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.
The answer is Wetland and Stream
Option (i) would have the highest 2nd Ionization Energy.
Option (i) is Sodium.
Can be Written as 2, 8 , 1
For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.
Now After Removing that Electron...
Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.
Neon has a very stable Octet.
It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.
So
Option (i) [Sodium] has the highest 2nd Ionization Energy
Answer:
c. HF can participate in hydrogen bonding.
Explanation:
<u>The boiling points of substances often reflect the strength of the </u><u>intermolecular forces</u><u> operating among the molecules.</u>
If it takes more energy to separate molecules of HF than of the rest of the hydrogen halides because HF molecules are held together by stronger intermolecular forces, then the boiling point of HF will be higher than that of all the hydrogen halides.
A particularly strong type of intermolecular attraction is called the hydrogen bond, <em>which is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond</em>, such as N-H, O-H, or F-H, and an electronegative O, N, or F atom.
Activity series of metals: K,Na,Mg,Al,Zn,Fe,Cu,Ag. Metals on the left are more reactive than metals on the right. For example Zn is more reactive than Fe and can displace him.
Reaction than can occur is: <span>CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s).</span>
