Which electron transition represents a gain of energy?
1 answer:
You might be interested in
<h2>Heptene formed is -</h2><h2>
</h2>
Explanation:
The two possibilities when the peroxide is not present
-
≡
→
+ HBr →
In presence peroxide,
≡
+ HBr →
- When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
- This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
- One mole of HBr adds to one mole of 1-heptane.
- The structure of heptene formed is -
Answer:
ksp = 0,176
Explanation:
The borax (Na₂borate) in water is in equilibrium, thus:
Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)
<em>When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ </em><em>(1)</em>
The ksp is defined as:
<em>ksp = [borate²⁻] [Na⁺]²</em>
Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:
B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻
The moles of HCl that reacts with B₄O₇²⁻ are:
0,500M×0,01200L = 6,00x10⁻³ mol of HCl
As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:
6,00x10⁻³ mol of HCl× = <em>3,00x10⁻³ mol of B₄O₇²⁻</em>
For (1), moles of Na⁺ are <em>3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺</em>
The [borate²⁻] is <em>3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = </em><em>0,353M</em>
And [Na⁺] is <em>6,00x10⁻³ mol of Na⁺ / 0,00850L = </em>0,706M
Replacing in the expression of ksp:
ksp = [0,353] [0,706]²
<em>ksp = 0,176</em>
<em></em>
I hope it helps!