Places where the seafloor is forced under continental plates.

Chemistry

2 answers:

Bas_tet [7]3 years ago

6 0

This place is called the subduction zone.

Natasha2012 [34]3 years ago

3 0

Subduction zones are where the seafloor is forced under continental plates

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Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.

Firdavs [7]

Answer:

Explanation:

a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .

C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O

O is provided by KMnO₄

b ) In this reaction isophthalic acid is formed .

C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂

4-Propyl-3-t-butyltoluene

In this oxidation , three side chains of ring are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .

The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )

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3 years ago

Which of the following is an ether?<br> A)<br> B)<br> C)<br> D)

koban [17]

B because it shows the COOCH

6 0

3 years ago

hat is the pressure of CO(g) in equilibrium with the CO2(g) and O2(g) in the atmosphere at 25 ????C? The partial pressure of O2(

Lubov Fominskaja [6]

Answer:

The partial pressure of CO is 5.54x10⁻⁴⁹atm. You shouldn't worry because it is very low pressure

Explanation:

First, the balanced reaction is:

CO + 1/2O₂ → CO₂

The energies of formation are:

ΔG(CO)=-137.168kJ/mol

ΔG(O₂)=0

ΔG(CO₂)=-394.359kJ/mol

The energy of the reaction is:

$delta-G_{reaction} =delta-G_{CO_{2} } -(delta-G_{CO} +1/2delta-G_{O_{2} } )\\delta-G_{reaction}=-394.359-(-137.168+0)=-257.191kJ/mol$

The expression for calculate the partial pressure of CO is:

$p_{CO} =\frac{p_{CO2} }{p_{O_{2} }^{1/2}*exp^{-\frac{delta-G}{RT} } } \\p_{CO}=\frac{3x10^{-4} }{0.2^{1/2}*exp(-\frac{-257.191*1000}{8.314*298} ) } \\p_{CO}=5.54x10^{-49} atm$