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Nastasia [14]
3 years ago
8

Water has a density of 0.997 g/cm^3 at 25 degrees C; ice has a density of 0.917 g/cm^3 at -10 degrees C. (question part a) If a

soft drink bottle whose volume is 1.50L is completely filled with water and then frozen to -10 degrees C, what volume does the ice occupy? (question part b) Can the ice be contained within the bottle?
Chemistry
2 answers:
KatRina [158]3 years ago
8 0
<span>1.5L x 0.05
= 0.075
= 1.425L

</span>Once melted ice will then take the same volume as before (10cm^3<span>), but it was dispersing only </span>9.5cm^3<span>, so the water level will rise to account for the additional .</span>5cm^3<span>. This is a fairly small amount (only about 5% of the volume of the melted water), but it's notable.
</span><span>
Therefore yes it can be contained.




</span>
Luden [163]3 years ago
4 0
Mass of water added:
0.997 x 1500
= 1495.5 grams

a) Volume = mass / density
Volume = 1495.5 / 0.917
Volume = 1630 cm³ = 1.63 L

b) The ice cannot be contained in the bottle as its volume exceeds that of the bottle.
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Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

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