Answer:
To fulfill octet of B atom in borane, nucleophilic attack by pi electrons of alkene takes place with electrophilic B center in borane.
Explanation:
In Borane (), B atom has energetically vacant 2p orbital and thereby octet is incomplete.Therefore B center in borane act as an electrophilic center.
In alkene, pi bonding electrons are loosely bound together due to side on overlap of two constituent p-orbitals. Hence this pi bond can be easily broken. Alternatively, we can say that pi bond in alkene act as a potential nucleophile.
The first step of hydroboration occurs in a concerted manner where pi electrons of alkene first attack vacant 2p orbital of B in borane to fulfill it's octet (represented by an arrow drawn from alkene to B atom) and forms a 4 memebered cyclic intermediate. Simulaneously, a B-H bond in borane is donated to alkene through 3c-2e bond (3 center-2 electron bond).
Full mechanism has been shown below.
Answer:
5.0 × 10⁻⁶
Explanation:
Step 1: Write the balanced equation for the solution of chromium(III) iodate
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
Step 2: Calculate the solubility product constant (Ksp)
To relate Ksp and the solubility (S), we will make an ICE chart.
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product constant is:
Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶
V ( H2SO4) = 35 mL / 1000 => 0.035 L
M ( H2SO4) = ?
V ( NaOH ) = 25 mL / 1000 => 0.025 L
M ( NaOH ) = 0.320 M
number of moles NaOH:
n = M x V
n = 0.025 x 0.320 => 0.008 moles of NaOH
Mole ratio:
<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??
0.008 x 1 / 2 => 0.004 moles of H2SO4 :
Therefore:
M ( H2SO4) = n / V
M = 0.004 / 0.035
= 0.114 M
hope this helps!
Visit this page and read the Radium series part
https://en.wikipedia.org/wiki/Uranium-238
Sodium:
moles = mass/RFM = 8.3/23 = 0.3608..
Chlorine:
moles = mass/RFM = 4.5/71 = 0.06
Ratio of Na to Cl is 2:1
So 0.06 x 2 = 0.12
So we know that the Na is in excess
Now we have to figure out by how much:
Mass = moles x RFM = 0.12 x RFM of 2NaCl = 0.12 x 58.5 = 7.4 (roughly)
Now to figure out by how much do 8.3 + 4.5 = 12.7 and do 12.7 - 7.4 = 5.3 (roughly)
So your answer is A