This reaction is most likely to fall under SN2 because the
thing called carbonication does not occur in SN1. The carbon forms a partial
bond with the nucleophile during the intermediate phase and the leaving group.
So for this question the reaction will fall under SN2.
Answer:
Group 2A (or IIA) of the periodic table are the alkaline earth metals: beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). They are harder and less reactive than the alkali metals of Group 1A.
Explanation:
Answer
B I’m pretty sure
Explanation
Answer:
When the results of a new experiment to test atomic theory do not agree with the theory, scientist will repeat the experiment to make sure that his results are reliable.
Explanation:
In the scientific context, each new experiment must be performed with precision and following the steps of the scientific method.
An experiment that does not provide reliable data to demonstrate a theory must be reviewed in detail and performed again to confirm the data obtained in the first attempt.
A theory is a postulate that tries to explain a natural phenomenon, but whose argument can be discussed or does not have the acceptance of a law. When the theory is proven and there are no arguments against it, it can be universally accepted and becomes a law.
The other options are not valid due to:
- <u><em>Scientists worldwide will reject atomic theory because of the new results</em></u><em>. A theory cannot be discarded without solid arguments or evidence in order to dismiss it and establish a new one.</em>
- <u><em>The scientist will change his results to agree with the accepted theory</em></u><em>. This would be an unethical procedure and unacceptable to the scientific community.</em>
- <em><u>Other scientists will reject the results because they do not agree with the theory</u></em><em>. The opinion of other scientists is not enough to dismiss a theory, if it has a valid scientific basis.</em>
Answer:
ΔE = -2661 KJ/mole
ΔH = -2658 KJ/mole
Explanation:
ΔH = q - PΔV
ΔE = q + w
<u>First, to find ΔE:</u>
The reaction PRODUCES 2658 kJ of h (q), and does 3 kJ of work (w).
2658 kJ(q) + 3 kJ(w) = 2661 kJ, BUT the reaction <u><em>PRODUCES</em></u> heat, which means ΔE is negative.
ΔE = -2661 KJ/mole
<u>Second, to find ΔH:</u>
ΔH = q - PΔV
ΔH = 2658 kJ(q) - PΔV
Now, the question states that butane burns at a constant pressure; that just translates to the pressure of the reaction is equal to 0.
ΔH = 2658 KJ(q) - (0)ΔV
ΔH = 2658 KJ - 0
ΔH = 2658 kJ, BUT, like before, the reaction PRODUCES heat, which also mean ΔH is negative.
ΔH = -2658 KJ/mole
I hope this helped! Have a nice week.