Answer:
d. The energy required to evaporate 1 kg of liquid water equals the energy released when 1 kg of water vapor condenses into liquid.
Explanation:
Hello,
Since we're considering the same amount of water, the vapor phase has a higher energy content than the liquid phase, thus, for the specified amount of water particles (those contained in the given 1 kg) the energy MUST be same when taking them either to a gaseous phase or to a liquid phase, the only difference is the sign which is negative from gaseous to liquid (heat withdrawal) and positive from liquid to gaseous (heat adding).
Best regards.
Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.
Normality comes out to be 8.11
<h3>
Given </h3>
- Mass of solute: 1000g
- Volume of solution (V): 5000 ml = 5 liters
- Equivalent mass of solute (E) = molar mass / n-factor
n-factor for
is 6 and molar mass is 148g
So, on calculating equivalent mass is equal to 24.66g
FORMULAE of Normality (N) = (Mass of the solute) / (Equivalent mass of the solute (E) × Volume of the solution (V)
N=
<u> N=8.11</u>
Therefore, normality of 1 kg aluminum sulfide is 8.11
Learn more about normality here brainly.com/question/25507216
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In this order.
protons. 6. carbon atom. different. isotopes. atomic masses. same. chemical reaction. reactions. electrons. neutrons.
Hello!
The concentration of the final solution when a<span> chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water is
0,3 MTo calculate that, you'll need to use the dilution law, where initial and final concentrations are M1 and M2 respectively, and initial and final volumes are V1 and V2, as shown below.
Keep in mind that the final volume is the sum of the 200 mL of water and the 50 mL of H</span>
₂SO₄ that were added by the teacher. 
Have a nice day!
Molality= moles NaCl/ Kg H2O
250 g (1 Kg/ 1000 grams)= 0.250 Kg
Molality= 0.611 moles/ 0.250 Kg= 2.44 molal