Answer: The equation showing conservation of atoms is
Explanation:
Every balanced chemical equation follows law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.
For the given chemical equations:
The given chemical equation follows:
There are 2 hydrogen atoms and 2 oxygen atoms on the reactant side and 2 hydrogen and 1 oxygen atom on product side.
So, this equation does not show conservation of atoms.
The given chemical equation follows:
There are 2 hydrogen atoms and 2 oxygen atoms on the reactant side and 4 hydrogen atoms and 2 oxygen atoms on product side.
So, this equation does not show conservation of atoms.
The given chemical equation follows:
There are 4 hydrogen atoms and 2 oxygen atoms on the reactant side and 4 hydrogen atoms and 2 oxygen atoms on product side.
So, this equation shows conservation of atoms.
The given chemical equation follows:
There are 4 hydrogen atoms and 4 oxygen atoms on the reactant side and 4 hydrogen atoms and 2 oxygen atoms on product side.
So, this equation does not show conservation of atoms.
Hence, the equation showing conservation of atoms is
(B) lattice energy
(C) kinetic energy
(D) activation energy
(E) ionization energy
Cl₃ is formed by thecombination of 3 chlorine atoms. Hence, not a single element and not a compound. Instead Cl₃ is a molecule of chlorine.
Molecules are the combination of atoms of same element or are the basic units of a compound. A compound is formed by the combination of two or more atoms of different elements.
The combination of atoms can be different bond types, such as ionic bond, covalent bond, hydrogen bonding etc. The type of bond is dependent on the nature of atoms.
Combination of same atoms in any number are forming the molecules of that element and not a compound. Hence, Cl₃ is a molecules and not element or compound.
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Answer:
CI3 is a compound called carbon tetra iodide
Answer: 181 kJ
Explanation:
The balanced chemical reaction is;
To calculate the moles, we use the equation:
According to stoichiometry:
1 mole of on complete combustion give= 555 kJ
Thus 0.326 moles of on complete combustion give=
Thus the enthalpy change for combustion of 15.0 g of ethanol is 181 kJ
b. more easily compressed.
c. less organized.
d. more likely to flow.
The balanced chemical equation for the combustion of octane is 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O. From the stoichiometry of the reaction, 3.50 mol of CO2 will need approximately 50g of octane.
The combustion of octane (C8H18) in the presence of oxygen (O2) produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O.
To calculate the mass of octane needed to release 3.50 mol CO2, we need to understand the stoichiometry of the reaction. From the balanced equation, we can see that 16 mol of CO2 is produced from 2 mol of C8H18. So, 1 mol of C8H18 produces 8 mol of CO2.
Therefore, to produce 3.50 mol of CO2, we would need 3.50/8 = 0.4375 mol of C8H18. The molar mass of octane is approximately 114 g/mol, so the required mass would be 0.4375mol x 114g/mol = approximately 50g.
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The combustion of octane produces carbon dioxide and water, as described by the balanced chemical equation C8H18 + 12.5O2 → 8CO2 + 9H2O. To produce 3.50 mol of CO2, approximately 50g of octane is needed.
The combustion of octane (C8H18) in the presence of oxygen (O2) produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is: C8H18 + 12.5O2 → 8CO2 + 9H2O.
Considering the stoichiometry of the reaction, we can see that 1 mol of octane produces 8 moles of CO2. Therefore, to produce 3.50 mol of CO2, the amount of octane required would be 3.50/8 = 0.4375 mol. Converting this to mass using the molar mass of octane (114.22 g/mol), we get 0.4375 mol * 114.22 g/mol ≈ 50 g. Thus, approximately 50g of octane is needed to produce 3.50 mol of CO2.
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