4. How many grams of ethylene glycol (C2H6O2) must be added to 1.00kg of water to produce a solution that frees at -50c?

To produce a solution that frees at -50°C we need 1668.5 g of ethylene glycol

Explanation:

Formula for freezing point depression is:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m . i

Our solute is the ethylene glycol, then water is our solvent.

As ethylene glycol is a non ionic compound, i = 1 (Van't Hoff factor, numbers of ions dissolved)

Kf = Cryoscopic constant. For water is 1.86 °C/m

We replace data:

0°C - (-50°C) = 1.86 °C/m . m . 1

To determine grams of ethylene glycol needed, we need to find m (molality)

50°C / 1.86 m/°C = m → 26.8 mol/kg

As this moles of solute are contained in 1kg, definetely we need 26.8 moles of ethylene glycol.

To find the answer, we convert moles to mass:

26.8 mol . 62.07g / mol = 1668.5 g

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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