11.2
Step-by-step-explanation
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:
1.18 × 10²⁴ particles Mg
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
47.7 g Mg
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of Mg - 24.31 g/mol
<u>Step 3: Convert</u>
<u />
= 1.18161 × 10²⁴ particles Mg
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
1.18161 × 10²⁴ particles Mg ≈ 1.18 × 10²⁴ particles Mg
<u>Answer:</u> The standard potential of the cell is 0.77 V
<u>Explanation:</u>
We know that:

The substance having highest positive
reduction potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u>
( × 2)
To calculate the
of the reaction, we use the equation:

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation follows:

Hence, the standard potential of the cell is 0.77 V