Answer:
The PH of the mixture is 4.74
Explanation:
The number of millimoles of acetic acid is calculated using the formula:
No of millimoles= Molarity * Volume( in ml)
= 0.25M * 30ml = 7.5 moles
Number of millimoles of KOH is calculated using:
Number of millimoles = Molarity * Volume ( in ml)
=0.05M * 75ml
= 3.75 moles
The PH of the solution is derived using:
pH = pKa + log [salt] / acid
= ![-log [ 1.8 * 10^5 ] + log [ 3.75 mmoles/ 3.75 mmoles]](https://tex.z-dn.net/?f=%20-log%20%5B%201.8%20%2A%2010%5E5%20%5D%20%2B%20log%20%5B%203.75%20mmoles%2F%203.75%20mmoles%5D%20)
=4.74
Explanation:
Lone pairs are present on nitrogen atoms of both pyridine and pyrrole.
In pyrrole, the electron pair on nitrogen is involved in resonance and hence does not available for bonding. Therefore, pyrrole does not possess basic character and remains unreactive towards HCl.
In case of pyridine, as 6 pi electrons are present in the ring, so, electron pair of nitrogen does not participate in resonance. So, because of presence of lone pair, pyridine possess basic character and hence, reacts readily reacts with HCl.
Answer:
Neutral (no net charge)
Explanation:
Protons - Positive
Electrons - Negative
Neutrons - No charge
<em>Why? </em>
Neutrons are the particle in an atom with a (as the name suggests) neutral charge. They are neither positive nor negative, unlike protons or electrons.
Answer:
209.98 g of NaOH
Explanation:
We are given;
- Volume of HCl as 3 L
- Molarity of HCl as 1.75 M
We are required to calculate the mass of NaOH required to completely neutralize the acid given.
First, we write a balanced equation for the reaction between NaOH and HCl
That is;
NaOH + HCl → NaCl + H₂O
Second, we determine the number of moles of HCl
Number of moles = Molarity × Volume
= 1.75 M × 3 L
= 5.25 moles
Third, we use the mole ratio to determine the moles of NaOH
From the reaction,
1 mole of NaOH reacts with 1 mole of HCl
Therefore;
Moles of NaOH = Moles of HCl
= 5.25 moles
Fourth, we determine the mass of NaOH
Molar mass of NaOH = 39.997 g/mol
Mass of NaOH = 5.25 moles × 39.997 g/mol
= 209.98 g
Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl
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