Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
Answer:
A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:
C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol
The value given for ΔH∘rxnΔHrxn∘ means that:
a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.
b. the reaction is endothermic.
c. the enthalpy of formation of propane is 2202 kJ/mol.
d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.
e. None of these.
Answer:
Newton 3rd Law of Motion or the Law of Force Pairs
(An applied force)
Atoms of different elements can be identical
Answer:
a) 1,6%
b) 64,775 g/mol
c) 3,6×10⁻² M
d) 2,3×10⁻³ g/mL
Explanation:
a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:
- Helium: 0,25 moles ×
= 1 g of Helium - Argon: 0,25 moles ×
= 10 g of Argon - Krypton: 0,25 moles ×
= 20,95 g of krypton - Xenon: 0,25 moles ×
= 32,825 g of Xenon
Total grams: 1g+10g+20,85g+30,825g= 62,675 g
Mass fraction of helium: 
× 100 = <em>1,6%</em>
<em />
<em>The mass fraction of Helium is 1,6%</em>
<em />
<em>b)</em><em> </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:
= 64,775 g/mol
c) The molar concentration is possible to know ussing ideal gas law, thus:
= M
Where:
P is pressure: 150 kPa
R is gas constant: 8,3145
T is temperature: 500 K
And M is molar concentration. Replacing:
M = 3,6×10⁻² M
d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:
3,6×10⁻² 
× 
× 
=
2,3×10⁻³ g/mL
I hope it helps!
