half life = 35 hours
what about 105 hours = ? half-lives
= (1 half life x105 hours) /35 hours = 3 half-lives has passed after 105 hours
Answer:
Explanation:
-log(3.5 * 10^-11)
= 10.4559
Be careful how you put this into your calculator. I had to use Exp to get it to work properly.
-log
(3.5 * 10 exp -11)
=