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LenaWriter [7]
1 year 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]1 year 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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Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2
nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔  2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2  [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ]  ' ²

Substituting the new concentration terms ,

K ' = 1/2  [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ]  ²

K ' = 1/4  [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ]  ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' =   [ NO ] ² [ Br₂ ] /  [ NoBr ]  ²

Hence ,

K' = K

5 0
3 years ago
Metallic elements can be recovered from ores that are oxides, carbonates, halides, or sulfides. Give an example of each type.
vredina [299]

Metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

Ores which is deposited in earth's crust which contain minerals and metals. Metals can be obtained economically and sold commercially.

<h3>How metals obtained from sulphide or carbonate ore? </h3>

As we get to know that it is easy to obtain metals from their oxides. So, firstly ores which is found in the form of carbonates and sulphide are converted into their oxides by using the process of calcination and roasting.

The metals which is in the middle of the activity series are moderately reactive. These metals are found in the crust of the earth is mainly found as oxides, sulphides, or carbonates.

A metal which can occur in the form of sulphide ore is lead.

A metal which can occur in the form of oxide ore is iron.

A metal which can occur in the form of carbonate ore is zinc.

A metal which can occur in the form of halides ore is silver.

Thus, we concluded that the metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

learn more about ore:

brainly.com/question/10306443

#SPJ4

6 0
1 year ago
a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is
fiasKO [112]

Answer: The empirical formula is CH_2.

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.142}{0.142}=1

For H =\frac{0.287}{0.142}=2

The ratio of C: H = 1: 2

Hence the empirical formula is CH_2.

4 0
3 years ago
Three different atoms or atomic cations with 3 electrons.
sveticcg [70]

Answer:

Li, Be^+, B^{2+}

Explanation:

To answer this question successfully, we need to remember that atoms are neutral species, since the number of protons, the positively charged particles, is equal to the number of electrons, the negatively charged particles. That said, we may firstly find an atom which has 3 electrons (and, as a result, 3 protons, as it should be neutral).

The number of protons is equal to the atomic number of an element. We firstly may have an atom with 3 protons and 3 electrons (atomic number of 3, this is Li).

Similarly, we may take the atomic number of 4, beryllium, and remove 1 electron from it. Upon removing an electron, it would become beryllium cation, Be^+.

We may use the same logic going forward and taking the atomic number of 5. This is boron. In this case, we need to remove 2 electrons to have a total of 3 electrons. Removal of 2 electrons would yield a +2-charged cation: B^{2+}.

6 0
2 years ago
______ break down sugars to produce energy.
Mariana [72]

Answer:

mitochondria

Explanation:

I took the test

8 0
2 years ago
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