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MariettaO [177]
3 years ago
11

Which of the following contains most of the water on earth?

Chemistry
2 answers:
lions [1.4K]3 years ago
8 0
The answer is A because most of the earth is made up of oceans.
mrs_skeptik [129]3 years ago
4 0

Answer:

a

Explanation:

most of the world is made of water because the oceans.

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What causes water to become denser when it is carried to the poles by surface currents? increased temperature and increased sali
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The thermohaline circulation moves the water slowly. This water moves mainly due to differences in its relative density. Much denser water sinks over water that is less dense. Two factors impact the density of seawater: temperature and salinity.

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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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