First, we need to get moles of NaOH:
when moles NaOH = volume * molarity
= 0.02573L * 0.11 M
= 0.0028 moles
from the reaction equation:
H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)
we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH
∴ X mol H3PO4 reacts with → 0.0028 moles NaOH
∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol
now we can get the concentration of H3PO4:
∴[H3PO4] = moles H2PO4 / volume
= 9.4 x 10^-4 / 0.034 L
= 0.028 M
Answer: O fluorine to complete the octet rule
that right
The correct answer is letter A: <span>Unlike junction diodes, point-contact diodes are enclosed in a suitable casing and have terminals for connecting them to a circuit.</span>
Answer:
Br- Withdraws electrons inductively
Donates electrons by resonance
CH2CH3 - Donates electrons by hyperconjugation
NHCH3- Withdraws electrons inductively
Donates electrons by resonance
OCH3 - Withdraws electrons inductively
Donates electrons by resonance
+N(CH3)3 - Withdraws electrons inductively
Explanation:
A chemical moiety may withdraw or donate electrons by resonance or inductive effect.
Halogens are electronegative elements hence they withdraw electrons by inductive effect. However, they also contain lone pairs so the can donate electrons by resonance.
Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.
-NHCH3 and contain species that have lone pair of electrons which can be donated by resonance. Also, the nitrogen and oxygen atoms are very electron withdrawing making the carbon atom to have a -I inductive effect.
+N(CH3)3 have no lone pair and is strongly electron withdrawing by inductive effects.
D is the answer since it is changing the element.
