First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
Answer: The ratio of carbon dioxide molecules to oxygen molecules is 20 :31
Explanation:
Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The balanced combustion reaction is,:
Thus the ratio of carbon dioxide molecules to oxygen molecules is 20 : 31
Answer:
13.94moles of Na₂O
Explanation:
The balanced reaction expression is given as:
4Na + O₂ → 2Na₂O
Given parameters:
Number of moles of O₂ = 6.97moles
Unknown:
Number of moles of Na₂O
Solution:
To solve this problem;
1 mole of O₂ will produce 2 moles of Na₂O ;
6.97 moles of O₂ will produce 6.97 x 2 = 13.94moles of Na₂O
The symbol, the atomic mass, the number of protons and electrons
Answer:
The correct answer is 160.37 KJ/mol.
Explanation:
To find the activation energy in the given case, there is a need to use the Arrhenius equation, which is,
k = Ae^-Ea/RT
k1 = Ae^-Ea/RT1 and k2 = Ae^-Ea/RT2
k2/k1 = e^-Ea/R (1/T2-1/T1)
ln(k2/k1) = Ea/R (1/T1-1/T2)
The values of rate constant k1 and k2 are 3.61 * 10^-15 s^-1 and 8.66 * 10^-7 s^-1.
The temperatures T1 and T2 are 298 K and 425 K respectively.
Now by filling the values we get:
ln (8.66*10^-7/3.61*10^-15) = Ea/R (1/298-1/425)
19.29 = Ea/R * 0.001
Ea = 160.37 KJ/mol