Answer: increase in entropy
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
is positive when randomness increases and is negative when randomness decreases.
When the non polar solutes are placed in water, the hydrogen bonding network of water is disrupted, and there are fewer ways for water to hydrogen-bond with itself. That means the water molecules are more randomly arranged and thus have more entropy and thus is positive.
1 kpa = 0.0098692327 atm so just multiply that by 45.6
Answer:
no lo sé chico ve a morir en un agujero
Explanation:
The correct answer for the question that is being presented above is this one:
Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m
<span>delta Tf = Kfm Kf H2O = 1.86 degrees C/m
</span>
We need to know the formula for Molality.
molality = mol solute / kg solvent
<span>We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. </span>
<span>25 g NaCl / 58.5 g/mol = 0.427 mol </span>
<span>Then, use the formula for molality. </span>
<span>molality = mol solute / kg solvent </span>
<span>= 0.427 / 1 </span>
<span>= 0.427 m </span>
<span>Use now the formula to get the boiling point.</span>
<span>delta Tb = Kbm </span>
<span>= (0.52)(0.427) </span>
<span>= 0.22C </span>
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
