Answer: pH=12.69
Explanation:



Initial 0.12 0 0
Eqm 0.12-x x x
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
(neglecting small value of x in comparison to 0.12)

Moles of 



0.06 moles of NaOH will give 0.06 moles of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
Now
moles of
will be neutralized by
moles of
and
moles of
will be left.
Molarity of 
![pOH=-\log[OH^-]=-\log[0.049]=1.31](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B0.049%5D%3D1.31)
pH = 14 - pOH= 14 - 1.31 = 12.69
The weight in grams = 7.93 g
Given volume = 2.00
Given density = 0.242 g/
We need to find the Mass(weight) in grams.
To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.
The unit of given volume is
and unit of volume in density is
, so first we need to change the unit of volume from
to
so that the volume units will cancel out, leaving only the mass units.
1
= 16.39
(given conversion)

units get cancel out leaving the
unit.

Mass = Density X Volume.
Density = 0.242 g/
and Volume = 32.78 

Mass = 7.93 grams (g)
A. Fossil fuels take a very long time to form so we mine it faster than it can replenish.
Answer:
The value of the heat capacity of the Calorimeter
= 54.4 
Explanation:
Given data
Heat added Q = 4.168 KJ = 4168 J
Mass of water
= 75.40 gm
Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c
From the given condition
Q =
ΔT +
ΔT
Put all the values in above equation we get
4168 = 75.70 × 4.18 × 11.24 +
× 11.24
611.37 =
× 11.24
= 54.4 
This is the value of the heat capacity of the Calorimeter.
<span>Answer:
It depends on what came after "0.5440 M H...".
If it was a monoprotic acid, like HCl, the calculation would go like this:
(55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) =
122.7 mL KOH
If it was a diprotic acid, like H2SO4, like this:
(55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) =
245.4 mL KOH
If it was a triprotic acid, like H3PO4, like this:
(55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) =
368.0 mL KOH</span>