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!!!
During combustion, chemical energy stored in fossil fuels is converted into KINETIC energy.
Answer:
Limiting reactant is NiSO₄
Explanation:
The reaction of aluminum metal with aqueous nickel(II) sulfate to produce aqueous aluminum sulfate and nickel is:
2 Al(s) + 3 NiSO₄ → Al₂(SO₄)₃ + 3 Ni
<em>That means 2 moles of Al react with 3 moles of nickel sulfate.</em>
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Moles of Al and NiSO₄ are:
Al: 108g × (1mol / 26.98g) = 4.00 moles of Al
NiSO₄: 464g × (1mol / 154.75g) = 3.00 moles of NiSO₄
For a complete reaction of aluminium there are necessary:
4.00mol Al ₓ ( 3 moles NiSO₄ / 2 moles Al) = 6 moles of NiSO₄
As you have just 3.00 moles of NiSO₄, the <em>limiting reactant is NiSO₄</em>
Answer:
B. a H+ ion is the answer dear.
Explanation:
ㅗㅐㅔㄷ ㅅㅗㅑㄴ ㅗㄷㅣㅔ ㅛㅐㅕ.
Answer:
d = 0.9 g/L
Explanation:
Given data:
Number of moles = 1 mol
Volume = 24.2 L
Temperature = 298 K
Pressure = 101.3 Kpa (101.3/101 = 1 atm)
Density of sample = ?
Solution:
PV = nRT (1)
n = number of moles
number of moles = mass/molar mass
n = m/M
Now we will put the n= m/M in equation 1.
PV = m/M RT (2)
d = m/v
PM = m/v RT ( by rearranging the equation 2)
PM = dRT
d = PM/RT
The molar mass of neon is = 20.1798 g/mol
d = 1 atm × 20.1798 g/mol / 0.0821 atm. L/mol.K × 273K
d = 20.1798 g/22.413 L
d = 0.9 g/L
