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kotykmax [81]
3 years ago
12

Which water resource is most likely to dry up for part of the year?

Chemistry
2 answers:
Pavel [41]3 years ago
8 0
Ponds are most likely to dry up 

laila [671]3 years ago
6 0

Answer:

Ponds

Explanation:

Ponds are defined as small and large water bodies that are surrounded by land on all of its sides. In simple words, they are comprised of stagnant water bodies. They receive water mostly from the rainfall and the nearby rivers.

When there occurs changes in the weather and climatic conditions, such as low or absence of rainfall, then rapid evaporation takes place, as a result of which the water is released into the atmosphere, making the pond dry.

So, the ponds are most likely to be dry for some part of a year.

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Which of the following is an example of a chemical change?
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A chemical change changes the bonds by forming or breaking them, and cannot be brought back to it's original form. 

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Answer:

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Answer:

CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

Explanation:

According to this question, sodium carbonate reacts with sulfuric acid to form aqueous sodium sulfate, carbon dioxide and water. The balanced chemical equation is as follows:

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- Next, we cancel out the spectator ions, which are ions that remain the same in the reactants and products side of a chemical reaction. The spectator ions in this equation are 2Na+(aq) and SO42-(aq).

CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

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CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

3 0
3 years ago
Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °
Mila [183]
This problem is to use the Claussius-Clapeyron Equation, which is:

ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]

Where p2 and p1 and vapor pressure at estates 2 and 1

ΔH is the enthalpy of vaporization

R is the universal constant of gases = 8.314 J / mol*K

T2 and T1 are the temperatures at the estates 2 and 1.

The  normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa

Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol 

=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]

=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x =  0.000157 + 1/330.95 = 0.003179

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