Answer:
33.3 kg of air
Explanation:
This is a problem of conversion unit.
Density is mass / volume
Therefore we have to calculate the volume in the room, to be multiply by density. That answer will be the mass of air.
Volume of the room → 9 ft . 11 ft . 10 ft = 990 ft³
Density is in g/L, therefore we have to convert the ft³ to dm³ (1 dm³ = 1L)
990 ft³ . 28.3 dm³ / 1ft³ = 28017 dm³ → 28017 L
This is the volume of the room, if we replace it in the density formula we can know the mass of air in g.
1.19 g/L = Mass of air / 28017 L
Mass of air = 28017 L . 1.19 g/L → 33340 g of air
Finally, let's convert the mass in g to kg → 33340 g . 1kg / 1000 g = 33.3 kg
To calculate percent composition, you first need to find the molar mass of C (carbon), H (hydrogen) and O (oxygen).
C is 12.01
H is 1.00
O is 16
Then multiply each by the number of atoms of each element in the formula (the number that comes after each element in the equation for example C6 means 6 carbon atoms.
C: 12.01 x 6= 72.06
H: 1x12= 12
O: 16x6= 96
Then add them up.
72.06+ 12+ 96= 180.06
Now find the percent composition of carbon.
72.06/ 180.06 x 100= 40.01%
So the answer is C 40%.
Conduction is heat tranfer through physical contact. Hope this helps. :)
Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
