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djyliett [7]
3 years ago
6

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.48 L against a constant pressure

of 0.819 atm?
Chemistry
1 answer:
Dahasolnce [82]3 years ago
5 0
Given:
ΔV = 5 - 1.48 L = 3.52 L = 0.00352 m³, th chane in volume
p = 0.818 atm = 0.819*101325 Pa = 82985.2 Pa

By definition, the work done is
W = \int pdV = (82985.2 \,  \frac{N}{m^{2}} )*(0.00352 \, m^{3}) = 292.1 \, J

Answer: 292.1 J
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Using the reagents below, list in order (by letter, no period) those necessary to transform 1-chlorobutane into 1-butyne. Note:
Andru [333]

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.

3 0
3 years ago
What is the number of atoms in a mole of any element?
ozzi

6.022×10^23 should be correct. Are there any options to choose from?


<u>Avogadros number</u>

4 0
3 years ago
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What is the minimum number of moles of potassium permanganate that would be required to completely oxidize 1.0 moles of benzalde
makvit [3.9K]
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!


7 0
3 years ago
What is the molarity of a solution in which 250 grams of NaCl are dissolved in 2.9 L of water?
Maksim231197 [3]

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>

3 0
2 years ago
I really need help with these three questions PLUS 32 PONITS
raketka [301]
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>
8 0
3 years ago
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