The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of </span>C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂
the answer is A......
it is supported by practical evidence and examples. this is the answer because he tried and tested many different ways to see what would happen so he is happy with the conclusion that what he tested is what he gets.
125 mile *1gallon/35 mi = 135/35 = (27/7) gallon gasoline
27/7 gallon * 1 L/0.264 gallon = 14.6 L gasoline
14.6 L gasoline * 2.5kg CO2/1L gasoline= 36.5 kg CO2
36.5 kg CO2 * 1lb/0.454 kg = 80.4 lb
Answer: 80.4 lb CO2
The fluorine atom has a neutron number of 10 it also have 10 valence electrons
Answer:
Explanation:
Given : Density - 2.41 g/liter
Temperature - 25° C
Pressure : 770 mm Hg
R = 0.0821 L atm mol-¹K-¹
Find : Molecular mass of gas
Solution : Ideal gas equation with respect to density will be : PM = dRT. In the formula, P is pressure, M is molecular mass, d is density, R is gas constant and T is temperature.
Keeping the values in equation-
Pressure : 770 mm Hg = 1 atm
Temperature : 273 + 25 = 298 K
M = dRT/P
M = (2.41*0.0821*298)/1
M = 58.96 gram/mol
Thus, the molecular mass of gas is 58.96 gram/mol.
