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.
The concentration of the solution is 8 M
<u><em>calculation</em></u>
<em> </em>Concentrati<em>on = moles/volume in liters</em>
<em> </em><em>step 1: find moles of HF</em>
<em>moles of HF =mass/molar mass</em>
<em>molar mass of HF = 1+ 19 )= 20 g/mol</em>
<em> moles is therefore = 32.0 g/ 20 g/mol= 1.6 moles</em>
<em>Step 2: convert ml to L</em>
<em>volume in liters = 2.0 x 10^2 / 1000 =0.2 l</em>
<em>step 3: find the concentration</em>
<em>concentration = 1.6 mol / 0.2 l = </em><em>8 M</em>
Answer:
Conjugate base of naphthol is highly stable but conjugate base of ethanol is unstable.
Explanation:
Naphthol gives acid base reaction with NaOH where naphthol acts as an acid and NaOH as a base.
Reactivity of naphthol towards NaOH is quite high because the conjugate base of naphthol is highly stable. Stability of the conjugate base arises due to delocalization of negative charge on oxygen atom into aromatic naphthalene ring.
Ethanol is not so reactive towards NaOH because the ethoxide ion is highly unstable due to lack of delocalization of negative charge on oxygen atom.
Resonance structures of conjugate base of naphthol has been shown below.
Answer:
no it is not light sensitive
Explanation:
Water is not light sensitive