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3 years ago

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,

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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

dangina [55]

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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