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.
To know more about solubility, visit the below link:
brainly.com/question/23946616
Answer:
The answer is: <u>Al2O3</u>
Explanation:
The data they give us is:
To find the empirical formula without knowing the grams of the compound, we find it per mole:
- 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
- 0.485 g O * 1 mol O / 16 g O = 0.03 mol O
Then we must divide the results obtained by the lowest result, which in this case is 0.02:
- 0.02 mol Al / 0.02 = 1 Al
- 0.03 mol O / 0.02 = 1.5 O
Since both numbers have to give an integer, multiply by 2 until both remain integers:
Now the answer is given correctly:
Answer:
10.945 x 10^-4
Explanation:
Balanced equation:
Mn(OH)2 + 2 HCl --> MnCl2 + H2O
it takes 2 moles HCL for each mole Mn(OH)2
Next find the molarity of the Mn(OH)2 solution
= (1 mole Mn(OH)2 / 2 mole HCl) X (0.0020 mole HCl / 1000ml) X (4.86 ml)
= 4.86 x 10^-3 mole
this is now dissolved in (70 + 4.86) = 74.86 ml or 0.07486 L
thus [Mn(OH)2] = 4.86 x 10^-3 mole / 0.07486 L = 0.064921 M
Ksp = [Mn2+][OH-]^2 = 4x^3 = 4(0.064921)^3 = 10.945 x 10^-4
Answer:
84.8%
Explanation:
Step 1: Given data
Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.
Na + 1/2 Cl₂ ⇒ NaCl
According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.
Step 2: Calculate the percent yield.
We will use the following expression.
%yield = real yield / theoretical yield × 100%
%yield = 3.45 g / 4.07 g × 100% = 84.8%
Answer:
The answer is given below.
Explanation:
We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.
AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M
HA → H+ + A-
Initial concentration: 0.1 → 3.98 ^10-3 + 0
equilibrium concentration: 0.1(1-α) → 3.98 * 10-3 + 0.1α 0.1α
pKa of chloroacetic acid is 2.9
-log(Ka) = 2.9
Ka = 1.26 * 10-3
From the equation, Ka = [H+] * [A-] / [HA]
1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)
Since α<<1, we assume 1-α = 1
Solving the equation, we have: α = 0.094
Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%
