Answer:
Explanation:
Given parameters:
pH = 3.50
Unknown:
concentration of [H₃0⁺] = ?
concentration of [OH⁻] = ?
Solution:
In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.
pH = -log₁₀[H₃O⁺]
[H₃O⁺] = inverse log₁₀ (-pH) = 
= 
[H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³
For the [OH⁻]:
we use : pOH = -log₁₀ [OH⁻]
Recall: pOH + pH = 14
pOH = 14 - pH = 14 - 3.5 = 10.5
Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]
[OH⁻] = 
[OH⁻] = 
= 3.2 x 10⁻¹¹moldm⁻³
The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.
Answer:
A. The balloons will increase to twice their original volume.
Explanation:
Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:
P ∝ 1/V
P = k/V
PV = k (constant)
P = pressure, V = volume.
Let the initial pressure of the balloon be P, i.e. 
, initial volume be V, i.e. 
. The pressure is then halved, i.e. 
Therefore the balloon volume will increase to twice their original volume.
Answer:
44.9g
Explanation:
You have to convert grams of CH4 to moles, use the mole-to-mole ratio of CH4 to water, and convert back to grams.
(20.0g CH4)(1 mol CH4/16.04g)(2 mol H2O/1 mol CH4)(18.01 g H2O/ 1 mol) = 44.9127 g
Hope this helps!
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>
<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
<h2>pH = 8.34</h2>
Answer:
Because sodium and potassium are very reactive metals so they react explosively with HCL and H2SO4 evolving a large amount of heat.
Explanation:
