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
Kf is large, so x ≪ 6.106×10⁻³. Then
Answer:
Be the anode
Explanation:
The standard hydrogen electrode is regarded as the standard reference electrode and it has been assigned an electrode potential of 0.0V.
If any substance has an electrode potential that is more negative than hydrogen, then that half cell will function as the anode when connected to the standard hydrogen electrode.
Similarly, any substance that has a more positive electrode potential than hydrogen will serve as the cathode when its half cell is connected to the standard hydrogen electrode.
Answer:
Mg(s) + 2NaF (aq) —> MgF₂ (aq) + 2Na(s)
Explanation:
Magnesium => Mg
Sodium fluoride => NaF
Magnesium fluoride => MgF₂
Sodium metal => Na
The equation can be written as:
Mg + NaF —> MgF₂ + Na
Thus, the above equation can be balance as illustrated below:
Mg + NaF —> MgF₂ + Na
There are 2 atoms of F on the right side and 1 atom on the left side. It can be balance by writing 2 before NaF as shown below:
Mg + 2NaF —> MgF₂ + Na
There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by writing 2 before Na as shown below:
Mg + 2NaF —> MgF₂ + 2Na
Mg(s) + 2NaF (aq) —> MgF₂ (aq) + 2Na(s)
Now, the equation is balanced.
C10H8N2O2S2
C=12,H=1,N=14,O=16,S=32
R. F. M (Relative Formula Mass)=(12*10)+8+(2*14)+(2*16)+(2*32)
Number of moles =given mass/formula mass
=29/252
=0.1151 moles