The amine here is the easiest to spot since there’s only one structure that has a nitrogen atom, which would be the first (the first structure is a primary amine).
The distinguishing functional group of an alcohol is the hydroxy group (—OH). Both the second and third structures have an —OH group, but the —OH in the third structure is part of a carboxyl group (—COOH or —C(=O)OH). A carboxyl group takes priority over hydroxy group. Thus, the second structure would be an alcohol and the third structure would be a carboxylic acid.
That leaves us with the fourth structure, a hydrocarbon with a halogen substitutent, or, aptly named, a halocarbon.
Answer: 2.73g of CH3CH2OH Will be consumed
Explanation:
CH3CH2OH + O2 —> CH3COOH + H2O
MM of CH3CH2OH = 12 + 3 +12 + 2 16 +1 = 46g/mol
MM of O2 = 16 x2 = 32
Mass conc. Of O2 = 1.9g
From the equation,
32g of O2 consumed 46g of CH3CH2OH .
Therefore, 1.9g of O2 will consume Xg of CH3CH2OH i.e
Xg of CH3CH2OH = (1.9 x 46)/32 = 2.73g
Answer:
42672 kJ
Explanation:
Chemical reaction:
C₂H₄ + H₂ → C₂H₆ + heat
Now we calculate the number of moles of ethane C₂H₆:
number of moles = mass / molecular weight
number of moles of ethane = 10.1 / 30 = 0.336 kmoles = 336 moles
And determine the heat released:
if 127 kJ are released when 1 mole of C₂H₆ is produced
then X kJ are released when 336 moles of C₂H₆ is produced
X = (127 × 336) / 1 = 42672 kJ
Answer:
1. A
2. B
Explanation:
1. Most of the answers don't make that much sense, so using process of elimination I resulted in this answer. I recommend you also try to check with others.
2. This question also was pretty confusing but when using process of elimination I resulted in either A or B. To me B makes more sense.
The half cell in which the electrode gains electrons is where reduction occurs, and the half cell in which the electrode loses electrons is where oxidation occurs.
<h3><u>What is a Galvanic cell ?</u></h3>
Voltaic or galvanic cells are electrochemical devices that use spontaneous oxidation-reduction events to generate electricity. In order to balance the overall equation and highlight the actual chemical changes, it is frequently advantageous to divide the oxidation-reduction reactions into half-reactions while constructing the equations.
Two half-cells make up most electrochemical cells. The half-cells allow electricity to pass via an external wire by separating the oxidation half-reaction from the reduction half-reaction.
<h3><u>
Oxidation:</u></h3>
The anode is located in one half-cell, which is often shown on the left side of a figure. On the anode, oxidation takes place. In the opposite half-cell, the anode and cathode are linked.
<h3><u>Reduction:</u></h3>
The second half-cell, cathode, which is frequently displayed on a figure's right side. The cathode is where reduction happens. The circuit is completed and current can flow by adding a salt bridge.
To know more about processes in Galvanic cell, refer to:
brainly.com/question/13031093
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