Answer:
Molarity = 0.809 M
mole fraction = 0.047
Explanation:
The complete question is
Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Solution -
Solution for molarity:
1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.
1)
Mass of 1.09 mole of acetone
= 1.09 mol x 58.0794 g/mol = 63.306 g
Density of acetone = 0.788 g/cm3
Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3
For ethanol
1000 g divided by 0.789 g/cm3 = 1267.427 cm3
Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3 = 1.347 L
a) Molarity:
1.09 mol / 1.347 L = 0.809 M
Mole Fraction
a) moles of ethanol:
1000 g / 46.0684 g/mol = 21.71 mol
b) moles of acetone:
1.09 / (1.09 + 21.71) = 0.047
Answer:
I'm pretty sure its 1.96g
Explanation:
The kinetic energy of a gas molecule is given by :
It means, whenever the temperature of gas is increased, its kinetic energy will increase and the space between the particles gets increase. So, the space in which particles can collide with each other increases i.e. its volume gets high.
Hence, the correct option is (A). "higher temperature more kinetic energy → more space between particles – higher volume".
Answer:
Increases temperature
Supplies energy to a reaction
Answer:
the potential with respect to a silver–silver chloride electrode is 0.715
Explanation:
the potential to a silver chloride electrode can be derived from the question above by saying that
the potential with respect to a silver–silver chloride electrode? = 0.9120v - 0.197 V
= 0.715
therefore, the potential with respect to a silver–silver chloride electrode is 0.715
