Answer:
gde
Explanation:
We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.
Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.
6.022×10^23 should be correct. Are there any options to choose from?
<u>Avogadros number</u>
Mole ratio :
<span>5 C</span>₆<span>H</span>₆<span>CHO + 2 KMnO</span>₄<span> + 6 H</span>⁺ <span>= 5 C</span>₆<span>H</span>₆<span>COOH + 2 Mn</span>²⁺<span> + 3 H</span>₂<span>O + 2 K</span>⁺
5 moles C₆H₆CHO ------------------ 2 moles KMnO₄
1.0 moles C₆H₆CHO ---------------- ( moles of KMnO₄ )
moles of KMnO₄ = 1.0 x 2 / 5
moles of KMnO₄ = 2 / 5
= 0.40 moles of KMnO4
hope this helps!
Answer:
About 1.48 M.
Explanation:
The formula for molarity is mol/L.
So firstly, you must find the amount of moles in 250 grams of NaCl.
I do this by using stoichiometry. First, I find how nany grams are in a single mole of NaCl. This is around 58.44 grams/mole. Now that I know this, I can now use a stoich table. (250 g NaCl * 1 mol NaCl / 58.44 g NaCl). I plug this into my calculator.
I get that 250 grams of NaCl is equal to about 4.28 moles.
Now I just plug into the formula!
4.28 moles/2.9 L = about 1.48
<em><u>I've</u></em><em><u> </u></em><em><u>attached</u></em><em><u> </u></em><em><u>a </u></em><em><u>picture </u></em><em><u>of </u></em><em><u>my </u></em><em><u>personal </u></em><em><u>notes </u></em><em><u>below </u></em><em><u>which </u></em><em><u>shows </u></em><em><u>work </u></em><em><u>I </u></em><em><u>have </u></em><em><u>done</u></em><em><u> </u></em><em><u>in </u></em><em><u>similar </u></em><em><u>equations.</u></em>
1, neutralization of hydrochloric acid with sodium hydroxide
<span>,2. balanced </span>
<span>3. reactants are sodium hydroxide also called lye and hydrochloric acid which as a gas is hydrogen chloride, also stomach acid </span>
<span>products are water and sodium chloride, also called table salt</span>