Answer:
a. pH = 13.50
b. pH = 13.15
Explanation:
Hello!
In this case, since the undergoing chemical reaction between KOH and HBr is:

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:
![pOH=-log([OH^-])=-log(0.320)=0.50](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.320%29%3D0.50)
Thus, the pH is:

Which is the same answer for a and b as they ask the same.
Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):
![[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BKOH%5D%3D%5Cfrac%7B0.00425mol%7D%7B0.03000L%7D%3D0.142M)
So the pOH and the pH turn out:

Best regards!
Answer:
10.5L
Explanation:
The volume in this question can be calculated by using the formula for gas law equation as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the provided information in this question:
P = 6.18 atm
V = ?
n = 2.35 moles
R = 0.0821 Latm/molK
T = 63°C = 63 + 273 = 336K
Using PV = nRT
V = nRT/P
V = 2.35 × 0.0821 × 336/ 6.18
V = 64.83/6.18
V = 10.49
V = 10.5L
Since a pH of 3 is three numbers higher than a pH of 6, we can find the change in acidity by taking 10 to the third power. The solution with a pH of 3 is 1000 times more acidic than the solution with a pH of 6.
Spectroscopy be used to distinguish between the following is the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.
<h3>What is spectroscopy?</h3>
Spectroscopy is the study of emission or absorption of light. It is used to study the structure of atoms and molecules.
The three types of spectroscopy are:
- atomic absorption spectroscopy (AAS)
- atomic emission spectroscopy (AES)
- atomic fluorescence spectroscopy (AFS)
Thus, the correct option is B, the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.
Learn more about spectroscopy
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Answer:
a) Xbenzene = 0.283
b) Xtoluene = 0.717
Explanation:
At T = 20°C:
⇒ vapor pressure of benzene (P*b) = 75 torr
⇒ vapor pressure toluene (P*t) = 22 torr
Raoult's law:
∴ Pi: partial pressure of i
∴ Xi: mole fraction
∴ P*i: vapor pressure at T
a) solution: benzene (b) + toluene (t)
∴ Psln = 37 torr; at T=20°C
⇒ Psln = Pb + Pt
∴ Pb = (Xb)*(P*b)
∴ Pt = (Xt)*(P*t)
∴ Xb + Xt = 1
⇒ Psln = 37 torr = (Xb)(75 torr) + (1 - Xb)(22 torr)
⇒ 37 torr - 22 torr = (75 torr)Xb - (22 torr)Xb
⇒ 15 torr = 53 torrXb
⇒ Xb = 15 torr / 53 torr
⇒ Xb = 0.283
b) Xb + Xt = 1
⇒ Xt = 1 - Xb
⇒ Xt = 1 - 0.283
⇒ Xt = 0.717