<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.
<u>Explanation:</u>
Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.
The given redox reaction follows:
To balance the given redox reaction in basic medium, there are few steps to be followed:
- Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons
Oxidation half-reaction: 
Reduction half-reaction: 
- Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions
Oxidation half-reaction: ( × 3)
Reduction half-reaction: ( × 2)
The half-reactions now become:
Oxidation half-reaction: 
Reduction half-reaction: 
- Add the equations and simplify to get a balanced equation
Overall redox reaction: 
As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution
Answer:
N- 1s2 2s2 2p3
Mg- 1s2 2s2 2p6 3s2
O- 1s2 2s2 2p4
F- 1s2 2s2 2p5
Al-1s2 2s2 2p6 3s2 3p1
Explanation:
Order of decreasing atomic radius
Mg,Al, N,O,F
Order of increasing ionization energy
Mg,Al, N,O,F
Reason:
Atomic radius decreases with increase in nonmetallic character. Looking at the electronic configurations, as effective nuclear charge increases, the atom becomes smaller and the attractive force between the nucleus and the outermost electrons increases. Hence, the radius of the atom decreases and ionization energy increases. Note that the addition of more orbital electrons implies addition of more nuclear charge since the both must exactly balance for the atom to remain electrically neutral. The more the electrons in the outermost shell, the higher the first ionization energy.
Answer:
CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Explanation:
To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:
pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2
Also, Kw / Kb = Ka
Thus:
pKa of CH3CH2NH3+/CH3CH2NH2 is:
Kw / kb = Ka = 1.79x10⁻¹¹
-log Ka = pKa
pKa = 10.75
pKa of C6H5NH3+/ C6H5NH2 is:
Kw / kb = Ka = 2.5x10⁻⁵
-log Ka = pKa
pKa = 4.6
That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
First of all, the problem says that you have to convert the acid salt to its acidic form. If you take it to the neutral pH, that won't be acidic at all. As simple as that, you don't take it to neutral pH because it would lose its definition of being acidic afterall.
