Molybdenum in periodic table
or
Molarity definition
Answer:
400 mL
Explanation:
Given data:
Mass of barium = 2.17 g
Pressure = 748 mmHg (748/760 = 0.98 atm)
Temperature = 21 °C ( 273+ 21 = 294k)
Milliliters of H₂ evolved = ?
Solution:
chemical equation:
Ba + 2H₂O → Ba(OH)₂ + H₂
Number of moles of barium:
Number of moles = mass/ molar mass
Number of moles = 2.17 g / 137.327 g/mol
Number of moles = 0.016 mol
Now we will compare the moles of barium with H₂.
Ba : H₂
1 : 1
0.016 : 0.016
Milliliters of H₂:
PV = nRT
V = nRT/P
V = 0.016 mol × 0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm
V = 0.39 atm. L/0.98 atm
V = 0.4 L
L to mL
0.4 × 1000 = 400 mL
Answer:
2.2 moles of Fe will be produced
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 3.3 moles
Number of moles of iron oxide = 1.5 moles
Step 2: The balanced equation
3H2 + Fe2O3 → 2Fe + 3H2O
Step 3: Calculate the limiting reactant
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
Hydrogen gas is the limiting reactant. It will completely be consumed (3.3 moles). Fe2O3 is in excess. There will react 3.3 / 3 = 1.1 moles
There will remain 1.5 - 1.1 = 0.4 moles Fe2O3
Step 4: Calculate moles Fe
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
For 3.3 moles H2 we'll have 2/3 * 3.3 = 2.2 moles Fe
2.2 moles of Fe will be produced
Answer:
-372000 J or -372 KJ
Explanation:
We have the electrochemical reaction as;
Mg(s) + Fe^2+(aq)→ Mg^2+(aq) + Fe(s)
We must first calculate the E∘cell from;
E∘cathode - E∘anode
E∘cathode = -0.44 V
E∘anode = -2.37 V
Hence;
E∘cell = -0.44 V -(-2.37 V)
E∘cell = 1.93 V
n= 2 since two electrons were transferred
F=96,500C/(mol e−)
ΔG∘=−nFE∘
ΔG∘= -( 2 * 96,500 * 1.93)
ΔG∘= -372000 J or -372 KJ
