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%
Answer:
Explanation:The pi-molecular orbitals in propene (CH3-CH=CH2) are essentially the ... This central carbon thus provides two p-orbitals – one for each pi bond – and these two different p-orbitals have to be perpendicular, leading to a twisted structure as shown: ... It all comes down to where the location of the electron-deficient carbon
The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
A pi bond is a bond formed by the overlap of orbitals in a side-by-side fashion with the electron density concentrated above and below the plane of the nuclei of the bonding atoms. ... Three sigma bonds are formed from each carbon atom for a total of six sigma bonds total in the molecule.
