•3.9g of ammonia
•molar mass of ammonia = 17.03g/mol
1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:
(3.9 g)/ (17.03g/mol) = 0.22900763mols
Then convert the moles to molecules by multiplying it with Avogadro’s number:
Avogadro’s number: 6.022 x 10^23
0.22900763mols x (6.022 x 10^23 molecs/mol)
= 1.38 x 10^23 molecules
The matter will be consumed by other living organisms and the blood will settle to the bottom of the body
Answer:
0.22 mol HClO, 0.11mol HBr.
0.25mol NH₄Cl, 0.12 mol HCl
Explanation:
A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.
Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.
Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em> will produce NH₄⁺. 0.25mol HCl consume all NH₃.
The pH of a 0.0115 m aqueous formic acid solution is mathematically given as
pH=2.8424
This is further explained below.
<h3>What is the ph of a 0.0115 m aqueous formic acid solution?</h3>
Generally, the equation for the chemical equation is mathematically given as
HCOOH H^+ + HCOO


![&\left[\mathrm{H}^{+}\right]=\mathrm{C \alpha}\\\\&=0.125 \times 0.0115\\\\&=1.4375 \times 10^{-9}\\\\&P=-\log \left[H^{+}\right]\\\\&=-\log \left[1.4375 \times 10^{-3}\right]\\\\&P H=2.8424](https://tex.z-dn.net/?f=%26%5Cleft%5B%5Cmathrm%7BH%7D%5E%7B%2B%7D%5Cright%5D%3D%5Cmathrm%7BC%20%5Calpha%7D%5C%5C%5C%5C%26%3D0.125%20%5Ctimes%200.0115%5C%5C%5C%5C%26%3D1.4375%20%5Ctimes%2010%5E%7B-9%7D%5C%5C%5C%5C%26P%3D-%5Clog%20%5Cleft%5BH%5E%7B%2B%7D%5Cright%5D%5C%5C%5C%5C%26%3D-%5Clog%20%5Cleft%5B1.4375%20%5Ctimes%2010%5E%7B-3%7D%5Cright%5D%5C%5C%5C%5C%26P%20H%3D2.8424)
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