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3 years ago

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A container holds 2.0 of liquid water. The water absorbs 477 kJ of energy. If the water's initial temperature was 298 K,what is

it's final temperature?

1 answer:

tester [92]3 years ago

5 0

Answer:

$T_{final}=82C$

Explanation:

Here mass of water not given directly but volume of water is given and we already know the density of water so we will use the volume density relationship to find mass.

volume=2L

density=1kg/L

$density=\frac{mass}{volume}$

$mass=density\times volume$

mass=2kg=2000gram

initial temperature=298K=25C

specific heat of water (C)=4.186 joule/gram

heat energy lose or gain= $mC\Delta T$

$477\times 10^3 J=2000g\times 4.186J/g \times \Delta T$

$\Delta T=57 C$

$T_{final}-T{initial}=57$

$T_{final}=82C$

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Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

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From the given information:

The equation for the reaction can be represented as:

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The I.C.E table can be represented as:

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Initial: 14 2.6 0

Change: -2x -x +2x

Equilibrium: 14 - 2x 2.6 - x 2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

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= 2.6 - 1.6

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Thus;

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Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction $2SO_2 + O_2 \to 2SO_3$ = 0.8325;

If we want to find:

$SO_2 + \dfrac{1}{2}O_2 \to SO_3$

Then:

$K_c = (0.8325)^{1/2}$

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Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

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Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O

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Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

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The balanced equilibrium reaction will be:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

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Let the solubility will be, 's'

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Now put the value of $K_[sp}$ in this expression, we get the solubility.

$5.02\times 10^{-6}=(4s)^3$

$s=1.079\times 10^{-2}M$

The concentration of $Ca^{2+}$ ion = s = $1.079\times 10^{-2}M$

The concentration of $OH^-$ ion = 2s = $2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M$

First we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.158\times 10^{-2})$

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Now we have to calculate the pH.

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Therefore, the pH of a saturated solution is, 12.33

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