Estructura molecular del óxido nitroso

What is the major organic product obtained from the sequence of reactions 2-phenyl-4 bomobutane and NaN3?

Anna11 [10]

Answer:

(E)-1-phenylbut-1-ene

Explanation:

2-phenyl-4 bromobutane is an amphetamine that contains a phenyl group. It forms a major stable product with other reacting agents.

The major organic product that is obtained from the sequence of the reactions of the 2-phenyl-4 bomobutane when it reacts with $NaN_3$ is the (E)-1-phenylbut-1-ene.

Thus the answer is 2-phenyl-4 bromobutane is an amphetamine that contains a phenyl group. It forms a major stable product with other reacting agents.

The major organic product that is (E)-1-phenylbut-1-ene.

3 0

3 years ago

K + Li2O what is the product

shutvik [7]

Answer:

2K + Li2O → 2Li + K2O

3 0

3 years ago

Methods: Part A: Preparation of Buffers Make two buffers starting with solid material, which is the most common way to make buff

Alecsey [184]

Answer:

0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.

0,856 g of Tris-HCl and 0,553 g of Tris-base

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀ $\frac{A^{-} }{HA}$

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for phosphate buffer is:

pH = 7,21 + log₁₀ $\frac{HPO4^{2-} }{H2PO4^{-} }$

If desire pH is 7,0 you will obtain:

0,617 = $\frac{HPO4^{2-} }{H2PO4^{-} }$ (1)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] (2)

Replacing (1) in (2) you will obtain:

[H₂PO₄⁻] = 0,0618 M

And with this value:

[HPO₄²⁻] = 0,0382 M

As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:

Na₂HPO₄ = 0,1 L× $\frac{0,0382mol}{1L}$ × $\frac{141,96g}{1mol}$ = 0,542 g of Na₂HPO₄

NaH₂PO₄ = 0,1 L× $\frac{0,0618mol}{1L}$ × $\frac{119,96g}{1mol}$ = 0,741 g of NaH₂PO₄

For tris buffer the equilibrium is:

Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075

Henderson–Hasselbalch equation for tris buffer is:

pH = 8,075 + log₁₀ $\frac{Tris-base }{Tris-H^{+} }$

If desire pH is 8,0 you will obtain:

0,841 = $\frac{Tris-base }{TrisH^{+} }$ (3)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [Tris-base] + [Tris-H⁺] (4)

Replacing (3) in (4) you will obtain:

[Tris-HCl] = 0,0543 M

[Tris-base] = 0,0457 M

As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:

Tris-base = 0,1 L× $\frac{0,0457mol}{1L}$ × $\frac{121,1g}{1mol}$ = 0,553 g of Tris-base

Tris-HCl = 0,1 L× $\frac{0,0543mol}{1L}$ × $\frac{157,6g}{1mol}$ = 0,856 g of Tris-HCl

I hope it helps!

8 0

3 years ago

a student balanced the chemical equation for the reaction of magnesium with oxygen by writing Mg + O2 ----> MgO2. What is wro

egoroff_w [7]

Answer:

Explanation:

Magnesium when it oxidizes has a valence of 2.

Oxygen, when it mixes with something, has a valence of - 2

So Mg and O2 will form something, but what? The answer is MgO

Mg + O2 ===> MgO

To balance this equation, you need 2 Magnesiums on the right and 2 Oxygens also on the right. The left will need a two Magnesiums.

The balanced equation will be

2Mg + O2 ====> 2MgO

6 0

4 years ago

Helpppppp please please

polet [3.4K]

What do we do? hehehehe

6 0

3 years ago

Estructura molecular del óxido nitroso​

Estructura molecular del óxido nitroso