Answer:
Explanation:
Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.
1. Moles of Ni²⁺
2. Moles of NH₃
3. Initial concentrations after mixing
(a) Total volume
V = 135 mL + 190 mL = 325 mL
(b) [Ni²⁺]
(c) [NH₃]
3. Equilibrium concentration of Ni²⁺
The reaction will reach the same equilibrium whether it approaches from the right or left.
Assume the reaction goes to completion.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0
C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0
E/mol·L⁻¹: 0 0.1095 6.106×10⁻³
Then we approach equilibrium from the right.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 0 0.1095 6.106×10⁻³
C/mol·L⁻¹: +x +6x -x
E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x
![K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bf%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BNi%28NH%24_%7B3%7D%24%29%24_%7B6%7D%5E%7B2%2B%7D%24%5D%7D%7D%7B%5Ctext%7B%5BNi%24%5E%7B2%2B%7D%24%5D%7D%5Ctext%7B%5BNH%24_%7B3%7D%24%5D%7D%5E%7B6%7D%7D%20%3D%202.0%20%5Ctimes%2010%5E%7B8%7D)
Kf is large, so x ≪ 6.106×10⁻³. Then
![K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bf%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BNi%28NH%24_%7B3%7D%24%29%24_%7B6%7D%5E%7B2%2B%7D%24%5D%7D%7D%7B%5Ctext%7B%5BNi%24%5E%7B2%2B%7D%24%5D%7D%5Ctext%7B%5BNH%24_%7B3%7D%24%5D%7D%5E%7B6%7D%7D%20%3D%202.0%20%5Ctimes%2010%5E%7B8%7D%5C%5C%5C%5C%5Cdfrac%7B6.106%20%5Ctimes%2010%5E%7B-3%7D%7D%7Bx%5Ctimes%200.1095%5E%7B6%7D%7D%20%3D%202.0%20%5Ctimes%2010%5E%7B8%7D%5C%5C%5C%5C6.106%20%5Ctimes%2010%5E%7B-3%7D%20%3D%202.0%20%5Ctimes%2010%5E%7B8%7D%5Ctimes%200.1095%5E%7B6%7Dx%3D%20345.1x%5C%5Cx%3D%20%5Cdfrac%7B6.106%20%5Ctimes%2010%5E%7B-3%7D%7D%7B345.1%7D%20%3D%201.77%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5C%5Ctext%7BThe%20concentration%20of%20Ni%24%5E%7B2%2B%7D%24%20at%20equilibrium%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.77%20%5Ctimes%2010%5E%7B-5%7D%7D%5Ctextbf%7B%20mol%2FL%7D%7D%24%7D)
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
Answer:
Yes chemistry. Try to add then multiply the top. Get the moles and you will find it.
Explanation:
Try to add then multiply the moles in the equation
We need to find the number of moles that each of the given substances. The molar mass of magnesium is 24.3 g/mol and Bromine has a molar mass of 79.90 g/mol. From the given mass of each species, the moles of Mg and Br are 0.018 and 0.369 respectively. There are 2 Br in one Mg. The sample is MgBr2.
