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alexandr1967 [171]
3 years ago
13

A substance occupies one half of an open container. The atoms of the substance are closely packed but are still able to slide pa

st each other.
What is most likely the phase of the substance?

O: Gas
O: liquid
O: Solid and gas
O: liquid and solid

Pls help ;-;
Chemistry
1 answer:
SCORPION-xisa [38]3 years ago
8 0

For a substance to occupy one half of an open container with the atoms being able to slide past each other, the most likely phase of the substance would be liquid.

Atoms of gases would not occupy just one-half of an open container because atoms of gases will diffuse and spread out from an open container.

Atoms of solids are fixed about a particular position and will most likely not be able to slide past each other in an open container.

Only the atoms of liquid take the shapes of their containers and do not have the capacity to diffuse out ordinarily. They are also able to flow despite the closeness of the atoms and would occasionally slide past each other.

More on states of matter can be found here: brainly.com/question/9402776

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The freezing point depression of a solution containing 30.7 g of glycerin  is  calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

K_{f}or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

             ΔT_{f} = i ×K_{f} ×m

T⁰-T_{s}  = i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water     in     kg}

in which,

ΔT_{f} = change in freezing point

ΔT_{s} = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T_{s} = 1 ×1.86°C/m ×\frac{30.7}{92.09g/mol * 0.376kg}

T_{s} = -1.65°C

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<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

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