Answer:
Solve the following problems (assuming constant temperature). Assume all numbers are 3 sig figs. 1. A sample of oxygen gas occupies a volume of 250 mL at 740 torr pressure. ... the gas exert if the volume was decreased to 2.00 liters? ... A 175 mL sample of neon had its pressure changed from 75.0 kPa to 150 kPa.
Explanation:
Anything that has mass and volume (takes up space) is called matter.
COVALENT BOND IS THE BOND EXISTING BETWEEN 2 ATOMS THAT SHARE 6 ELECTRONS
Answer:
The entropy change for a real, irreversible process is equal to <u>zero.</u>
The correct option is<u> 'c'.</u>
Explanation:
<u>Lets look around all the given options -:</u>
(a) the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.
(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .
(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.
(d) impossible to tell. This option is invalid , thus incorrect .
<u>Hence , the correct option is 'c' that is zero.</u>
B. 3.0 mol·L⁻¹ NaCl
Explanation:
Freezing point is a colligative property: it depends only on the number of particles in solution.
The for freezing point depression ΔT_f is
ΔT_f = iK_fb
where
i = the number of moles of particles available from one mole of solute
K_f = the molal freezing point depression constant
b = the molal concentration of the solute
All your solutions are aqueous NaCl. They differ only in their concentrations.
Thus, the most concentrated solution will have the greatest freezing point depression and the lowest freezing point.
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