Answer:
b) 3.10
Explanation:
HF ⇄ H
+ + F
Using Henderson-Hasselbalch Equation:
pH = pKa + log [A-]/[HA].
Where;
pKa = Dissociation constant = -log Ka
Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266
[A-] = concentration of conjugate base after dissociation = moles of base/total volume
= 0.15 x 0.3/0.8
= 0.05625 M
[HA] = concentration of the acid = moles of acid/total volume
= 0.10 x 0.5/0.8
= 0.0625 M
Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>
pH = 3.14266 + log [0.05625/0.0625]
= 3.14267 + (-0.04575749056)
= 3.09691250944
<em>From all the available options below:</em>
<em>a) 2.97
</em>
<em>b) 3.10
</em>
<em>c) 3.19
</em>
<em>d) 3.22
</em>
<em>e) 3.32</em>
The correct option is b.
Answer:
4.) 9, 1, and 4 5.) 4, 1, and 4
Explanation:
I am not quite sure about this because I cannot remember if the coefficient (the number before the elements) is applied to every element in the compound. If it is then your number of atoms are as follows: CORRECTION: you do not have to apply the coefficient to every element only the one that is after it. So when you back and fix the error your number of atoms will be as follows:
number 4
H: 9
P: 1
O: 4
number 5:
H: 4
S: 1
O: 4
you can calculate the number of atoms present in this compound by multiplying the coefficient and the subscripts of each atom.
hope this helped you :)
Answer:
The traditional electrolyte for aluminium electrolysis is based on molten cryolite (Na3AlF6), acting as solvent for the raw material, alumina (Al2O3).Metals are found in ores combined with other elements. Electrolysis can be used to extract a more reactive metal from the ore.
Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions. As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere. In an aqueous solution, if the potential is high enough, OH− and O2− are generated at the anode, which can then react with the aluminum to produce aluminum oxide. Al^3+ can also be generated directly. The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the Al^3+ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode. In this way, coatings thicker than the normal passivation in air can be produced. However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made). This is the normal behaviour in aqueous solutions at near-neutral pH (5–7).
However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface. One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed OH−, leaving pores in the oxide layer through which the ions can travel and continue to react. These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.
Global winds, coriolis effect and <span>continental deflections. Hope this helps!</span>
In the physical properties, it is mentioned that the element has 4 valence electrons. The elements in the periodic table are arranged such that, the number of valence electrons present in the neutral atoms belonging to a particular group is equal to the group number.
Thus, the unidentified element can be best classified as a nonmetal in period 4
Ans C)