The final step in the scientific method is the conclusion. The conclusion will either clearly support the hypothesis or it will not. If the results support the hypothesis a conclusion can be written
Answer:
Option B.
Explanation:
As any reaction of combustion, the O₂ is a reactant and the products are CO₂ and H₂O. Combustion reaction for ethane is:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.
Then 2 moles of ethane will produce 4 moles of CO₂
Answer:
Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}
Explanation:
First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:
NO(g) + H₂O(g) → NH₃(g) + O₂(g)
To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):
4 NO(g) + 6 H₂O(g) → 4 NH₃(g) + 5 O₂(g)
All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:

The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
Answer:
1. London dispersion
Explanation:
Sulphur trioxide ( SO₃ ) -
The chemical compound SO₃ is planar in structure , the only intermolecular forces shown by SO₃ is the London forces .
dipole - dipole is not observed in this compound , as it is not possible to generate poles between the sulfur and oxygen atom due to very less difference in the electronegativity .
Hydrogen bonding is also not observed , because there is not hydrogen atom .
Hence , only London forces are observed in SO₃ .