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3 years ago

"Seeing is believing."

Chemistry

1 answer:

s344n2d4d5 [400]3 years ago

3 0

Answer:

stoopid

Explanation:

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Ketones undergo a reduction when treated with sodium borohydride, NaBH4. The product of the above reaction has the following spe

Anna71 [15]

Answer:

The product is cyclohexanol

Explanation:

Firstly,

A ketone undergo a borohydride reduction reaction to form an alcohol as below,

R-CO-R' ⇒ R-CO(OH)-R'

IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹
From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.
Check with other spectroscopic properties,

3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-CH₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-CH₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-CH(CH₂-)-OH).

1H NMR confirms,

1.56 δ (4H, triplet) - (-CH₂-CH₂-CH-OH) ; triplet as coupling with 2 H,

1.78 δ (4H, multiplet) - (-CH₂-CH₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH

3.24 δ (1H, quintet); - (-CH₂-CH₂-CH(CH₂-)-OH), coupling with4 H of 2 group of CH₂

3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-OH), hydrogen of alcohol group, not tend to coupling with other hydrogen

4 0

3 years ago

What is the fate of glucose 6‑phosphate, glycolytic intermediates, and pentose phosphate pathway intermediates in this cell? Gly

madreJ [45]

Answer:

The Phosphorylated glucose(glucose +inorganic phosphate), with the energy supplied from ATP hydrolysis formed glucose 6- phosphate, which is later converted to 2 molecules of fructose 6-phosphate- this is phosphorylation.And represented the fate of glucose -6-phosphate.

The fructose 6-phosphate are converted to triose phosphate- which is a 2-molecules of 3C compound. The latter is oxidized by NAD→ NADH+ to form intermediates in the glycolytic pathways .

These intermediates are converted to ribose 5-phosphates in the presence of transketolase and transaldolase enzymes.And they are finally converted to pyruvate in the glycolytic pathway with the production of 2ATPs per molecule of glucose.

Basically the phosphate pathway reaction is very slow due to enzyme catalysis.

3 0

3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago

Which is the correct equilibrium constant expression for the following reaction? 2brcl 3 (g) br 2 (g) 3cl 2 (g)

marishachu [46]

Kc = concentrations of product / concentrations of reactant

Kc = [Br₂] [Cl₂]₃ / [BrCl₃]₂

What is the equilibrium constant?

The relationship between a reaction's products and reactants with regard to a certain unit is expressed by the equilibrium constant(K) This article introduces the mathematics needed to determine the partial pressure equilibrium constant as well as how to formulate expressions for equilibrium constants. By allowing a single reaction to reach equilibrium and then measuring the concentrations of each chemical participating in that reaction, one can determine the numerical value of an equilibrium constant. it is the ratio of product concentrations to reactant concentrations. The equilibrium constant for a given reaction is unaffected by the initial concentrations because the concentrations are measured at equilibrium.

To learn more about the equilibrium constant, visit:

brainly.com/question/19340344

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6 0

1 year ago

What do we call genes

qwelly [4]

Hi there so i need one more brainliest mark so if i get it right can i get brainliest please :)

A gene is the basic physical and functional unit of heredity. Genes are made up of DNA. Some genes act as instructions to make molecules called proteins. However, many genes do not code for proteins. In humans, genes vary in size from a few hundred DNA bases to more than 2 million bases

4 0

2 years ago