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pogonyaev
3 years ago
9

a sample of water with a mass of 648.00 kg at 298 K is heated with 87 kh of energy. the specific heat of water is 1 J-1 kg K-1.

what is the final temperature of the water
Chemistry
2 answers:
Keith_Richards [23]3 years ago
8 0

Answer : The final temperature of water is, 432.26 K

Solution :

Formula used :

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

Q = heat supply = 87 kJ = 87000 J

m = mass of water = 648.00 kg

c = specific heat of water = 1J/kg.K      

\Delta T=\text{Change in temperature}  

T_{final} = final temperature = ?

T_{initial} = initial temperature = 298 K

Now put all the given values in the above formula, we get the final temperature of water.

87000J=648.00kg\times 1J/kg.K\times (T_{final}-298K)

T_{final}=432.26K

Therefore, the final temperature of water is, 432.26 K

Stels [109]3 years ago
3 0

q = mCΔT

The correct specific heat capacity of water is <em>4.187 kJ/(kg.K)</em>.

ΔT = q/mC = 87 kJ/[648.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2713 kJ/K) = 0.032 K

Tf = Ti + ΔT = 298 K + 0.032 K = 298.032 K


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Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
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No precipitate is formed.

Explanation:

Hello,

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

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

And the undergoing chemical reaction:

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:

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

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

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

Regards.

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