I think the correct answers from the choices listed above are the first, third and the last option. Ionic compounds are compounds that dissociates into ions when in aqueous solution. From the list, NH4Cl, KF and MgO are the ionic compounds. Hope this answers the question.
Answer:
m = 700 g
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
Density of octane = 0.700 g/cm³
Volume = 1 L
Mass = ?
Formula:
D=m/v
D= density
m=mass
V=volume
First of all we will convert the volume in cm³ because density is given in g/cm³ unit.
1 L = 1000 cm³
Now we will put the values in formula:
d= m/v
m = v × d
m = 1000 cm³ × 0.700 g/cm³
m = 700 g
Answer: the boiling point is = 137.325°C
Explanation:
From the formula: ∆Tb= Kb*m
From the question, Kb= 0.95, m= 27.5, T1= 111.2°C
Substitute into ∆Tb= Kb*m
∆Tb= 0.95*27.5= 26.125
∆Tb= T2-T1
Hence
T2- 111.2=26.125
T2= 26.125+ 111.2= 137.325°C
The answer is C. because <span>particles settle out over time ,can block light and scatter light .</span>
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.