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3 years ago

1. Classify each of the following as a homogeneous or heterogeneous substance.

Chemistry

2 answers:

Olegator [25]3 years ago

7 0

a. iron ore - heterogeneous substance

(it contains a mixture of iron oxides)

b. quartz - homogeneous substance

(it is the crystalline silicon dioxide SiO₂, which is a very pure form of SiO₂)

c. granite - heterogeneous substance

(it as mixture of different oxides, mainly aluminium and silicon oxides)

d. energy drink - heterogeneous substance

(a mixture of water, sugar, carbon dioxide and different substances that may boost concentration, like caffeine)

e. oil-and-vinegar salad dressing - heterogeneous substance

(this salad dressing of oil and vinegar is composed of unsaturated triglycerides, acetic acid and water)

f. salt - homogeneous substance

(sodium chloride NaCl)

g. rainwater - homogeneous substance

(assuming the the atmosphere is clean, it is just pure water H₂O)

h. nitrogen - homogeneous substance

(it is the main component of air and it is a diatomic molecule N₂)

Rashid [163]3 years ago

4 0

Answer:

Explanation:

A homogenous substance is one that has a uniform composition in all its parts.

A heterogenous substance has different composition in their parts.

a. iron ore

Iron ore is a heterogenous substance in which the composition of its part are not uniform.

b. quartz

Quartz is a homogenous substance because it is made up of silica alone joined in a framework pattern.

c. granite

Granite is heterogenous because it is an aggregate of several minerals like quartz, feldspars e.t.c

d. energy drink

Energy drink is homogenous due to the uniformity of the mixture.

e. oil-and-vinegar salad dressing

This is a heterogenous substance

f. salt

Salt is a homogenous substance because all its parts are uniform

g. rainwater

Rainwater is homogeous substance

h. nitrogen

Nitrogen is homogeous substance

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2 years ago

1) 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution. 2) 734 grams of lithium sulfate are dissolved to m

OLEGan [10]

Answer:

1)Molarity of 0.5 Moles of Sodium Chloride in 0.05 Liters of solution is 10 M.

2)Molarity of 734 grams of lithium sulfate are dissolved to make 2500 mL of solution is 2.68 M.

3)Molarity of a solution is made by adding 83 grams of sodium hydroxide to 750 mL of water is 2.77 M.

4)Molarity of a solution that contains .500 mol HC₂H₃O₂ in 0.125 kg H₂O is 4 M.

5)Molarity of a solution that contains 63.0 g HNO₃ in 0.500 kg H₂O is 2 M.

6)The mass of water required to form 3.0 M solution by dissolving 0.5kg of C₂H₅OH is 3.62 kg.

Explanation:

Molarity is given as

$M=\dfrac{n}{V}$

Here

n is number of moles
V is the volume of the solution in liters

Using this all the values are calculated as follows:

1)Molarity of 0.5 Moles of Sodium Chloride in 0.05 Liters of solution is

$M=\dfrac{n}{V}\\M=\dfrac{0.5}{0.05}\\M=10$

So the molarity is 10 M.

2)Molarity of 734 grams of lithium sulfate are dissolved to make 2500 mL of solution is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of lithium sulphate is as follows:

$MM of Li_2SO_4=2\times Li+S+4\times O\\MM of Li_2SO_4=2\times 7+32+4\times 16\\MM of Li_2SO_4=110$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{734}{110}\\n=6.67$

V is given as

$V_{L}=\dfrac{V_{mL}}{1000}\\V_{L}=\dfrac{2500}{1000}\\V_{L}=2.5\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{6.67}{2.5}\\M=2.68\ M$

So the molarity is 2.68 M.

3)Molarity of a solution is made by adding 83 grams of sodium hydroxide to 750 mL of water is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of Sodium hydroxide is as follows:

$MM of NaOH=Na+O+H\\MM of NaOH=23+16+1\\MM of NaOH=40$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{83}{40}\\n=2.075$

V is given as

$V_{L}=\dfrac{V_{mL}}{1000}\\V_{L}=\dfrac{750}{1000}\\V_{L}=0.75\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{2.075}{0.75}\\M=2.77\ M$

So the molarity is 2.77 M.

4)Molarity of a solution that contains .500 mol HC₂H₃O₂ in 0.125 kg H₂O is

$M=\dfrac{n}{V}$

V is given as

$V_{L}=\dfrac{mass}{density}\\V_{L}=\dfrac{0.125}{1 kg/L}\\V_{L}=0.125\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{0.5}{0.125}\\M=4.0\ M$

So the molarity is 4.00 M.

5)Molarity of a solution that contains 63.0 g HNO₃ in 0.500 kg H₂O is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of HNO₃ is as follows:

$MM of HNO_3=H+N+3\times O\\MM of HNO_3=1+14+3\times 16\\MM of HNO_3=63$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{63}{63}\\n=1.00$

V is given as

$V_{L}=\dfrac{mass}{density}\\V_{L}=\dfrac{0.5}{1 kg/L}\\V_{L}=0.5\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{1}{0.5}\\M=2.00\ M$

So the molarity is 2.00 M.

6)Mass of water must be used to dissolve 0.500 kg C₂H₅OH to prepare a 3.00 m solution is calculated as follows

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of C₂H₅OH is as follows:

$MM of C_2H_5OH=2\times C+6\times H+O\\MM of C_2H_5OH=2\times 12+6\times 1+16\\MM of C_2H_5OH=46$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{500}{46}\\n=10.87$

V required is given as

$M=\dfrac{n}{V}\\V=\dfrac{n}{M}\\V=\dfrac{10.87}{3}\\V=3.62\ L$

So the mass of water is given as

$mass=density\times Volume\\mass=1 kg/L \times 3.62 L\\mass=3.62\ kg$

So the mass of water required to form 3.0 M solution by dissolving 0.5kg of C₂H₅OH is 3.62 kg.

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