Answer:
2.90 x 10⁻¹¹moldm⁻³
Explanation:
Given parameters:
[H⁺] = 3.5 x 10⁻⁴mol/dm³
Unknown
[OH⁻] = ?
Solution;
The ionic product of water can be used to solve this problem. It has been experimentally determined to be 1 x 10⁻¹⁴mol² dm⁻⁶
[H⁺] [OH⁻] = 1 x 10⁻¹⁴
Therefore;
[OH⁻] = 
= 
= 0.29 x 10⁻¹⁰moldm⁻³
= 2.90 x 10⁻¹¹moldm⁻³
Number 3. c.
Number 4. a.
Number 5. d.
Use PV =nRT. Rearrange it to n = PV/RT.
P = 202.6 kPa
V = 4.0L
R = 8.314 kPa*L/mol*K
T = 127 °C + 273 = 400 K
Plug it in and solve. I got 0.24 moles of H2.
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Na and Cl,
E.N of Chlorine = 3.16
E.N of Sodium = 0.93
________
E.N Difference 2.23 (Ionic Bond)
For C and O,
E.N of Oxygen = 3.44
E.N of Carbon = 2.55
________
E.N Difference 0.89 (Polar Covalent Bond)
For N and Cl,
E.N of Chlorine = 3.16
E.N of Notrogen = 3.04
________
E.N Difference 0.12 (Non-Polar Covalent Bond)
For B and O,
E.N of Oxygen = 3.44
E.N of Boron = 2.04
________
E.N Difference 1.40 (Polar Covalent Bond)
The answer is B. Chemical change
