<span>One problem with his atomic theory was that it claimed that all matter is composed of tiny indivisible atoms.</span>
There should be 2 since there’s 2 D atoms in the reaction.
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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<u> Increasing pH will increase the solubility of the Hg2(CN)2 by shifting </u><u>equilibrium </u><u>to right side.</u>
What is the meaning of OH in chemistry?
The chemical group, ion, or radical OH that consists of one atom of hydrogen and one of oxygen and is neutral or negatively charged.
Hg2(CN)2 + 2OH- ----> 2HgO(s) + 2HCN
adding OH- to the mercury(l) cyanide will cause the formation of the solid HgO.
therefore increasing pH will increase the solubility of the Hg2(CN)2 by shifting equilibrium to right side.
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Answer:
Kp = 0.049
Explanation:
The equilibrium in question is;
2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)
Kp = p SO₃² / ( p SO₂² x p O₂ )
The initial pressures are given, so lets set up the ICE table for the equilibrium:
atm SO₂ O₂ SO₃
I 3.3 0.79 0
C -2x -x 2x
E 3.3 - 2x 0.79 - x 2x
We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:
p SO₂ = 3.3 -0.47 atm = 2.83 atm
p O₂ = 0.79 - (0.47/2) atm = .56 atm
Now we can calculate Kp:
Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )
Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.
