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bearhunter [10]
3 years ago
8

A 40 g gold coin is cooled from 50°C to 10°C (CAu is 0.13 J/g-°

Chemistry
1 answer:
Debora [2.8K]3 years ago
4 0
When the Heat gain or lose = the mass * specific heat * ΔT

and when we have the mass of gold coin= 40 g 

and the specific Heat of gold=  0.13 J/g°

and ΔT = (Tf- Ti) = 10°C - 50°C = -40 °C

so by substitution:

∴Heat H = 40 g * 0.13 J/g° * -40
               = - 208 J
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Answer:

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Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

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Explanation:

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Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

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Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

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