Yes. bromine and sodium iodide can react to form sodium bromine and free iodine
Transition metals are from group 3 to group 12.
Answer:
Explanation:
As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)
ΔG = ΔG° + RTInQ
Q = 1
ΔG = ΔG°
ΔG = =nFE°
n=no of electrons transfered.
E° = 1.1v
ΔG° = -2 * 96500 * 1.10
= -212300J
ΔG° =-212.3kJ/mol
<h3>Therefore, the ΔG° = -212.3kJ/mol</h3>
Answer:
Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)
Explanation:
First, we will write the molecular equation because it is the easiest to balance.
FeCl₃(aq) + 3 KOH(aq) → Fe(OH)₃(s) + 3 KCl(aq)
The full ionic equation includes all the ions and the molecular species.
Fe³⁺(aq) + 3 Cl⁻(aq) + 3 K⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s) + 3 K⁺(aq) + 3 Cl⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)
Answer:
Percentage mass of copper in the sample = 32%
Explanation:
Equation of the reaction producing Cu(NO₃) is given below:
Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)
From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.
Molar mass of copper = 64 g/mol
mass of copper = number of moles * molar mass
mass of copper = 0.01 mol * 64 g/mol = 0.64 g
Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%
Percentage mass of copper in the sample = 32%
