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Alenkinab [10]
2 years ago
7

If a helium filled ballon has a volume of 3.4l at 25°c and 1.18atm., what is its volume at STP?​

Chemistry
1 answer:
vladimir2022 [97]2 years ago
5 0

Answer:

3.711 L

Explanation:

The formula you need to use is the following:

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\

3.4L / 298 K = V2 / 273 K

V2 = 3.711 L

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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

3 0
3 years ago
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