I would go with C because that seems like the best answer choice
A word equation is a chemical reaction described using words.
A common example is the act of photosynthesis - the process plants use to make glucose (sugar) to use as 'food'.
Plants convert water and carbon dioxide into oxygen and glucose.
A word equation to express this is:
Water + Carbon Dioxide → Glucose + Oxygen
The other type of equation is a symbol equation - this uses the symbols of the elements instead of the common names:
H₂O + CO₂ → C₆H₁₂O₆ + O₂
There is also a balanced version:
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
<em>If you want information on the balanced symbol equations, feel free to PM me.</em>
Answer:
amount, pH value.
Explanation:
The buffer range is the pH range in which the buffer performs optimally, i.e., neutralizes even when a strong acid or base is introduced to it and resists any major change in its pH value.
The buffer capacity is the amount of acid or base that can be added before the pH of the buffer solution changes significantly.
Thus, the final statement becomes,
Buffer capacity is the amount of acid or base a buffer can handle before pushing the pH value outside of the buffer range.
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.