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hichkok12 [17]
3 years ago
15

A 15.6 grams of ethanol absorb 868 J as it is heated. The initial temperature is 21.5 degrees Celsius. What is the final tempera

ture if the specific heat of ethanol is 2.41 Joules/ grams Celsius?
Chemistry
1 answer:
77julia77 [94]3 years ago
3 0
You have to use the equation q=mcΔT and solve for T(final).  
T(final)=(q/mc)+T(initial)
q=the amount of energy absorbed or released (in this case 868J)
m=the mass of the sample (in this case 15.6g)
c= the specific heat capacity of the substance (in this case 2.41 J/g°C)
T(initial)=the initial temperature of the sample (in this case 21.5°C)

When you plug everything in, you should get 44.6°C.
Therefore the final temperature of ethanol is 44.6°C

I hope this helps.  Let me know if anything is unclear.
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No precipitate is formed.

Explanation:

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In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

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MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

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Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

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In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

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