Answer:
a) KI and Na₃PO₄
b) HF
c) CaBr₂
Explanation:
The depression in freezing point, relative lowering of vapor pressure and elevation in boiling point are colligative properties.
The collegative properties depend upon the number of solute particles.
Let us see the concentration of particles in each of the given solutions:
a) 0.010 m Na₃PO₄ in water : each molecule will give three sodium ions and one phosphate ion.
So the concentration of particles in the solution = 4 X 0.01 = 0.04 m
b) 0.020 m CaBr₂ in water : each molecule will give one calcium ion and two bromide ion thus the concentration of particles in the solution
= 3 X 0.02 = 0.06 m
c) 0.020 m KCl in water: each molecule gives two ions thus the concentration of solution = 2 X 0.02 = 0.04 m
d) 0.020 m HF (weak acid) in water: the acid is weak, so each molecule will not give two ions completely (complete dissociation will not be observed). concentration of solution will be more than 0.02 but less than 0.04.
a) Both KCl and Na₃PO₄ will have same boiling point as 0.04 C₆H₁₂O₆.
b) the solution with least concentration of solute will have highest pressure. Thus HF is the answer,
c) the largest depression will be in the solution with highest solute concentration: Answer: CaBr₂.
Hey there!
In a scientific experiment, there are three types of variables: dependent, independent, and control.
Independent variables are the variables that are changed and doesn't depend on another.
Dependent variables are the variables that are depending on the independent variable. When the independent variable changes, it affects the dependent variable.
Control variables are the variables that are kept the same throughout the entire experiment. These don't change at all.
Hope this helps!
I think the correct answers from the choices would be options B and C. You need the chemical symbols of the atoms so you will know how the atoms share the bond. Also, you need the electrons shared between atoms since it is what makes the covalent bond between the atoms.
<h2>Answer: 105 mL</h2>
<h3>Explanation:</h3>
We can use the formula M₁V₁ = M₂V₂ the total volume of water that will yield 0.800 M HCl. We can then subtract that from the initial volume to get how much water needs to be added.
Since M₁V₁ = M₂V₂ (starting mole× starting volume = end mole × end volume)
⇒ (5 mol)(20 mL) = (0.80 mol) V₂
100 mol/mL = 0.8 mol × V₂
⇒ V₂ = 100 mol/mL ÷ 0.8 mol
= 125 m
Since we began with 20.0 mL of water,
water needed to add = 125 mL - 20 mL = 105 mL
