Answer:
Volume of 
= 15.31 mL
Volume of 
= 4.69 mL
Explanation:
Given that:
the density of the mixture = 1.82 g/mL
From the density of the pure samples
The density of = 1.492 g/mL
The density of = 2.890 g/mL
The total volume of the liquid mixture = 20.0 mL
Suppose the volume of = P ml
and the volume of = Q ml
the sum of their volumes should be equal to the total volume of the mixture
----- (1)
However, we know that Density = mass/volume
∴ mass = density × volume
The equation can now be expressed as:
1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL ---- (2)
From equation (1) ;
let Q = 20 - P
The replace the value of P into equation (2)
1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL
1.492 P g + 57.8g - 2.890 P g = 36.4g
1.492 P g - 2.890 P g = 36.4g - 57.8g
-1.398 P g = -21.4g
P = -21.4g/-1.398g
P = 15.31 mL
Q = 20 - P
Q = (20 - 15.31) mL
Q = 4.69 mL
∴
Volume of = 15.31 mL
Volume of = 4.69 mL
Answer:
7.12 mm
Explanation:
From coulomb's law,
F = kqq'/r².................... Equation 1
Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.
Make r the subject of the equation,
r = √(kqq'/F).......................... Equation 2
Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute into equation 2
r = √[ (75×10⁻⁹ )²9.0×10⁹/1]
r = 75×10⁻⁹.√(9.0×10⁹)
r = (75×10⁻⁹)(9.49×10⁴)
r = 711.75×10⁻⁵
r = 7.12×10⁻³ m
r = 7.12 mm
Hence the distance between the point charge = 7.12 mm
Answer:
4.96E-8 moles of Cu(OH)2
Explanation:
Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.
Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.
![pH= -log[H]\\pH= -log (\frac{kw}{[OH]})](https://tex.z-dn.net/?f=pH%3D%20-log%5BH%5D%5C%5CpH%3D%20-log%20%28%5Cfrac%7Bkw%7D%7B%5BOH%5D%7D%29)
![8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}](https://tex.z-dn.net/?f=8.23%20%3D%20-%20log%28%5Cfrac%7BKw%7D%7B%5BOH%5D%7D%20%5C%5C10%5E%7B-8.23%7D%20%3D%20Kw%2F%5BOH%5D%5C%5C%20%5BOH%5D%20%3D%20Kw%2F10%5E%7B-8.23%7D)
![[OH]=1.69E-6](https://tex.z-dn.net/?f=%5BOH%5D%3D1.69E-6)
This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:
In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":
The expression for Kps is:
![Kps= [Cu^{2+}] [OH]^2](https://tex.z-dn.net/?f=Kps%3D%20%5BCu%5E%7B2%2B%7D%5D%20%5BOH%5D%5E2)
The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.
"s" is the soluble quantity of Cu(OH)2.
The solution for this third grade equation is 
Now, let us calculate the moles in 1 L:
