The compound solubility which will not be affected by a low pH in solution is AgBr.
<h3>What is pH?</h3>
pH is a measure of the acidity or basicity of any solution and according to the pH scale 0 to 6.9 shows the acidity, 7 is neutral and 7.1 to 14 shows the basicity of any solution.
- AgBr is sparingly soluble in water and not soluble in acids, so if we low the pH of the solution towards the acidity its solubility not affected.
- NiCO₃ is a basic salt and and shows solubility in the acidic medium so change in pH will affect its solubility.
- Co(OH)₂ it is also a basic compound and shows its solubility in the acidic medium and get affected when change in pH takes place.
- PbF₂ is a strong base and also shows solubility in the acidic medium easily, so get affected when change in pH takes place.
- In CuS, sulphide is basic ion and whole compound shows solubility in the acidic medium and get affected when low pH of solution takes place.
AgBr is not affected by a low pH in solution.
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Answer: The mass of electrons is mostly ignored because electrons are extremely small compared to neutrons and protons.
Explanation: A proton is about 1,836 times the size of an electron.
On the periodic table, the atomic number for each element can be found. This number is found by measuring the weight of 6.02 x 10^23 atoms of the element in grams. Electrons aren't ignored when finding exact math, but for the sake of simplification high school teachers will generally have you only count the number of protons and neutrons when calculating the mass of atoms.
Answer:
- <u>First choice: 0.042</u>
Explanation:
Given decomposition reaction:
- 1PCl₅ (g) ⇄ 1PCl₃ + 1Cl₂(g)
Equilibrium constant:
Stoichiometric coefficients and powers equal to 1 are not usually shown as they are understood, but I included them in order to shwow you how they intervene in the equilibrium expressions: each concentration is raised to a power equal to the respective stoichiometric coefficient in the equilibrium equation.
So, your calculations are:
