A metallic bond since both are pure metals but are not ionic
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
Answer:
0.5 mole of CO₂.
Explanation:
We'll begin by calculating the number of mole in 42 g of baking soda (NaHCO₃). This can be obtained as follow:
Mass of NaHCO₃ = 42 g
Molar mass of NaHCO₃ = 23 + 1 + 12 + (16×3)
= 23 + 1 + 12 + 48
= 84 g/mol
Mole of NaHCO₃ =?
Mole = mass / molar mass
Mole of NaHCO₃ = 42/84
Mole of NaHCO₃ = 0.5 mole
Next, balanced equation for the reaction. This is given below:
NaHCO₃ + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O + CO₂
From the balanced equation above,
1 mole of NaHCO₃ reacted to produce 1 mole of CO₂
Finally, we shall determine the number of mole of CO₂ produced by the reaction of 42 g (i.e 0.5 mole) of NaHCO₃. This can be obtained as follow:
From the balanced equation above,
1 mole of NaHCO₃ reacted to produce 1 mole of CO₂.
Therefore, 0.5 mole of NaHCO₃ will also react to produce 0.5 mole of CO₂.
Thus, 0.5 mole of CO₂ was obtained from the reaction.
Answer:
4.94g of material
Explanation:
Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:
Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>
100mL of a 5% solution contains 5g of material in 100mL of water. Thus:
8 = X / 100mL / (5g-X) / 100mL
<em>Where X is the amount of material in grams that comes to the organic phase.</em>
8 = X / 100mL / (5g-X) / 100mL
8 = 100X / (500-100X)
4000 - 800X = 100X
4000 = 900X
4.44g = X
<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>
And will remain 5g-4.44g = 0.56g.
In the second extraction:
8 = X / 100mL / (0.56g-X) / 100mL
8 = 100X / (56-100X)
448 - 800X = 100X
448 = 900X
0.50g = X
<em>In the second extraction, you will extract 0.50g of material</em>
Thus, after the two extraction you will lost:
4.44g + 0.50g = <em>4.94g of material</em>
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Give more points, dont even feel like reading this for 5 pointsssss