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3 years ago

Arrange the processes of the water cycle in the correct order, starting with the heat from the Sun.

Chemistry

1 answer:

tensa zangetsu [6.8K]3 years ago

3 0

There are four processes in water cycle. Starting from the heat from the sun is the process of evaporation.

Evaporation: In this process, light from the sun evaporate water from oceans, rivers, lakes, ice and soil into water vapor. Water vapor molecules combine to form clouds.

Condensation: In this process, water vapor from clouds cool down and turns back into liquid water in the form of water droplets. These water droplets stays in the air. Precipitation: In this process, water droplets they combine in the air, they become too heavy to stay in the air. Therefore, they fall in the form of rain, snow or other form.

Collection: In this process, water that falls as rain, snow or other form comes back in the ocean, lakes, river or any other water body. Water will also be absorbed by soil and will be collected as ground water.

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1 year ago

Tara predicts the number of paper clips attracted by the magnet

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The responding variable refers to the variable that changes as the independent variable is being manipulated. In this case, the responding variable is the number of paper clips attracted by the magnet.

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What is the economic importance of phosphorus in Canada or elsewhere in the world?

vladimir1956 [14]

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3 years ago

Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an

ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

Ti $Cl_{4(l)}$ ⇒ Ti $Cl_{4(g)}$

ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -804.2 kJ/mol

ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) = -763.2 kJ/mol

Therefore,

ΔH° $_{f}$ = ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) - ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(Ti $Cl_{4(l)}$ ) = 221.9 J/(mol*K)

s°(Ti $Cl_{4(g)}$ ) = 354.9 J/(mol*K)

Therefore,

s° = s° (Ti $Cl_{4(g)}$ ) - s°(Ti $Cl_{4(l)}$ ) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH° $_{f}$ /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

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