Explanation:
pOH = ? (Data) molarity = 2.1 x 10 -6 M (formula) pOH = log /[OH] substitution pOH = -log [2.1 x 10-6] pOH = 5.7 and pH =14 - pOH ph -14 -5.7 pH = 8.3
Answer:
A) 31.22
Explanation:
The reaction of sulfuric acid with NaOH is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O
To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.
<em>Moles H₂SO₄ (Molar mass: 98.08g/mol):</em>
0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =
3.7289x10⁻³moles H₂SO₄
And moles of NaOH that you require to neutralize the acid are:
3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =
7.4578x10⁻³ moles NaOH
Using a 0.2389M NaOH solution:
7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL
Right answer is:
<h3>A) 31.22
</h3>
he would not be able to tell the world about the presence of positive charges that is protons, in the center of an atom.
Answer:
a) The flame test of Na, at 589nm will show a golden yellow color when the sodium ion is emitted. These is due to the electropositive nature of the alkali metals.
While the flame test for K at 404nm will ignite a violet color when the Potassium ion is emitted. This properties are due to the high electropositive nature of the group1 elements which also indicate their strong reducing agent.
b) The cobalt glass filter act as hindrance during the flame test to seperate or filter the golden yellow color caused as a result of the presence of the sodium, as it makes the violet color to be more visible.
c) These is due to the Oxidizing ability of KClO₄ or KClO₃ compared to the salts of sodium. Also is the low solubility of the two salts and their solubility constant (Ksp) compared to sodium salts.
Explanation:
a) The flame test of Na, at 589nm will show a golden yellow color when the sodium ion is emitted. These is due to the electropositivity nature of the alkali metals.
While the flame test for K at 404nm will ignite a violet color when the Potassium ion is emitted. This properties are due to the high electropositivity nature of the group1 elements which also indicate their strong reducing agent.
b) The cobalt glass filter act as hindrance during the flame test to seperate or filter the golden yellow color caused as a result of the presence of the sodium, as it makes the violet color to be more visible.
c) These is due to the Oxidizing ability of KClO₄ or KClO₃ compared to the salts of sodium. Also is the low solubility of the two salts and their solubility constant (Ksp) compared to sodium salts.