STRONGEST METALL ON EARTH

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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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.

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.

Answer:

major will be propan-2-ol

minor will be propan-1-ol

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.

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%

Answer:

Answer is in the explanation.

Explanation:

Before the equivalence point, the pH of the solution of HCl that is titrated with NaOH has a pH <<< 7. When you are adding more NaOH nearing, thus, to the equivalence point the change in pH occurs quickly, and, with 1 drop of excess of NaOH after equivalence, the pH of the solution change to a pH >>> 7

That means the volume added at pH 5 or pH 9 is, almost, the same doing the indicator work just as well as an indicator with change color at pH 7

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.

Answer: 98.08 g/mol

Explanation:

i just took this test and got it right

Answer:

A

Explanation:

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]

Answer:

60ug

Explanation:

If it has experienced two half lives, that means it has been halved twice. in that case, to undo it, just multiply it by two twice. 0.15ug * 2 = 0.30ug. 0.30ug * 2 = 0.60ug. Hope this helps.

Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

We calculate [H⁺] from the pH:

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

We calculate [HA]:

Finally we calculate the Ka:

Answer:

Neutrons are all identical to each other, just as protons are. Atoms of a particular element must have the same number of protons but can have different numbers of neutrons.

Explanation:

Since the vast majority of an atom's mass is found its protons and neutrons, subtracting the number of protons (i.e. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. In our example, this is: 14 (atomic mass) – 6 (number of protons) = 8 (number of neutrons).

Answer:

AgCl

Explanation:

The reaction that takes place is:

The precipitate (meaning a solid substance) formed is silver chloride, AgCl.

All salts formed with silver and a halogen (F, Cl, Br, I) are insoluble in water, meaning that when working with aqueous solutions they will be precipitates.

Answer:

H 4so8 is the answer of balance the reaction

Answer:

Concentration and reaction rate is directly proportional

Explanation:

Basically, concentration is directly proportional to the concentration, so the higher the concentration, the higher the reaction rate.

Once the concentration is increased, more chemical will be added hence the reaction rate will increase

Answer: The number of cations are [tex]3.24 \times 10^{22}[/tex] and number of anions are  [tex]3.24 \times 10^{22}[/tex] in 6.42 g of KBr.

Explanation:

The molar mass of KBr is (39.10 + 79.90) g/mol = 119.00 g/mol

Now, the dissociation equation for KBr is as follows.

[tex]KBr \rightarrow K^{+} + Br^{-}[/tex]

This means that 1 mole of KBr is forming 1 mole of [tex]K^{+}[/tex] (cation) and 1 mole of [tex]Br^{-}[/tex] (anion).

According to mole concept, 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms. Hence, number of cations present in 6.42 g KBr is calculated as follows.

[tex]No. of cations = Moles \times 6.022 \times 10^{23}\\= \frac{mass}{molar mass} \times 6.022 \times 10^{23}\\= \frac{6.42 g}{119.00 g/mol} \times 6.022 \times 10^{23}\\= 3.24 \times 10^{22}[/tex]

As according to the equation, there are equal number of moles of both cation and anions.

This means that the number of anions are also [tex]3.24 \times 10^{22}[/tex].

Thus, we can conclude that the number of cations are [tex]3.24 \times 10^{22}[/tex] and number of anions are  [tex]3.24 \times 10^{22}[/tex] in 6.42 g of KBr.

The number of cations and anions present in potassium bromide is 3.24 × 10²².

In one mole of any substance 6.022 × 10²³ atoms of that substance is present and this is known as Avogadro's number.

KBr is a strong electrolyte means it fully dissociates into their constitute ions. So, the number of moles of produced ions is equal to the moles of KBr and dissociation is represented as:

KBr → K⁺ + Br⁻

From this it is clear that 1 mole of cation and 1 mole of anion is produced from 1 mole of KBr. Moles of KBr will be calculated as:

n = W/M, where

W = given mass = 6.42g

M = molar mass = 119

n = 6.42 / 119 = 0.053 moles

No. of cations and anions present in 0.053 moles = 0.053 × 6.022 × 10²³ = 3.24 × 10²².

Hence, 3.24 × 10²² is the no. of cations and anions.

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Answer:

(a): Precipitate of calcium carbonate will form.

(b): Precipitate of barium sulfate will form.

(c): Precipitate of silver iodide will form.

(d): Precipitate of sodium chromate will form.

Explanation:

Complete ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

(a):

The balanced molecular equation is:

[tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]

The complete ionic equation follows:

[tex]Ca^{2+}(aq)+2Cl^-(aq)+2K^+(aq)+CO_3^{2-}(aq)\rightarrow 2K^+(aq)+2Cl^-(aq)+CaCO_3(s)[/tex]

As potassium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

[tex]Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)[/tex]

Precipitate of calcium carbonate will form.

(b)

The balanced molecular equation is:

[tex]BaCl_2(aq)+MgSO_4(aq)\rightarrow MgCl_2(aq)+BaSO_4(s)[/tex]

The complete ionic equation follows:

[tex]Ba^{2+}(aq)+2Cl^-(aq)+Mg^{2+}(aq)+SO_4^{2-}(aq)\rightarrow Mg^{2+}(aq)+2Cl^-(aq)+BaSO_4(s)[/tex]

As magnesium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Precipitate of barium sulfate will form.

(c):

The balanced molecular equation is:

[tex]AgNO_3(aq)+KI(aq)\rightarrow KNO_3(aq)+AgI(s)[/tex]

The complete ionic equation follows:

[tex]Ag^{+}(aq)+NO_3^-(aq)+K^+(aq)+I^{-}(aq)\rightarrow K^+(aq)+NO_3^-(aq)+AgI(s)[/tex]

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

[tex]Ag^{+}(aq)+I^{-}(aq)\rightarrow AgI(s)[/tex]

Precipitate of silver iodide will form.

(d):

The balanced molecular equation is:

[tex]2NaCl(aq)+(NH_4)_2CrO_4(aq)\rightarrow 2NH_4Cl(aq)+Na_2CrO_4(s)[/tex]

The complete ionic equation follows:

[tex]2Na^{+}(aq)+2Cl^-(aq)+2NH_4^+(aq)+CrO_4^{2-}(aq)\rightarrow 2NH_4^+(aq)+2Cl^-(aq)+Na_2CrO_4(s)[/tex]

As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

[tex]2Na^{+}(aq)+CrO_4^{2-}(aq)\rightarrow Na_2CrO_4(s)[/tex]

Precipitate of sodium chromate will form.