<span>a. 0.325 g / 63.55 g/mol = 5.11 X 10^-3 moles Cu. SHould form 5.11 X 10^-3 mol Cu2+
b. Should form 5.11 X 10^-3 mol Cu(OH)2
c. 1 g Zn / 65.4 g/mol = 0.0153 mol Zn
Excess Zn = 0.0153 - 0.0051 = 0.0102 moles excess zinc
d. 5.11 X 10^-3 mol Mg X 24.3 g/mol = 0.124 grams Mg</span>
Answer : The volume in mL of the sodium carbonate stock solution is 364 mL.
Explanation :
According to dilution law:
where,
= molarity of aqueous sodium carbonate
= molarity of aqueous sodium carbonate stock solution
= volume of aqueous sodium carbonate
= volume of aqueous sodium carbonate stock solution
Given:
= 1.00 M
= 1.58 M
= 575 mL
= ?
Now put all the given values in the above formula, we get:
Therefore, the volume in mL of the sodium carbonate stock solution is 364 mL.
The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
D 4 grams
i need to write a couple more character for explanation so there
N2 + 3 H2 >> 2 NH3
moles NH3 = 11.50 g /17.0307 g/mol=0.6753
the ratio between H2 and NH3 is 3 : 2
moles H2 needed = 0.6753 x 3/2 =1.013
mass H2 = 1.013 mol x 2.106 g/mol=2.042 g
