I think it is the third one.
Answer:
1.84 L
Explanation:
Using the equation for reversible work:
Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( - 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L
Answer:
The molarity of the solution is 0,31 M
Explanation:
We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)
Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g
58, 5 g-----1 mol NaCl
13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl
727 ml solution------ 0, 224 mol NaCl
1000ml solution------x= (1000ml solutionx0, 224 mol NaCl)/727 ml solution
x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>
Answer:
you have to attach the diagram
Explanation:
Answer: Colligative properties are those properties of solutions that are dependent on the concentration of the solutes in the solution.
Colligative properties has to do with solutions, that is, solutes that are dissolved in solvents. Examples of colligative properties are: freezing point depression, vapour pressure lowering, boiling point elevation and osmotic pressure. Colligative properties do not depend on the identity of the solutes, this implies that the effect of colligative properties are uniform across all solutions. For example, the freezing point depression of any solution will depend on the concentration of solutes that are dissolve in solution.
