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almond37 [142]
3 years ago
5

A 28.5 gram piece of iron is added to a graduated cylinder containing 45.5 mL of water. The water in the cylinder rises to the 4

9.1 mark. Calculate the density of the iron piece.
Chemistry
1 answer:
yan [13]3 years ago
4 0

7.91 g/ml is the density of the iron piece of 28.5 gms.

Explanation:

The density of a substance is defined as the volume it occupies.  It tells the matter present in a substance.

The density is mass per unit volume and is denoted by p.

The formula for density is given by:

density (p) = \frac{mass}{volume}

Data given is :

mass= 28.5 grams

V1 = 45.5 ml

V2= 49.1 ml

The initial volume of water was 45.5 ml, when iron piece of 28.5 grams was added the final volume was 49.1 ml.

Putting the values in the equation of density

p = \frac{28.5}{49.1-45.5}

p = 7.91 g/ml

Since iron is a dense material it will occupy less volume

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What are the boiling points and freezing points (in oC) of a solution of 50.3 g of I2 in 350 g of chloroform? The kb = 3.63 oC/m
patriot [66]

Answer:

Boiling point: 63.3°C

Freezing point: -66.2°C.

Explanation:

The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:

<em>Boiling point increasing:</em>

ΔT = kb*m*i

<em>Freezing point depression:</em>

ΔT = kf*m*i

ΔT are the °C that change boiling or freezing point.

m is molality of the solution (moles / kg)

And i is Van't Hoff factor (1 for I₂ in chloroform)

Molality of 50.3g of I₂ in 350g of chloroform is:

50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:

0.198 moles / 0.350kg = 0.566m

Replacing:

<em>Boiling point:</em>

ΔT = kb*m*i

ΔT = 3.63°C/m*0.566m*1

ΔT = 2.1°C

As boiling point of pure substance is 61.2°C, boiling point of the solution is:

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<em>Freezing point:</em>

ΔT = kf*m*i

ΔT = 4.70°C/m*0.566m*1

ΔT = 2.7°C

As freezing point is -63.5°C, the freezing point of the solution is:

-63.5°C - 2.7°C = -66.2°C

7 0
3 years ago
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