Answer:
Given: 42 g of N2
Solve for O2 mass that contains the same number of molecules to 42 g of N2.
Solve for the number of moles in 42 g of N2
1 mole of N2 = (14 * 2) g = 28 g so the number of moles in 42 g of N2 is equal to 42 g / 28 g per mole = 1.5 moles
Solve for mass of 1 mole of oxygen
1 mole of O2 = 16 g * 2 = 32 g per mole
Solve for the mass of 1.5 moles of oxygen
mass of 1.5 moles of O2 = 32 g per mole * 1.5 moles
mass of 1.5 moles of O2 = 48 g
So 48 g of O2 contains the same number of molecules as 42 g of N2
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
Answer:
An example of a phase is the final part of a multiple part landscaping project. An example of a phase is when you can the various shapes of the moon. An example of phase is a period of time during which you are a teenager when you fight with your parents all the time.
Explanation:
Answer:
0.1832 moles of ethyl acetate (
)
Explanation:
1. Find the balanced chemical equation:
In the production of ethyl acetate, the acetic acid 
reacts with ethanol to produce ethyl acetate and water, that is:
2. Find the theoretical maximum moles of ethyl acetate :
As the problem says that the acetic acid is the limiting reagent, use stoichiometry to find the moles of ethyl acetate produced:
