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natta225 [31]
3 years ago
11

if you pour out some of the water in a bucket,does the density of the remaining water change? explain.

Chemistry
2 answers:
Aleksandr [31]3 years ago
5 0
The density does not change because it is still the same liquid as before
Fynjy0 [20]3 years ago
3 0
Yes, because if you assume that you have a full bucket of water and when you try to pick it up, it too heavy to pick up without a struggle so when you pour half the water out of the bucket, the bucket of water is a lot easier to pick up, thus the density of the bucket of water changes.
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Comparing weak acids and strong acids of equal concentrations which of the following statements is true The pH of the weak acid
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Answer:

the true statement is... The pH of the weak acid will be higher than the pH of the strong acid

Explanation:

pH is a measured of the extent to which acids dissociate into ions when plced in aqueous solution.

Strong acid dissociate near-completely, and weak acids barely dissociate.

At equal concentrations, a strong acid will have a lower pH than a weak acid, since the strong one will donate more proton to the solution.

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The standard reduction potentials of lithium metal and chlorine gas are as follows:Reaction Reduction potential(V)Li+(aq)+e−→Li(
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Answer:

A) E° = 4.40 V

B) ΔG° = -8.49 × 10⁵ J

Explanation:

Let's consider the following redox reaction.

2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)

We can write the corresponding half-reactions.

Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq)      E°red = 1.36 V

Anode (oxidation):  2 Li(s) → 2 Li⁺(aq) + 2 e⁻         E°red = -3.04

<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>

The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V

<em>B) Calculate the free energy ΔG° of the reaction.</em>

We can calculate Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n.F.E°

where,

n are the moles of electrons transferred

F is Faraday's constant

ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J

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