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LenaWriter [7]
2 years ago
8

How might the yield of 1-bromobutane be affected if water was not added, and what product(s) would be favored?

Chemistry
1 answer:
jek_recluse [69]2 years ago
5 0
  • Due to the inability of the reaction to take place, the yield of 1-Bromobutane would drop.
  • Since 1-Butanol won't react with the additional sodium bromide, bromination won't happen.
  • If water had been supplied, the equilibrium would have shifted extremely far to the left, preventing the reactants from interacting with the acid and favoring the yield of 1-bromobutane instead.
<h3>What is Bromination?</h3>
  • When a substance undergoes bromination, bromine is added to the compound as a result of the chemical reaction.
  • After bromination, the result will have different properties from the initial reactant.
  • For example, an alkene is brominated by electrophilic addition of Br_{2}.
  • Benzene ring bromination by electrophilic aromatic substitution.

Learn more about Bromine here:

brainly.com/question/862562

#SPJ4

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Asprin can be made by adding 200.0 g of salicylinc
damaskus [11]
Correct, Was that a question.
7 0
3 years ago
Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.
madreJ [45]

Answer:

Molar \ solubility=3.12x10^{-5}M

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-

The equilibrium expression is:

Ksp=[Ca^{2+}][F^-]^2

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent x is computed as follows:

3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M

Thus, the molar solubility equals the reaction extent x, therefore:

Molar \ solubility=3.12x10^{-5}M

Regards.

4 0
3 years ago
(a) Given that Ka for acetic acid is 1.8 X 10^-5 and that for hypochlorous acid is 3.0 X 10^-8, which is the stronger acid? (b)
Gala2k [10]

Answer:

HOAc is stronger acid than HClO

ClO⁻ is stronger conjugate base than OAc⁻

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

Explanation:

Assume 0.10M HOAc => H⁺ + OAc⁻  with Ka = 1.8 x 10⁻⁵

=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺

Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸

=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺

HOAc delivers more H⁺ than HClO and is more acidic.

Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

4 0
3 years ago
Based on the activity series, which metals could X represent in the reaction below? (Note: The equation is not balanced.)
Nikitich [7]
Answer is: A) Sr (strontium).
The reactivity series<span> is a series of metals from highest to lowest reactivity.</span><span> Metal higher in the reactivity series will displace another.
</span>Strontium is only higher in this group from magnesium. Strontium is stronger reducing agent than magnesium, gives electrons easier.
3 0
3 years ago
Manganese, Mn, forms two ions, one with a 2+ charge and one with a 3+ charge. What is the formula for manganese (II) sulfide?
maria [59]
It would be MnSO4

The (II) lets you know it’s the form with a 2+ charge and Sulfate has a 2- charge

These will cancel out making it plain MnSO4

If it was manganese (iii) sulfide the answer would be Mn2(SO4)3
4 0
3 years ago
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