3-ethyl-2.4-dimethyl-octanoic acid
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:
Molar mass of acetone = 58 g/mol
Number of moles = 
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have = 
To calculate the entropy change for the system, we use the formula:
Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
(Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,
Answer:
315.
Explanation:
Hello.
In this case, since the given number has five significant figures as the zero is to the right of the first nonzero digit (3), if it is required to report it with three significant figures, it is necessary to "cut" it at the first five without any rounding since the subsequent zero is less than five.
Thus the number turns out:
315
Best regards.
Answer:
I don't really get the options but it favoures the reactant side.
Explanation:
Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g
If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.
With the explanations I have made, I hope the question is now clear to you.
Answer:
15.2 g H2
Explanation:
2H2O -> 2H2 + O2
9.06 x 10^24 molecules x (1 mol/6.022 x 10^23 molecules) x (2 mol H2/2 mol H2O) x (1.008 g/1 mol) = 15.2 g H2
