Answer:
The major product is 2-methyl-2-pentene [ CH₃-CH₂-CH=C(CH₃)₂ ] and a minor product 2-methyl-1-pentene [ CH₃-CH₂-CH₂-C(CH₃)=CH₂ ].
Explanation:
Dehydration reaction is a reaction in which a molecule loses a water molecule in the presence of a dehydrating agent like sulfuric acid (H₂SO₄).
<u>Dehydration reaction of 2-methyl-2-pentanol</u> gives a major product 2-methyl-2-pentene and a minor product 2-methyl-1-pentene.
CH₃-CH₂-CH₂-C(CH₃)₂-OH (2-methyl-2-pentanol)→ CH₃-CH₂-CH=C(CH₃)₂ (2-methyl-2-pentene, major) + CH₃-CH₂-CH₂-C(CH₃)=CH₂ (2-methyl-1-pentene, minor)
<u>Since more substituted alkene is more stable than the less substituted alkene. So, the trisubstituted alkene, 2-methyl-2-pentene is more stable than the disubstituted alkene, 2-methyl-1-pentene.</u>
<u>Therefore, the trisubstituted alkene, 2-methyl-2-pentene is the major product and the disubstituted alkene, 2-methyl-1-pentene is the minor product.</u>
This is due to the difference in density. The rock is
denser than the leaf. And also, the rock is denser than the liquid in the pond.
If the material is denser than the other material, it will sink. The same holds
true for the rock, it sinks. But when the material is less dense than the other
material, it floats. And it holds true for the leaf, it floats.
Hi!
1) electrons
2) Chadwick
3) J. J. Thompson
4) Bohr
5) Rutherford
6) Dalton
7) Nucleus
I hope this helps!
Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543
Al is the reducing agent.That is answer B is the above answer
Al acts as a strong reducing agent. It reduces crO3 to form cr while Al is oxidized to Al2O3. Al is capable to reduce cr since Al is higher in reactivity series than cr.
