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Nastasia [14]
3 years ago
8

Water has a density of 0.997 g/cm^3 at 25 degrees C; ice has a density of 0.917 g/cm^3 at -10 degrees C. (question part a) If a

soft drink bottle whose volume is 1.50L is completely filled with water and then frozen to -10 degrees C, what volume does the ice occupy? (question part b) Can the ice be contained within the bottle?
Chemistry
2 answers:
KatRina [158]3 years ago
8 0
<span>1.5L x 0.05
= 0.075
= 1.425L

</span>Once melted ice will then take the same volume as before (10cm^3<span>), but it was dispersing only </span>9.5cm^3<span>, so the water level will rise to account for the additional .</span>5cm^3<span>. This is a fairly small amount (only about 5% of the volume of the melted water), but it's notable.
</span><span>
Therefore yes it can be contained.




</span>
Luden [163]3 years ago
4 0
Mass of water added:
0.997 x 1500
= 1495.5 grams

a) Volume = mass / density
Volume = 1495.5 / 0.917
Volume = 1630 cm³ = 1.63 L

b) The ice cannot be contained in the bottle as its volume exceeds that of the bottle.
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3 years ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)  + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

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Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)  + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

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There is a reaction would take place between magnesium and sodium chloride.

The chemical reaction between these two metals is given below.

It is Mg + 2NaCl >> MgCl2. + 2Na.

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