2C6H14 + 13O2 ---> 6CO2 +14H2O
M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol
86.17 g C6H14 is 1 mole.
2C6H14 + 13O2 ---> 6CO2 +14H2O
from reaction 2 mol 6 mol
from the problem 1 mol 3 mol
M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Answer : 132.0 g CO2
a.
Acids react with bases and give salt and water and the products.
Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
To balance the reaction equation, both sides hould have same number of elements.
Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1
Hence, the reaction equation is already balanced.
b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
= 5.681 x 10⁻³ mol
Stoichiometric ratio between NaOH and HCl is 1 : 1
Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol
5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.
Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M
The number of molecules in 16 grams of oxygen gas is 3.01*10^23, equal to half of Avogadro's number. ... How many molecules of oxygen (o2) are present in 16.0 g of o2…
Explanation:
According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.
That is, 
Hence, it is given that 
is 3.50 liters, 
is 20 degree celsius, and 
is 100 degree celsius.
Therefore, calculate 
as follows.
= 17.5 liter
Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.
Answer:
a) if the liquid is not vaporized completely, then the condensed vapor in the flask contains the air which is initially occupied before the liquid is heated. When calculating the molar mass of the vapor the moles of air which are initially present are not excluded, so that the molar mass of the vapor would be an increase in value.
b) While weighing the condensed vapor, the flask should be dried. If the weighing flask is not dried then the water which is layered on the surface of the flask is also added to the mass of the vapor. Therefore, the mass of the vapor that is calculated would be increase.
c) When condensing the vapor, the stopper should not be removed from the flask, because the vapor will escape from the flask and a small amount of vapor will condense in the flask. Therefore, the mass of the condensed vapor would be In small value.
d) If all the liquid is vaporized, when the flask is removed before the vapor had reached the temperature of boiling water, then the boiling
temperature of that liquid would be lower than that of the boiling temperature of the water.Therefore, the liquid may have more volatility.
