For this problem, let's use the Planck's equation which is written below:
E = hv
where
E is the energy
h is Planck's constant equal to 6.626×10⁻³⁴ J·s
v is frequency
Solving for E,
E = (6.626×10⁻³⁴ J·s)(7.14×10¹⁴ s⁻¹)
<em>E = 4.73×10⁻¹⁹ J</em>
Answer:
25.0%
A solution of benzene (C6H6) and toluene (C7H8) is 25.0% benzene by mass. At 25 °C the vapor pressures of pure benzene and pure toluene are 94.2 torr and 28.4 torr, respectively.
Answer:
1-bromo-1-tert-butylcyclohexane
Explanation:
The parent compound comprises of a cyclohexane to with a tertiary butyl carbon attached. We have been told that the reaction occurs by radical mechanism hence we must recall the order of stability of radicals: tertiary>a secondary> a primary. This implies that the reaction will occur at carbon 1 of the cyclohexane which is a tertiary carbon atom. This leads to the formation of a radical at the 1-position and bromination at that position hence the answer chosen above.
Answer: electrons
Explanation:
Oxidation is the loss of electrons during a reaction by a molecules, atom or ion. Oxidation occurs when the oxidation state a molecule, atom or ion is increased. The opposite process is called reduction , which occurs when there is a gain of electrons or the oxidation state of an atom, molecule, or ion decreases
Answer:
C. It does not participate in a decay series.
Explanation:
From this statement, we can deduce that a radioisotope that forms a stable isotope after it undergoes radioactive decay suggests that it does not participate in a decay series.
- It could have emitted any form of radioactive particles which can be alpha or beta.
- We do not know if it has a long or short half life because the value is not given.
- But since the radioactive decay in one step produces a stable isotope, we can conclude that it did not participate in a decay series.
- A decay series involves a radioactive decay in multiple steps.