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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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In which condition does barium chloride conduct electricity? And why
Anuta_ua [19.1K]

Answer:

When barium chloride (BaCl 2) is dissolved in water, the water conducts electricity. In what form will the dissolved BaCl 2 be found? a. as Ba 2+ and Cl - ions b. as Ba atoms and Cl 2 molecules

Explanation:

3 0
2 years ago
The CNO cycle produces energy in which of the following?
d1i1m1o1n [39]
Stars on the main sequence fuse hydrogen into helium via a six-stage sequence of reactions
7 0
3 years ago
The equilibrium constant (Kp) for the decomposition of phosphorus pentachloride (PCl5) to phosphorus trichloride (PCl3) and mole
Oksanka [162]

Answer: The partial pressure of Cl_2 is 1.86 atm

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

The given balanced equilibrium reaction is,

                            PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

Pressure at eqm.     0.973 atm                 0.548atm      x atm

The expression for equilibrium constant for this reaction will be,

K_c=\frac{(p_{PCl_3}\times (p_{Cl_2})}{(p_{PCl_5})}

Now put all the given values in this expression, we get :

1.05=\frac{(0.548)\times (x)}{(0.973)}

By solving the term 'x', we get :

x = 1.86 atm

Thus, the partial pressure of Cl_2 is 1.86 atm

4 0
3 years ago
Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous-secreting cells. The cells secrete
neonofarm [45]

Answer:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = -RTInK_(eq)

where:

R= universal gas constant

T= temperature

K_eq= equilibrum constant for the process

Similarly, free energy change and cell potentia; are related to each other as follows;

ΔG= -nFE°

from above;

F = faraday's constant

n = number of electrons exchanged in the process; and  

E = standard cell potential

∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:

ΔG° = -RTInK_(eq)

where;

[texK_eq[/tex]=\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}

For transport of ions to an internal pH of 7.4, the transport taking place can be given as:

H^+_{inside} ⇒ H^+_{outside}

Equilibrum constant for the transport is given as:

K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}

=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}

[H^+]_{cell}= 10⁻⁷⁴

=3.98 * 10⁻⁸M

[H^+]_{stomach lumen} = 10⁻²¹

=7.94 * 10⁻³M

Hence;

K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}

=\frac{3.98*10^{-8}}{7.94*10{-3}}

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = -RTInK_(eq)

If temperature T= 37° C ; in kelvin

=37° C + 273.15K

=310.15K; and

R-= 8.314 j/mol/k

substituting the values into the equation we have;

ΔG₁ = -(8.314J/mol/K)(310.15)TIn(5.0126*10^{-6})

= 31467.93Jmol⁻¹

≅ 31.47KJmol⁻¹

If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = -nFE°_{membrane}

ΔG₂ = -(1 mol)(96.5KJ/mol/V)(60*10^{-3})

= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

Now, we  can calculate the total amount of energyy required to transport H⁺ ions across the membrane:

Δ G_{total} = G_{1}+G_{2}

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

   The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹

5 0
3 years ago
During photosynthesis, light energy from the Sun causes a reaction to take place. The products of photosynthesis are used in a r
andrew11 [14]

Answer:

chemical reactions which proceed with the release of heat energy are called exothermic reactions

3 0
2 years ago
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