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%
Explanation:
A synthesis reaction: It is defined as a kind of reaction where one and more than one reactant attached and creates an individual product.
The formation of the water is an example of a synthesis reaction because here more than one reactants combine and create a single product (water). Water formation occurs when 2 hydrogens and an oxygen share electrons through covalent bonds.
2H + O ----> H2O.
Answer:
6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.
Explanation:
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given:
Concentration is decreased to 1.56 % which means that 0.0156 of
is decomposed. So,
= 0.0156
Thus,
kt = 4.1604
The expression for the half life is:-
Half life = 15.0 hours
Where, k is rate constant
So,

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>
Explanation:
The shapes and relative energies of the orbitals s,p,d and f orbitals are given by the principal quantum number and the azimuthal quantum number.
The principal quantum number gives the main energy level and the azimuthal quantum number denotes the shape of the orbitals.
- For the principal quantum number, they represent the energy levels in which the orbital is located or the average distance of the orbital from the nucleus. It takes the number n = 1,2,3,4,5,6,7......
- The azimuthal quantum number(L) shows the shape of the orbitals in subshells accommodating electrons. The number of possible shapes is limited by the the principal quantum number.
L Name of orbital shape of orbital
0 s spherical
1 p dumb-bell
2 d double dumb-bell
3 f complex
Principal Azimuthal Orbital
Quantum Quantum Designation of
Number (N) Number(l) Sublevel
1 0 1s
2 0 2s
1 2p
3 0 3s
1 3p
2 3d
4 0 4s
1 4p
2 4d
3 4f
Learn more:
Atomic orbitals brainly.com/question/9514863
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