There it is, I hope it gets to be helpful
Answer:
pH 8.89
Explanation:
English Translation
If the MgCl₂ solution of 0.2 M has its pH raised by adding NH₄OH, the precipitate will begin to form at a pH of approximately.
Given the solubility product (Ksp) of Mg(OH)₂ = 1.2 x 10⁻¹¹
Assuming all of the salts involved all ionize completely
MgCl₂ ionizes to give Mg²⁺ and Cl⁻
MgCl₂ ⇌ Mg²⁺ + 2Cl⁻
1 mole of MgCl₂ gives 1 moles of Mg²⁺
Since the concentration of Mg²⁺ is the same as that of MgCl₂ = 0.2 M
Mg(OH)₂ is formed from 1 stoichiometric mole of Mg²⁺ and 2 stoichiometric moles of OH⁻
Ksp Mg(OH)₂ = [Mg²⁺][OH⁻]²
(1.2 x 10⁻¹¹) = 0.2 × [OH⁻]²
[OH⁻]² = (6×10⁻¹¹)
[OH⁻] = √(6×10⁻¹¹)
[OH⁻] = 0.000007746 M
p(OH) = - log [OH⁻] = - log (0.000007746)
pOH = 5.11
pH + pOH = 14
pH = 14 - pOH = 14 - 5.11 = 8.89
Hope this Helps!!!
Answer:
The answer to your question is given below
Explanation:
1. C. 2NaCl + I2 —> 2NaI + Cl2 => C. Single displacement.
From the above equation, we can see that I2 replaces Cl in NaCl to produce NaI. This is simply called a single displacement reaction.
2. E. 2C4H10 + 13O2 —> 8CO2 + 10H2O => E. Combustion.
The above equation shows the burning of Hydrocarbon in the presence of O2. This is simply called Combustion as CO2 and H2O is produced.
3. D 2H2O —> 2H2 + O2 => D. Decomposition.
From the above equation, we can see that a single compound H2O produces two elements H2 and O2. This is simply called a decomposition reaction.
4. A. ZnS + 2HCl —> ZnCl2 + H2S => A. Double Decomposition.
From the above equation, we can see that Cl replaces S in ZnS to produce ZnCl2 and S replaces Cl in HCl to produce H2S. This is simply called double displacement reaction.
5. B. H2 + Br2 —> 2HBr => B. Synthesis.
From the above equation, we can see that two element H2 and Br2 combine to produce a single compound HBr. This is simply called a synthesis reaction.
Answer:
1) 90.0 mL
2) 11.25 M
3) 0.477 M
4) 144 mL
Explanation:
The main formula that will be used for all these calculations is:
C₁V₁ = C₂V₂
C stands for concentration and V stands for volume and the subscripts 1 and 2 indicate an initial concentration or volume and a final concentration or volume.
For each problem, it's best to start by figuring out what you have and what you need to find. Figure out if you're looking for an initial value or a final value.
1) We need to find the initial volume. So, take what values you have and plug them in and then solve for whatever variable:
5.00 M · V₁ = 500.0mL · 0.900 M - divide by 5.00
C₁ = 90.0 mL
2) This time we're finding the initial concentration:
20.0mL · C₁ = 150.0mL · 1.50 M - divide by 20.0mL
C₂ = 11.25 M
3) Now we're finding the final concentration:
12.00mL · 3.50 M = 88.0mL · C₂ - divide by 88.0mL
C₂ = 0.477 M
4) Finally, we're looking for the final volume:
9.0mL · 8.0 M = 0.50 M · V₂ - divide by 0.50 M
V₂ = 144mL
