Answer:
14.0067 g/mol
Explanation:
The molar mass of pure nitrogen is 14.0067 because when you look at the top right of an element on the periodic table of elements you can see the molar mass of an element. This can go for any other elements on the table, just look at the top right of the box. The molar mass of an element is also called the atomic weight. Hope this helps! :)
Answer:
The answer is option 3.
Explanation:
Option 3 shows a balanced equation.
Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
Answer:
4Cr + 3O2 —> 2Cr2O3
Explanation:
Information from the question include:
Chromium + oxygen -> chromium(III) oxide
From the word equation given above, the equation can be written as follow:
Cr + O2 —> Cr2O3
The equation can be balance by doing the following:
There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of Cr2O3 and 3 in front of O2 as shown below:
Cr + 3O2 —> 2Cr2O3
Now, we have 4 atoms of Cr on the right side and 1 atom on the left. It can be balance by putting 4 in front of Cr as shown below:
4Cr + 3O2 —> 2Cr2O3
Now the equation is balanced
