Answer:
0.55 mol Au₂S₃
Explanation:
Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.
1. Gather all the information in one place:
M_r: 34.08
Au₂S₃ + … ⟶ 3H₂S + …
m/g: 56
2. Calculate the moles of H₂S
Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)
= 1.64 mol H₂S
3. Calculate the moles of Au₂S₃
The molar ratio is 1 mol Au₂S₃/3 mol H₂S.
Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)
= 0.55 mol Au₂S₃
<h2>Question:- </h2>
A solution has a pH of 5.4, the determination of [H+].
<h2>Given :- </h2>
- pH:- 5.4
- pH = - log[H+]
<h2>To find :- concentration of H+</h2>
<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>
<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>
Take negative to other side
-pH = log H+
multiple Antilog on both side
(Antilog and log cancel each other )
Antilog (-pH) = [ H+ ]
New Formula :- Antilog (-pH) = [+H]
Now put the values of pH in new formula
Antilog (-5.4) = [+H]
we can write -5.4 as (-6+0.6) just to solve Antilog
Antilog ( -6+0.6 ) = [+H]
Antilog (-6) × Antilog (0.6) = [+H]
put the value in equation
![{10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]](https://tex.z-dn.net/?f=%20%7B10%7D%5E%7B%20-%206%7D%20%20%20%5Ctimes%204%20%3D%20%5BH%2B%5D%20%5C%5C%204%20%5Ctimes%20%20%20%7B10%7D%5E%7B%20-%206%7D%20%20%3D%20%5BH%2B%5D)
Answer: the concentration of H in the stomach acid is greater than that of the lemon juice
Explanation:Please see attachment for explanation
The correct answer is Applied Biochemistry.
In applied biochemistry the knowledge and methods of biochemistry is applied to solve real world problems like to discover effective medicine in the treatment of life threatening diseases such as cancer, to improve productivity in agriculture, to treat diseases caused by the mutation in the metabolic pathway and more.
