What is the value of Q when 100.0 mL of a 7.6 10-4 M Mg(NO3)2 is added to 100.0 mL of 7.1 10-4 M NaOH? (Ksp for Mg(OH)2 = 8.9 ✕ 10-12)

Q = 4.8x10⁻¹¹

Explanation:

The solubility of Mg(OH)₂ is:

Mg(OH)₂(s) → Mg²⁺(aq) + 2OH⁻(aq)

And Ksp is:

Ksp = [Mg²⁺] [OH⁻]²

Where [] are concentrations in equilibrium

Thus, Q could be defined as:

Q = = [Mg²⁺] [OH⁻]²

Where [] are actual concentrations

[Mg(NO₃)₂] = [Mg²⁺]

And [NaOH] = [OH⁻]

Both solutions are diluted twice (Because 100mL of each solution are diluted to 200mL). The equilibrium concentrations are:

[Mg²⁺] = 7.6x10⁻⁴M / 2 = 3.8x10⁻⁴M

[OH⁻] = 7.1x10⁻⁴ / 2 = 3.55x10⁻⁴M

That means Q is:

Q = [3.8x10⁻⁴M] [3.55x10⁻⁴M]²

linda: 3 ; laura: 15

step-by-step explanation:

note that linda + laura's age is 18, and that linda is 12 years younger than her sister.

let "linda" = x , "laura" = y:

x + y = 18

x = y - 12

first, plug in y - 12 for x in the first equation:

(y - 12) + y = 18

simplify.

2y - 12 = 18

isolate the variable (y). note the equal sign, what you do to one side, you do to the other. do the opposite of pemdas. first add, then divide.

2y - 12 (+12) = 18 (+12)

2y = 18 + 12

2y = 30

(2y)/2 = (30)/2

y = 30/2

y = 15

next, plug in 15 for y in the second equation (the other equation would work too).

x = y - 12

x = (15) - 12

x = 3

linda's age is 3, laura's age is 15.

~

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).