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wolverine [178]
3 years ago
5

Please help me with this question, if you can! :)

Chemistry
1 answer:
JulijaS [17]3 years ago
3 0

Answer: The volume of the solution be if it is diluted to 10% (v/v) is 6.0 L

Explanation:

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = concentration of pure acid solution = 15 %

V_1 = volume of pure acid solution = 4.0 L

C_2 = concentration of diluted acid solution= 10%

V_2 = volume of diluted acid solution= ?

Putting in the values we get:

15\times 4.0=10\times V_2

V_2=6.0L

Thus the volume of the solution be if it is diluted to 10% (v/v) is 6.0 L

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Well this has a good chance of being wrong but i wanna say,

When you change a physical property of something it doesn't affect the chemicals that make it up. Like Ice, you can freeze water to make ice. You change a physical property (state of matter) but it's chemical properties don't change because in the end it's still water.

However if you remove a chemical property from something you are changing what made the new substance with will also change the substance along with it.

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In the reaction 4Al+__O2→2Al2O3, what coefficient should be placed in front of the O2 to balance the reaction?
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Calculate the amount of heat energy in KJ required to convert 45.0 g of ice at -15.5'C to steam at 124.0°C. (Cwater 118 Jig'c, G
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Answer : The enthalpy change or heat required is, 139.28775 KJ

Solution :

The conversions involved in this process are :

(1):H_2O(s)(-15.5^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(124^oC)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change or heat required = ?

m = mass of water = 45 g

c_{p,s} = specific heat of solid water = 2.09J/g^oC

c_{p,l} = specific heat of liquid water = 4.18J/g^oC

c_{p,g} = specific heat of liquid water = 1.84J/g^oC

n = number of moles of water = \frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{45g}{18g/mole}=2.5mole

\Delta H_{fusion} = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

\Delta H_{vap} = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]

\Delta H=139287.75J=139.28775KJ     (1 KJ = 1000 J)

Therefore, the enthalpy change is, 139.28775 KJ

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