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.
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.
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%
help i’ll give u points
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.
A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka of the acid. g
Answer:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
HA ⇄ H⁺ + A⁻Ka = [H⁺][A⁻]/[HA]We calculate [H⁺] from the pH:
pH = -log[H⁺][H⁺] = [tex]10^{-pH}[/tex][H⁺] = 9.12x10⁻⁴ MKeep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
[HA] * 0.66/100 = [H⁺]We calculate [HA]:
[HA] = 0.138 MFinally we calculate the Ka:
Ka = [tex]\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}[/tex] = 6.02x10⁻⁶How does the neutron number in each compare
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).
When aqueous solutions of AgNO3 and MgCl2 are mixed, a precipitate forms. What is the correct formula for the precipitate
Answer:
AgCl
Explanation:
The reaction that takes place is:
2AgNO₃ (aq) + MgCl₂ (aq) → Mg(NO₃) (aq) + 2AgCl (s)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.
Which additional product balances the reaction h2so4+2naoh>na2so4
Answer:
H 4so8 is the answer of balance the reaction
what will happen to the initial rate of the reaction if the concentration of iron(III) chloride is tripled
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
Calculate the number of cations and anions in each of the following compounds. Enter your answers in scientific notation. (a) 6.42 g of KBr:
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.
How we calculate atoms from moles?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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Show the complete ionic equation and net ionic equation for all the equations below, then state whether or not a precipitate (insoluble compound) will form. To receive full credit, you must show ALL your work.
Cacl2(aq) + K2co3(aq) + -------->
Bacl2(aq) + MgSO4(aq) + -------->
AgNO3(aq) + Kl(aq) →
Nacl(aq) + (NH4)2Cro4(aq) →
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.
Please help me I need help with this questions I’m very confused fused as to what the answer is please
A student performs an experiment three times. The results are 4.52 g/mL, 4.54 g/mL, and 4.39 g/mL. Which of the following best describes these results with only the information given in this question?
a. high accuracy
b. high accuracy and high precision
c. high accuracy and low precision
d. high precision
This portion of the stratosphere if known as the _______.
A.
mesosphere
B.
tropopause
C.
thermosphere
D.
ozone layer
