Answer is 0.7 g
Reason: Given: mass of Fe = 5.60 gWe know that, half life of Fe-53 = t1/2 (Fe-53) = 8.51 min.
Thus, after 25.53 mins, number of half life passed = 25.53/8.51= 3
Now, Amount left after first half life = 5.60g/2 = 2.80 g
Amount left after second half life = 2.80g/2 = 1.40 g
Amount left after third half-life = 1.40g/2 = 0.7g.
Answer:
D.
Double replacement, CaO + Cl2O
Explanation:
Answer:
Cl⁻, Na⁺, OH⁻
Explanation:
The titration is:
CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)
In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.
Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.
From electronic configuration valence electron of Nitrogen is 5, oxygen 6x2 which 12 since it involve two molecules , that of is frulorine is 7, and that No2F is 24 which is gotten form adding (5,12,7 ).All resonance structure are as follows
F
.. I ..
: O : N :O:
..
OR : F:
I
N .. : F:
/ \ or I
.. .. N
:O : :O: / / \\
/ / \\
:O : : O:
