Answer: B.) I, II, and III
Explanation:
Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and for the reaction comes out to be negative.
Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and for the reaction comes out to be positive.
I) The temperature (of water) increases when calcium chloride dissolves in water : Thus the reaction is exothermic and for the reaction comes out to be negative.
II) Steam condenses to liquid water : The energy is released when bonds are formed when it coverts from gas to liquid and thus for the reaction comes out to be negative.
III) Water freezes : The energy is released when bonds are formed to get converted from liquid to solid and thus for the reaction comes out to be negative.
IV) Dry ice sublimes : The energy is absorbed when bonds are broken to get converted from solid to gas and thus for the reaction comes out to be positive.
Answer:
Explanation:
Hello,
In this case, since in a dilution process the moles of the solute must remain unchanged, we use the volumes and molarities as shown below:
Clearly, the concentrated solution is 12M and the diluted solution is 0.5 M, thus, the volume of the concentrated solution we should take is:
Best regards.
That would be convection. This is a test. ;)
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76