This is false. An alcohol does indeed have a polar C-O single bond, but what we should really be focusing on is the extraordinarily polar O-H single bond. When oxygen, fluorine, or nitrogen is bound to a hydrogen atom, there is a small (but not negligible) charge separation, where the eletronegative N, O, or F has a partial negative charge, and the H has a partial positive charge. Water has two O-H single bonds in it (structure is H-O-H). The partially negative charge on the O of the water molecule (specifically around the lone pair) can become attracted either a neighboring water molecule's partially positive H atom, or an alcohol's partially positive H atom. This is weak (and partially covalent) attraction is called a hydrogen bond. This is stronger than a typical dipole-dipole attraction (as would be seen between neighboring C-O single bonds), and much stronger than dispersion forces (between any two atoms). When the solvent (water) and the solute (the alcohol) both exhibit similar intermolecular forces (hydrogen bonding being the most important in this case), they can mix completely in all proportions (i.e. they are miscible) in water.
The difference between a mixture and a compound is that a mixture can be easily separated like a salad where you can pick things out and a compounds you are usually not able to undo
Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) = ![\frac{ΔvapH}{R}([tex]\frac{1}{T1}](https://tex.z-dn.net/?f=%5Cfrac%7B%CE%94vapH%7D%7BR%7D%28%5Btex%5D%5Cfrac%7B1%7D%7BT1%7D)
-
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ
vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P
1
=
100 mmHg
; T
1
=
34.7 °C
=
307.07 K
P
2
=
760mmHg
T
2
=T⁻²=?
Δ
vap
H
=
38.6 kJ/mol
R
=
0.008314 kJ⋅K
-1
mol
-1
ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K
1. Some differences are that they are all different kinds of dogs
2. Small dogs and big dogs
That is all I can answer
<span>361.4 pm is the length of the edge of the unit cell.
First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight.
Atomic weight copper = 63.546
Now divide the mass by the atomic weight, getting
8.94 g / 63.546 g/mol = 0.140685488 mol
And multiply by Avogadro's number to get the number of atoms:
0.140685488 * 6.022140857x10^23 = 8.472278233x10^22
Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume.
8.472278233x10^22 / 4 = 2.118069558x10^22
Now calculate the volume each unit cell occupies.
1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3
Let's get the cube root to get the length of an edge.
(4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm
Now let's convert from cm to pm.
3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm
Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
