Answer:
680 g/m is the molar mass for the unknown, non electrolyte, compound.
Explanation:
Let's apply the formula for osmotic pressure
π = Molarity . R . T
T = T° absolute (in K)
R = Universal constant gases
π = Pressure
Molarity = mol/L
As units of R are L.atm/mol.K, we have to convert the mmHg to atm
760 mmHg is 1 atm
28.1 mmHg is (28.1 .1)/760 = 0.0369 atm
0.0369 atm = M . 0.082 L.atm/mol.K . 293K
(0.0369 atm / 0.082 mol.K/L.atm . 293K) = M
0.0015 mol/L = Molarity
This data means the mol of solute in 1L, but we have 100mL so
Molarity . volume = mol
0.0015 mol/L . 0.1L = 1.5x10⁻⁴ mole
The molar mass will be: 0.102g / 1.5x10⁻⁴ m = 680 g/m
Answer: arrangement of constituent particles makes glass diff. from quartz. Glass-short range order of particles.
Quartz-Long range of particles.
If quartz is heated and cooled rapidly it can be converted to glass.
For i: 33mL
For ii: 87-88mL
For iii:22.3mL
Because K and Cl have such a large disparity in their electronegativities, KCl is a bipolar ionic molecule.
<h3>What exactly are polar and nonpolar bonds?</h3>
Polar covalent bonds develop when the distribution of electrons among atoms is uneven, whereas nonpolar side chains develop when the distribution of electrons is more even. The reason for the unequal sharing of electrons is because the atoms receiving them have various electronegativities.
<h3>How are polar bonds created?</h3>
Whenever a single pair of electrons is not shared equally, a polar molecule bond is created. This is caused by the electronegativity difference between the two elements. An unit of h as well as an unit of bromine share a pair of electrons, but not evenly.
To know more about polar bond visit:
brainly.com/question/10777799
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<h2>Hey there!</h2>
<h3>The correct option is (A) It has a partial negative charge on oxygen and a partial positive charge on hydrogen.</h3>
<h3>☆ Explanation:</h3>
¤ As water has the ability to form hydrogen bonds which makes it an excellent solvent.
¤ For this ability of water it can dissolve many different kinds of molecules.
<h2>Hope it helps </h2>