Answer:
B = - 0.0326 dm³/mol
Explanation:
virial eq until second term:
∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm
∴ T = 200°C = 473 K
∴ Vm = 3.90 dm³/mol
∴ R = 0.08206 dm³.atm/K.mol
⇒ PVm / RT = 1 + B/Vm
⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm
⇒ 0.99164 = 1 + B/Vm
⇒ B/Vm = - 8.357 E-3
⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )
⇒ B = - 0.0326 dm³/mol
Carbons starting from the left end:
- sp²
- sp²
- sp²
- sp
- sp
Refer to the sketch attached.
<h3>Explanation</h3>
The hybridization of a carbon atom depends on the number of electron domains that it has.
Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.
Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.
- A carbon atom with four electron domains is sp³ hybridized;
- A carbon atom with three electron domains is sp² hybridized;
- A carbon atom with two electron domains is sp hybridized.
Starting from the left end (H₂C=CH-) of the molecule:
- The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
- The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
- The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
- The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
- The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.
Answer:
True
Explanation:
The molecule CH20 contains two single bonds and one double bond.
Answer:
150
Explanation:
- C₄H₂OH + 6O2 → 4CO2 + 5H₂O
We can <u>find the equivalent number of O₂ molecules for 100 molecules of CO₂</u> using a <em>conversion factor containing the stoichiometric coefficients of the balanced reaction</em>, as follows:
- 100 molecules CO₂ *
= 150 molecules O₂
150 molecules of O₂ would produce 100 molecules of CO₂.
Answer:
The edge length of a face-centered cubic unit cell is 435.6 pm.
Explanation:
In a face-centered cubic unit cell, each of the eight corners is occupied by one atom and each of the six faces is occupied by a single atom.
Hence, the number of atoms in an FCC unit cell is:

In a face-centered cubic unit cell, to find the edge length we need to use Pythagorean Theorem:
(1)
Where:
a: is the edge length
R: is the radius of each atom = 154 pm
By solving equation (1) for "a" we have:
Therefore, the edge length of a face-centered cubic unit cell is 435.6 pm.
I hope it helps you!