Answer: 6.64 moles of carbon.
Explanation:
Given data:
Number of moles of C = ?
Number of moles of CCl₂F₂ = 6.64 mol
Solution:
In one mole of CCl₂F₂ there is one mole of carbon two moles of chlorine and two moles of fluorine are present.
In 6.6 moles of CCl₂F₂ :
Moles of carbon = 6.64 × 1 = 6.64 moles of carbon.
Moles of chlorine = 6.64× 2 = 13.28 moles of chlorine
Moles of fluorine = 6.64× 2 = 13.28 moles of fluorine
Read more on Brainly.com - brainly.com/question/15602143#readmore
1.0×10^−15/4.2×10^−7=<span>2.3809524e-23 Hoped I helped!</span>
Answer:
pKa = 3.675
Explanation:
∴ <em>C</em> X-281 = 0.079 M
∴ pH = 2.40
let X-281 a weak acid ( HA ):
∴ HA ↔ H+ + A-
⇒ Ka = [H+] * [A-] / [HA]
mass balance:
⇒<em> C</em> HA = 0.079 M = [HA] + [A-]
⇒ [HA] = 0.079 - [A-]
charge balance:
⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water
⇒ [H+] = [A-]
∴ pH = - log [H+] = 2.40
⇒ [H+] = 3.981 E-3 M
replacing in Ka:
⇒ Ka = [H+]² / ( 0.079 - [H+] )
⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )
⇒ Ka = 2.113 E-4
⇒ pKa = - Log ( 2.113 E-4 )
⇒ pKa = 3.675
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.
Answer:
hi im just trying to get points lol hope u got ur answer
Explanation:
