1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.
Let's consider the reaction for the combustion of Mg.
Mg + 1/2 O₂ ⇒ MgO
1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:
We can calculate the mass percent of O in MgO using the following expression.
You can learn more about mass percent here: brainly.com/question/14990953
Answer : The correct option is, (D) 3600 kJ
Explanation :
Mass of octane = 75 g
Molar mass of octane = 114.23 g/mole
Enthalpy of combustion = -5500 kJ/mol
First we have to calculate the moles of octane.
Now we have to calculate the heat released in the reaction.
As, 1 mole of octane released heat = -5500 kJ
So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)
= -3608 kJ
≈ -3600 kJ
Therefore, the heat released in the reaction is 3600 kJ
6 sodium and 6 Bromine in 6NaBr
Hello!
We know that by the Law of Avogrado, for each mole of substance we have 6.02 * 10²³ atoms, if:
The molar mass of water (H2O)
H = 2 * (1u) = 2u
O = 1 * (16u) = 16u
---------------------------
The molar mass of H2O = 2 + 16 = 18 g / mol
If:
1 mol we have 6.02 * 10²³ atoms
1 mole of H2O we have 18 g
Then we have:
18 g ------------- 6.02 * 10²³ atoms
5 g -------------- x
I Hope this helps, greetings ... DexteR! =)
