Answer:
Molality = 1.46 molal
The freezing point of the solution = -2.72 °C
Explanation:
Step 1: Data given
The freezing point of water H2O is 0.00°C at 1 atmosphere
urea = nonelectrolyte = van't Hoff factor = 1
Mass urea = 13.40 grams
Molar mass urea = 60.1 g/mol
Mass of water = 153.2 grams
Molar mass H2O = 18.02 g/mol
Kf = 1.86 °C/m
Step 2: Calculate moles urea
Moles urea = mass urea /molar mass urea
Moles urea = 13.40 grams / 60.1 g/mol
Moles urea = 0.223 moles
Step 3: Calculate the molality
Molality = moles urea / mass water
Molality = 0.223 moles / 0.1532 kg
Molality = 1.46 molal
Step 4: Calculate the freezing point of the solution
ΔT = i * Kf * m
ΔT = 1* 1.86 °C/m * 1.46 m
ΔT = 2.72 °C
The freezing point = -2.72 °C
A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water.
Answer: Hello!
first i believe we need a balanced equation to start...
i got 2H2 + 1O2 = 2H2O
This tells us that we need 2 moles of H2 for every 1 mole of O2 Since we only have 1 mole of H2 compared to the 5 moles of O2 hydrogen is the limiting reagent. For illustration, divide the balanced equation by 2 in order to get 1 mole of H2 If we start with 1.0 moles of H2 we'll produce 1.0 mole of H2O
Your welcome <3
Explanation:
Answer: option (a) is the correct answer
Explanation:
The complete questions says;
You observe an exothermic gaseous reaction that is not spontaneous in forward direction at 1 atm and 298K. Which of the following statements about this reaction is true? a. This reaction will become spontaneous in forward direction at some temperature below 298K. b. This reaction will be spontaneous in forward direction at a higher pressure at 298K. c. This reaction will become spontaneous in forward direction at some temperature above 298K. d. This reaction is never spontaneous. e. The reverse reaction is always spontaneous.
The Answer:
(a). This reaction will become spontaneous in forward direction at some temperature below 298K
Explanation: First of all, we can acknowledge that the reaction seen here is an exothermic one, i.e energy is released in the process outwardly and as a result temperature is reduced during this process of energy loss.
Having understood that scenario, say we reduce it's temperature by ourself than forward reaction favors and after reaching at particular temperature, therefore we can confirm this to be a spontaneous reaction.
Let us use this to confirm what we have been saying.
Given;
ΔG = ΔH - TΔS
here ΔH is negative
it is non spontaneous, which means ΔG is positive so we continuously decraeses it's temperature than at a particular temperature.
The Entropy change becomes positive and reaction becomes spontaneous and ΔG become negative.
cheers i hope this helped !!