Answer:
C.) HOCl Ka=3.5x10^-8
Explanation:
In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below
we Know that
pKa= -log(Ka)
therefore
A) pKa of HClO2 = -log(1.2 x 10^-2)
=1.9208
B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644
C) pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45
D) pKa of HCN = -log(4 x 1 0^-10)= 9.3979
If we consider the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution
The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.
So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.
Hence, HOCl will be chosen for buffer construction.
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
I think it's 2 I tried looking it up because I was not sure.
Use the clapeyron equation:
T in kelvin : 6.80 + 273 => 279.8 K
R = 0.082
n = 71.5 moles
P = 5.03 atm
Therefore:
P x V = n x R x T
5.03 x V = 71.5 x 0.082 x 279.8
5.03 x V = 1640.4674
V = 1640.4674 / 5.03
V = 326.13 L
hope ths helps!
Answer:
every method of removing heat from LED's should be considered. Conduction, convection, and radiation are the three means of heat transfer. Typically, LED's are encapsulated in a transparent resin, which is a poor thermal conductor. Nearly all heat produced is conducted through the back side of the chip. Heat is generated from the PN junction by electrical energy that was not converted to useful light, and conducted to outside ambiance through a long path, from junction to solder point, solder point to board, and board to the heat sink and then to the atmosphere. A typical LED side view and its thermal model are shown in the figures.
Explanation:
