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stellarik [79]
3 years ago
12

if you placed the contents of a packet of powdered iced tea mix into a bottle of water and shook it, which of the following woul

d be true? A. The iced tea mix would dissolve the water. B. The iced tea mix would be the solvent. C. The water would be the solvent. D. The solution would be homogenous if half of the powdered solute sat at the bottom of the bottle.
Chemistry
2 answers:
victus00 [196]3 years ago
3 0
If you placed the contents of a packet of powdered iced tea mix into a bottle of water and shook it, D. The solution would be homogenous if half of the powdered solute sat at the bottom of the bottle.
Sonbull [250]3 years ago
3 0

Answer: Option (C) is the correct answer.

Explanation:

A homogeneous solution is defined as the solution in which solute particles are evenly distributed in a solvent.

A homogeneous solution is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

On the other hand, a solution in which solute particles are unevenly distributed in a solvent then it is known as a heterogeneous solution.

In a solution, a substance present in small quantity is known as solute and a substance present in large quantity is known as a solvent.

Therefore, when powdered iced tea mix is added into a bottle of water then powdered iced tea is the solute and water is the solvent.

Thus, we can conclude that the statement, water would be the solvent is true.

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The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p
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Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

    Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g  

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}

(ii) Solve the relationship

 Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

    56x + 18(7.5 - x) = 267.3

      56x + 135 - 18x = 267.3

                        38x = 132.3

                            x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

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