Answer:
Two non bonded electron pairs and four bonded electron pairs
Explanation:
An image of the compound as obtained from chemlibretext is attached to this answer.
The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.
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 first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar
Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691
<u>Answer:</u>
P2 = 778.05 mm Hg = 1.02 atm
<u>Explanation:</u>
We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.
For this, we would use the equation:

where P1 = 724 mm hg
V1 = 3.05 L
T1 = 298 K
P2 = ?
V2 = 2.6 L
T2 = 173 K
Substituting the given values in the equation to get:

P2 = 778.05 mm Hg = 1.02 atm
<u>Answer:</u> The new concentration of lemonade is 3.90 M
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of lemonade solution = 2.66 M
Volume of solution = 473 mL
Putting values in equation 1, we get:

Now, calculating the new concentration of lemonade by using equation 1:
Moles of lemonade = 1.26 moles
Volume of solution = (473 - 150) mL = 323 mL
Putting values in equation 1, we get:

Hence, the new concentration of lemonade is 3.90 M