Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
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<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
I think the answer is peridotite
Answer: A balloon is charged by a process of frictional charging and the object is getting charged by the process of induction.
Explanation:
When two bodies are rubbed against each other, charging by friction or rubbing occurs. The electropositive object loses electrons to electronegative object. Thus, when balloon is rubbed on a wall, it becomes charged.
The charged balloon is able to attract an uncharged object by inducing charge on it without the two objects touching each other. Electrostatic force acts between two charged objects. Charged balloon causes electrons to move at one end thereby inducing opposite charge in the object and thus, charged balloon is able to attract uncharged object.
Answer: from the lies this is synthesis reaction
Explanation:
It's also oxidation-reduction reaction. Li is oxidised and
H is reduced
Answer:
Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group. ... The valence electrons of the larger atoms are farther from the nucleus and are easier to remove, so the metals near the bottom are more reactive than those at the top.
