Answer:
11x³ – 2x² + 3x + 11
Explanation:
From the question given above:
We are to find the sum of:
4x³ + 2x² – 4x + 3 and 7x³ – 4x² + 7x + 8
To add the above polynomials together we must recognise that we can only add like terms together. The sum of the above polynomials can be obtained as follow:
. 4x³ + 2x² – 4x + 3
+ 7x³ – 4x² + 7x + 8
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
= 11x³ – 2x² + 3x + 11
Thus, the sum of 4x³ + 2x² – 4x + 3 and 7x³ – 4x² + 7x + 8 is 11x³ – 2x² + 3x + 11
<span>Ag+(aq) + e– → Ag(s) Eo = +0.80 V
Pb2+(aq) + 2e– → Pb(s) Eo = –0.13 V
=>
</span>
<span>2Ag+(aq) + 2e– → 2Ag(s) Eo = +0.80 V
Pb(s) → Pb2+(aq) + 2e– - Eo = – (-0.13 V) = +0.13 V
----------------------------------
2Ag(+) (aq) + Pb(s) -------> 2Ag(s) + Pb(2+)
</span>
<span>E cell = E0 cathode – Eo anode = 0.80V - (- 0.13V) = 0.80V + 0.13V = 0.93V.
Answer: + 0.93V
</span>
Answer:
It is B
Explanation:
I was able to find my school notes and double-check. I hope you have a nice day!
Answer:
tertiary
Explanation:
tertiary halogenoalkanes are more reactive than primary and secondary as the carbocation is more stable due to alkyl groups( have high electron density) donating electrons to stabilise the carbocation
