Answer:
- 0.1852
- 0.0947
- 0.7201
- 3.0345 kg CO
/ Kg C
H
- 15.3848 Kg air / kg C H
Explanation:
Molar masses of each product are :
Butane = 58 kg /kmol
Oxygen = 32 kg/kmol
Nitrogen = 28 kg/kmol
water = 18 kg/kmol
<u><em>1) Calculate the mass fraction of carbon dioxide </em></u>
= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )
= 176 / 950.32
= 0.1852
<em><u>2) Calculate the mass fraction of water </u></em>
= ( 5 * 18 ) / (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))
= 90 / 950.32
= 0.0947
<em><u>3) Calculate the mass fraction of Nitrogen </u></em>
= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))
= 684.32 / 950.32
= 0.7201
<em><u>4) Calculate the mass of Carbon dioxide in the products</u></em>
Mco2 = ( 4 * 44 ) / 58 = 3.0345 kg CO / Kg C H
<u>5) Mass of Air required per unit of fuel mass burned </u>
Mair = ( 6.5 * 32 + 24.44 *28 ) / 58 = 15.3848 Kg air / kg C H
Answer:
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Explanation:
<u>Step 1:</u> Data given
ΔH∘=20.1 kJ/mol
ΔS is 45.9 J/K
<u>Step 2:</u> When is the reaction spontaneous
Consider temperature and pressure = constant.
The conditions for spontaneous reactions are:
ΔH <0
ΔS > 0
ΔG <0 The reaction is spontaneous at all temperatures
ΔH <0
ΔS <0
ΔG <0 The reaction is spontaneous at low temperatures ( ΔH - T*ΔS <0)
ΔH >0
ΔS >0
ΔG <0 The reaction is spontaneous at high temperatures ( ΔH - T*ΔS <0)
<u>Step 3:</u> Calculate the temperature
ΔG <0 = ΔH - T*ΔS
T*ΔS > ΔH
T > ΔH/ΔS
In this situation:
T > (20100 J)/(45.9 J/K)
T > 437.9 K
T > 164.75 °C
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
0.042 moles of Hydrogen evolved
<h3>Further explanation</h3>
Given
I = 1.5 A
t = 1.5 hr = 5400 s
Required
Number of Hydrogen evolved
Solution
Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.
Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)
Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻
Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)
So at the cathode H₂ gas is produced
Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C
Q = i.t
Q = 1.5 x 5400
Q = 8100 C
mol e⁻ = 8100 : 96500 = 0.084
From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042
Answer:
the difference in electronegativity is so large (2.04) that the bonding electrons spend almost all their time on the nitrogen atom.
Explanation:
Because calcium loses 2 electrons to become Ca2+, and nitrogen gains 3 electrons to become N3−, you need two calcium atoms and three nitrogen atoms in order to form a neutral compound.
Answer:
The statement is false. See the explanation below, please.
Explanation:
The hydrogen bond or bridge is a type of dipole-dipole interaction that is generated from the attraction of a hydrogen atom and a very electronegative atom (oxygen, fluorine or nitrogen). Examples of hydrogen bridge molecules: Water (H20), ammonia (NH3).
