Cu(OH)2 is insoluble in water
Answer:
Please check with the attachment posted.
Please do confirm the answer.
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:
The right answer is B) evaporation
Explanation:
Transpiration occurs at the leaf surface which is the loss of water due to the evaporation. This phenomenon works as trigger of water and mineral movement above to the xylem. Due to the evaporation of water at the leaf, negative pressure is created at the surface of leaf. Tension is produced which results in the pull of water from roots up to the xylem vessels.
Answer:
0.60g of the fungus can be allowed
Explanation:
The maximum concentration of the fungus that can be allowed is 25 ppb, that is 25mg/t.
As the peanuts are 24t, the mass of fungus that can be allowed is:
24t * (25mg/t) = 600mg of the fungus can be allowed. In grams are:
600mg * (1g/1000mg) =
<h3>0.60g of the fungus can be allowed</h3>
