Chemistry, 21.05.2021 23:40 cassi35

Question 9 (7 points) Question 3: Solids (7 points)
Identify the type of solid from the description. Select from the following choices: metallic solid, network solid,
molecular solid, ionic solid, or amorphous solid.
Column A
Column B
1.
a. metallic
2.
b
amorphous
3.
c. network
4.
d. solid
a. This solid is very hard and has a very high
melting point. (1 point)
b. This solid conducts heat well and has a range
of hardness. (1 point)
c. This solid is hard and depends on charge
attractions to hold it together. (1 point)
d. This solid does not have a regular crystal
structure. (1 point)
e. This solid has a very low melting point and is
soft. (1 point)
f. This solid has high melting point does not
conduct electricity, and does not dissolve in
water. (1 point)
g. This solid has a high melting point and
conducts electricity only when dissolved. (1
point)
e
5.
molecular
f. ionic
6.
7.
h

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air

in a tire you fill a tire with air

then a couple days later you check the air pressure in the tire

hope this mark me brainliest

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Adialysis unit requires 78.5 l of water.the estimated digit is 5.

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Um i think answer is a

Answer from: Quest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

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Question 9 (7 points) Question 3: Solids (7 points)
Identify the type of solid from the descr...

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