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ryzh [129]
3 years ago
14

Which mixture can be classified as a homogeneous mixture?

Chemistry
1 answer:
mart [117]3 years ago
7 0

Answer:

Tap water is homogeneous mixture because all minerals are dissolved and uniformly distributed init.

The lemonade is also consider the homogeneous mixture if pulp is separated.

Explanation:

The solution is always homogeneous mixture and transparent through which the light can travel. The mixture of water lemon and sugar without pulp (lemonade) is a solution because sugar and lemon is soluble in water and form homogeneous mixture. The tap water is also a homogeneous mixture because minerals are present in dissolved form. The sand and soil can not dissolve in water and particles scatter the light thus form heterogeneous mixture.

Homogeneous mixture:

"The homogeneous mixture are transparent through which the light can travel and called solution"

The mixture of water, lemon and sugar is a solution because sugar is soluble in water and form homogeneous mixture. The solubility of sugar is high as compared to the sand in water because the negative and positive ends of sucrose easily dissolve into the polar solvent i.e, water

Heterogeneous mixture:

"The heterogeneous mixture is a suspension  in which the solute particles settle down but does not dissolve"

The mixture of water and sand is suspension. The sand can not dissolve in water because it is mostly consist of quartz. The nonpolar covalent bonds of  sand are too strong and cannot be break by water molecules.

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The combustion reaction of isopropyl alcohol is given below: C 3 H 7 O H ( l ) + 9 2 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( g ) T
jek_recluse [69]

Answer:

the heat of formation of isopropyl alcohol is -317.82 kJ/mol

Explanation:

The heat of combustion of isopropyl alcohol is given as follows;

C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)

The heat of combustion of CO₂ and H₂O are given as follows

C (s) + O₂ (g) → CO₂(g) = −393.50 kJ

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Therefore we have

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as

3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3  =

4H₂ (g) + 2·O₂(g)   →  4H₂O (l) = −285.83 kJ × 4

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol

-1180.5 - 1143.32 +2006 = -317.82 kJ/mol

Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.

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