Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Answer:
- Acetic acid (CH₃COOH) and hydronium ion (H₃O⁺)
Explanation:
Hello,
In this case, based on the acid-base theory which states that acids are known as H⁺ donors, if we consider the direct reaction:
It is clear that the acetic acid is the first H⁺ donor as it losses one H⁺ to turn into the acetate ion. Moreover, if we consider the inverse reaction:
It is also clear that the hydronium ion is the second H⁺ donor as it losses one H⁺ to turn into water.
Best regards.
Answer: A substance that produces hydroxide ions when placed in water is base.
Explanation:
Bases are the substance:
- Which gives negatively charged hydroxide(
) ions in aqueous solution.
- Which have pH value ranging from 7 to 14.
Where as acid gives positively charged hydronium ion(
) in aqueous solution.
Answer: Metals form cations.
The alkali metals (the IA elements) lose a single electron to form a cation with a 1+ charge.
The alkaline earth metals (IIA elements) lose two electrons to form a 2+ cation.
Aluminum, a member of the IIIA family, loses three electrons to form a 3+ cation.
Therefore, metals in the s and p block of the periodic table have 1, 2 or 3 electrons in their outermost orbit (or valence shell). Now to gain a stable octet metals lose either 1, 2 or 3 electrons from the valence shell thus forming cation with +1, +2 or +3 charge.
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
