Answer:
Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions.
Answer:
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
Explanation:
Let volume of the 40% acid solution be x.
Let volume of the 60% acid solution be y.
Volume of solution formed after mixing both solution = 40 L
x + y = 40 L..[1]
Volume of acid 40% solution = 40% of x= 0.4x
Volume of acid 60% solution = 60% of y= 0.6y
Volume of acid formed = 45% of 40 L = 
..[2]
Solving [1] and [2]
x = 30 L , y = 10 L
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
You start by using proportions to find the number of liters of solution:
180 g of glucose / 1 liter of solution = 18 g of glucose / x liter of solution
=> x = 18 g of glucose * 1 liter of solution / 180 g of glucose = 0.1 liter of solution.
If you assume that the 18 grams of glucose does not apport volume to the solution but that the volume of the solution is the same volumen of water added (which is the best assumption you can do given that you do not know the how much the 18 g of glucose affect the volume of the solution) then you should add 0.1 liter of water.
Answer: 0.1 liter of water.
Answer:
A) he equilibrium concentration of PH3 = 0.0432M
B) he equilibrium concentration of BCl3 = 0.0432M
C) what is the minimum mass of PH3BCl3(s) that must be added to the flask to achieve equilibrium = 1.69g
Explanation:
The detailed steps and appropriate calculation is as shown in the attached file.