Answer:
Mass percent N₂ = 89%
Mass percent H₂ = 11%
Explanation:
First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:
- 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
So now we know that
- MolH₂ + MolN₂ = 0.307 mol
and
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g
So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:
Express MolH₂ in terms of MolN₂:
- MolH₂ + MolN₂ = 0.307 mol
Replace that value in the second equation:
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
- (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
- 0.614 - 2MolN₂ + 28molN₂ = 3.49
Now we calculate MolH₂:
- MolH₂ + MolN₂ = 0.307 mol
Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:
- N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
- H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂
Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%
Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%
False, covalent bonds result when atoms share electrons.
Answer:
It is true answer. Ozone = O3
Answer : The final equilibrium temperature of the water and iron is, 537.12 K
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of iron = 560 J/(kg.K)
= specific heat of water = 4186 J/(kg.K)
= mass of iron = 825 g
= mass of water = 40 g
= final temperature of water and iron = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the final equilibrium temperature of the water and iron is, 537.12 K
Answer:
Its basically a series of events that take place in a cell. A cell spends most of its time in what is called inter phase, and during this time it grows, replicates its chromosomes, and prepares for cell division. Then after that The cell then leaves interphase, undergoes mitosis, and completes its division.
Explanation:
-Hailey: )