Answer:
14 mol e⁻
Explanation:
Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese
8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)
Step 2: Calculate the moles corresponding to 110 g of manganese
The molar mass of Mn is 55 g/mol.
110 g × 1 mol/55 g = 2 mol
Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn
According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.
2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻
526 L O2 x 1 mol O2 / 22.4 L = 23.5 mol O2
Answer:
D. 34.5g
Explanation:
Using the following formula to calculate the mass of 1.5moles of sodium (Na);
Mole = mass/molar mass
Molar mass of Na = 23g/mol
mass = mole × molar mass
Mass of Na = 1.5mol × 23g/mol
Mass of Na = 34.5g
Answer:
Bromine displaced iodide to form iodine
Explanation:
In this experiment, liquid bromine was allowed to react with a solution of potassium iodide.
Initially, we had a colorless 
solution which was mixed with light orange 
. If we add a light orange solution to a colorless solution, liquid bromine will get diluted and we'll see a slight decrease in its color intensity, however, if no reaction takes place, no color change will occur.
In the experiment, however, we examine a formation of a deep brown solution. Remember that iodine solution would form a brown solution.
As a result, bromine displaced iodide anion from potassium iodide:
