Answer:
(i). C6H2COOH and Na2CO3(aq)
observation: <u>Bubbles</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>colourless</u><u> </u><u>gas</u><u> </u><u>(</u><u>carbon</u><u> </u><u>dioxide</u><u> </u><u>gas</u><u>)</u>
(ii) CH3CH2CH2OH and KMnO4 /H
observation: <u>The</u><u> </u><u>orange</u><u> </u><u>solution</u><u> </u><u>turns</u><u> </u><u>green</u><u>.</u>
[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>oxidation</em><em> </em><em>of</em><em> </em><em>propanol</em><em> </em><em>to</em><em> </em><em>propanoic</em><em> </em><em>acid</em><em> </em><em>occurs</em>]
(iii) CH3CH2OH and CH3COOH + conc. H2SO4
observation: <u>A</u><u> </u><u>sweet</u><u> </u><u>fruity</u><u> </u><u>smell</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>
[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>an</em><em> </em><em>ester</em><em>,</em><em> </em><em>diethylether</em><em> </em><em>is</em><em> </em><em>formed</em><em>]</em>
(iv) CH3CH = CHCH3 and Br2 /H2O
observation: <u>a</u><u> </u><u>brown</u><u> </u><u>solution</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>
+2
Barium has a positive charge of 2
The answer is 100 m/s = 223.694. %100(not part of question)
Answer:
b. 10 mL
Explanation:
First we <u>calculate the amount of H⁺ moles in the acid</u>:
- [H⁺] =
100 mL ⇒ 100 / 1000 = 0.100 L
- 1x10⁻⁵M * 0.100 L = 1x10⁻⁶ mol H⁺
In order to have a neutral solution we would need the same amount of OH⁻ moles.
We can use the pOH value of the strong base:
Then we <u>calculate the molar concentration of the OH⁻ species in the basic solution</u>:
- [OH⁻] =
= 1x10⁻⁴ M
If we use 10 mL of the basic solution the number of OH⁻ would be:
10 mL ⇒ 10 / 1000 = 0.010 L
- 1x10⁻⁴ M * 0.010 L = 1x10⁻⁶ mol OH⁻
It would be equal to the moles of H⁺ so the answer is b.
