Answer:
Both lone pairs reside in sp3 orbitals; they are co-planar with the CH3 group.
Explanation:
In the compound, methyl acetate, the lone-pair of electrons on the singly bonded oxygen atom is accommodated in sp3 hybridized orbitals.
If we look at the compound, we will notice the both lone pair of electrons on oxygen are accommodated in sp3 orbitals which are co-planar with the CH3 group in the molecule.
A pure copper cube has an edge length of 1.76 cm. How many copper atoms does it contain? (volume of a cube = (edge length)^3; density of copper = 8.96 g/cm^3 )
The Fischer esterification mechanism is examined in this question. The overall reaction is: Benzoic acid, C H 3 O H and H C l react to form a methyl ester, H 2 O and H C l. Benzoic acid is a carboxylic acid bonded to a benzene ring. Identify the results or mechanism of each step.
Answer:
See explanation and image attached
Explanation:
Fischer esterification is a type of reaction used to convert carboxylic acids to ester in the presence of excess alcohol and a strong acid which acts as a catalyst. Another final product formed in the reaction is water.
The mechanism for the fischer esterification of Benzoic acid and C H 3 O H in the presence of HCl as the catalyst is shown in the image attached to this answer.
The final products of the reaction are methyl benzoate, water and H^+ as shown in the image attached.
The methyl ester, water, and the acid catalyst (HCl) are byproducts of the Fischer esterification process, which involves protonation, nucleophilic attack, elimination, and deprotonation processes.
Carbonyl oxygen protonation: The carbonyl oxygen of the carboxylic acid (benzoic acid) is protonated by the acid catalyst (HCl) in the first step. The protonation of the carbonyl carbon increases its electrophilicity and promotes the alcohol's nucleophilic assault. Attack by the alcohol's nucleophilic oxygen (methanol, CH3OH) on the protonated carboxylic acid's carbonyl carbon results in the formation of a tetrahedral intermediate. The acid catalyst also helps with this phase. Elimination of water: In the following step, the water molecule must be removed from the tetrahedral intermediate. The hydroxyl group (-OH) from the carboxylic acid and a hydrogen from the hydroxyl group of the alcohol are removed to create this water molecule. Deprotonation: A deprotonation occurs after the removal of water.
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3 enzimas presentes en nuestro organismo e indique en que procesos actúan
Answer:
ATP asa, Helicasa, Proteasa, ARN polimerasa
Explanation:
Las enzimas son un tipo de biomoleculas que se corresponden con las proteinas.
Al momento de referirse a ellas, se utiliza la terminación asa.
ATPasa → Sintetizando ATP para el funcionamiento celular
Helicasa → Abre las hebras de ADN permitiendo el paso de la horquilla para el proceso de replicación de ADN.
Proteasas → Enzimas que degradan proteinas mal plegadas, rompen los enlaces peptídicos.
ARN polimerasa → Sintesis de ARN mensajero a partir de ADN en el proceso de la Transcripción. Se la puede conocer a veces, como primasa.
Select all the true statements.
a. Ionic bonding is more prevalent for the higher oxidation states and covalent bonding is more prevalent for the lower states.
b. The highest oxidation state of elements in Groups 3A through 7B is +3.
c. In the transition series, atomic size across a period decreases at first but then remains relatively constant.
d. The transition elements in a period show a steady increase in electronegativity.
Answer:
In the transition series, atomic size across a period decreases at first but then remains relatively constant.
The transition elements in a period show a steady increase in electronegativity.
Explanation:
In considering the transition series, we observe that atomic sizes of the elements decreases first and subsequently remain constant. The reason for the initial decrease in atomic size is the increase in nuclear charge across the period. After the first few elements in the period, the atomic size remains relatively constant due to shielding effect of the inner d electrons which opposes the increase in effective nuclear charge.
It is also observed that electro negativity increases smoothly across the period for the transition series. Consequently, the transition series become less electro positive across the period.
Sodium hydrogen carbonate, also called baking soda, is an active ingredient in some antacids used for the relief of indigestion. Determine the percent of carbon in sodium hydrogen carbonate.
Answer:
14.30%
Explanation:
Step 1: Given data
Chemical formula of sodium hydrogen carbonate: NaHCO₃
Step 2: Determine the mass of C in 1 mole of NaHCO₃
The is 1 atom of C in 1 molecule of NaHCO₃ and the molar mass of C is 12.01 g/mol. Then, there are 12.01 g of C in 1 mole of NaHCO₃.
Step 3: Determine the mnolar mass of NaHCO₃
M(NaHCO₃) = 1 × M(Na) + 1 × M(H) + 1 × M(C) + 3 × M(O)
M(NaHCO₃) = 1 × 22.98 g/mol + 1 × 1.01 g/mol + 1 × 12.01 g/mol + 3 × 16.00 g/mol = 84.00 g/mol
Step 4: Determine the mass percent of C in NaHCO₃
We will use the following expression.
%C = mC / mNaHCO₃ × 100%
%C = 12.01 g / 84.00 g × 100% = 14.30%
Some glucose produced by gluconeogenesis is stored in the body as glycogen. Order the steps of glycogen synthesis.
a. Pyrophosphatase converts PPi and water into two Pi
b. Glycogen synthase adds a glucose unit from UDP-glucose to glycogen, producing a larger glycogen molecule and UDP
c. Glvcogen synthase removes a glucose unit from a glycogen molecule producing a smaller glycogen molecule and IJDP
d. ADP-glucose pyrophosphorylase catalyzes the reaction of glucose-I -phosphate and ATP to ADP-glucose and PPi
e. UDP-glucose pyrophosphorylase catalyzes the reaction of glucose-I-phosphate and UTP to UDP-glucose and PPi
Answer:
e. UDP-glucose pyrophosphorylase catalyzes the reaction of glucose-I-phosphate and UTP to UDP-glucose and PPi
a. Pyrophosphatase converts PPi and water into two Pi
b. Glycogen synthase adds a glucose unit from UDP-glucose to glycogen, producing a larger glycogen molecule and UDP
Explanation:
Glycogen synthesis or glycogenesis is the process of synthesis of glycogen molecules from glucose molecules in living organisms. Glycogen is a polysaccharide storage form of glucose and helps to store excess glucose in the body form use when required by the body.
The synthesis of glycogen involves sugar nucleotides. Sugar nucleotides are compounds in which a sugar molecule is attached to a nucleotide through phosphate ester bond, resulting in the activation of the sugar molecule. The sugar nucleotides then are used as substrates for the polymerization of the monosaccharide sugars into disaccharides, oligosaccharides and polysaccharides.
In the synthesis of glycogen, glucose-6-phosphate from phosphorylation of free glucose by hexokinase is first isomerized to glucose-1-phosphate by phosphoglucomutase.
Glucose-1-phosphate is then converted to UDP-glucose by its reaction with UTP catalyse by UDP-glucose pyrophosphorylase. The reaction is favoured by the rapid hydrolysis of PPi produced to two molecules of inorganic phosphate by the enzyme pyrophosphatase.
Glycogen synthase then adds a glucose unit from UDP-glucose to a growing chain of glycogen, producing a larger glycogen molecule and free UDP.
What volume of water must be added to 10.5 mL of a pH 2.0 solution of HNO3 in order to change the pH to 4.0 g
Answer:
[tex]V'=1040ml[/tex]
Explanation:
From the question we are told that:
Initial Volume [tex]V_1=10.5mL[/tex]
Initial Aciditity of [tex]HN0_3 pH_1=2.0g[/tex]
Finial Aciditity of [tex]HN0_3 pH_2=4.0g[/tex]
Generally the equation for Acidity &Volume Relationship is mathematically given by
[tex]N_1V_1=N_2V_2[/tex]
Therefore
[tex]V_2=\frac{N_1V_2}{N_2}[/tex]
[tex]V_2=\frac{10^{-2}*10.5}{10^{-4}}V_2=1050ml[/tex]
Therefore
Change in Water volume is
[tex]V'=V_2-V_1[/tex]
[tex]V'=1050ml-10ml[/tex]
[tex]V'=1040ml[/tex]
