Answer:
Continue to move at 30 mphExplanation:
Newton's first law of motion :
"Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces exerted on it."
From the question we know that the net forces on the object were zero or that the there were no unbalanced forces on it.
Therefore we can assume that the object is moving along a straight line.
And the object was moving at a constant speed of 30 mph.
So it is clear from the Newton's first law that the object will remain in the state of motion as it was earlier.
That is the object will remain in motion at constant speed of 30 mph.
The flavor of anise is due to anethole, a compound with the molecular forumal C10H12O. Combustion of one mole of anethole produces 5541 kJ of thermal energy. If 0.950 g of anethole is combusted in a bomb calorimeter whose heat capacity (Ccalorimeter) is 7.854 kJ/°C, what is the change in temperature of the calorimeter?
Final answer:
The change in temperature of the calorimeter upon combusting 0.950 grams of anethole is approximately 24.90°C. This is calculated using the energy released in the combustion and the calorimeter's heat capacity.
Explanation:
The change in temperature of the calorimeter can be calculated using the relationship between the heat capacity of the calorimeter and the amount of heat released during combustion of the substance. Given the combustion of anethole releases 5541 kJ per mole, we first need to determine the amount of energy released during the combustion of 0.950 grams of anethole. Since the molecular weight of anethole (C10H12O) is approximately 148.2 g/mol, we can calculate q (the heat released) for the given mass, and then use this to determine the change in temperature (ΔT).
To calculate the energy released:
q = (0.950 g / 148.2 g/mol) * 5541 kJ/molq = 0.0353 mol * 5541 kJ/mol = 195.59 kJThen, using the calorimeter's heat capacity:
ΔT = q / CcalorimeterΔT = 195.59 kJ / 7.854 kJ/°C ≈ 24.90°CHence, the change in temperature of the calorimeter is approximately 24.90°C.
How is the atomic mass of an element calculated from isotope data
Answer:
[tex]aam=\frac{\Sigma m_{i} \times ab_{i} }{100}[/tex]
Explanation:
When an atom has 2 or more isotopes, the average atomic mass (aam) depends on the mass of each isotope (mi) and the percentual abundance in nature of each isotope (abi). The average atomic mass can be calculated using the following expression:
[tex]aam=\frac{\Sigma m_{i} \times ab_{i} }{100}[/tex]
A one or two letter abbreviation for a chemical element. True or false
Answer:
True
Explanation:
All the chemical elements have their abbreviations as one or two letters
The first letter in the abbreviation is a capital letter and the second letter is always a small letter
Majority of abbreviations are based on their english names
Final answer:
The statement regarding a chemical symbol being a one- or two-letter abbreviation for a chemical element is true. These symbols are a shorthand for elements, such as 'O' for oxygen or 'Fe' for iron, and are crucial for anyone studying chemistry.
Explanation:
The statement that a chemical symbol is a one- or two-letter abbreviation for a chemical element is true. Chemical symbols provide a concise way to represent elements in scientific writings and discussions. For instance, the symbol 'O' stands for oxygen, while 'Zn' represents zinc. The first letter of a chemical symbol is always capitalized, and if there is a second letter, it is lowercase.
Elements that have been known for a long time often have symbols derived from their Latin names. For example, the symbol for iron is 'Fe', which comes from its Latin name 'ferrum'. It is important for those studying chemistry to become familiar with the chemical symbols of the elements, most of which you can find on the periodic table.
Newton's 2nd Law says the acceleration of an object depends on its ______ and the amount of _______ acting on it.
Friction and inertia
Mass and force
Force and friction
Inertia and Gravity
Answer:
Newton's 2nd Law says the acceleration of an object depends on its mass and the amount of net force acting on it.
Explanation:
Definition of acceleration:
The acceleration is rate of change of velocity of an object with respect to time.
Formula:
a = Δv/Δt
a = acceleration
Δv = change in velocity
Δt = change in time
Units:
The unit of acceleration is m.s⁻².
Acceleration can also be determine through following formula,
F = m × a
a = F/m
This is the newton's second law:
"The acceleration of an object depends on its mass and the amount of net force acting on it"
The acceleration is depend directly on the force while inversely on the mass.
Answer:
Mass and Force
Explanation:
The following change occurs in acidic solution: S^2+ + Cr_2O_7^2+ S + Cr^3+ Complete and balance the foregoing equation. In the balanced equation, for every mole of Cr_2O_7^2 that reacts, moles of H+ are consumed. A. 5 B. 7 C. 8 D. 10 E. 14
Answer : The correct option is, (E) 14
Explanation :
Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.
Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.
Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.
Rules for the balanced chemical equation in acidic solution are :
First we have to write into the two half-reactions.Now balance the main atoms in the reaction.Now balance the hydrogen and oxygen atoms on both the sides of the reaction.If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion [tex](H^+)[/tex] at that side where the less number of hydrogen are present.Now balance the charge.The given chemical reaction is,
[tex]S^{2-}+Cr_2O_7^{2-}\rightarrow S+Cr^{3+}[/tex]
The oxidation-reduction half reaction will be :
Oxidation : [tex]S^{2-}\rightarrow S[/tex]
Reduction : [tex]Cr_2O_7^{2-}\rightarrow Cr^{3+}[/tex]
First balance the main element in the reaction.
Oxidation : [tex]S^{2-}\rightarrow S[/tex]
Reduction : [tex]Cr_2O_7^{2-}\rightarrow 2Cr^{3+}[/tex]
Now balance oxygen atom on both side.
Oxidation : [tex]S^{2-}\rightarrow S[/tex]
Reduction : [tex]Cr_2O_7^{2-}\rightarrow 2Cr^{3+}+7H_2O[/tex]
Now balance hydrogen atom on both side.
Oxidation : [tex]S^{2-}\rightarrow S[/tex]
Reduction : [tex]Cr_2O_7^{2-}+14H^+\rightarrow 2Cr^{3+}+7H_2O[/tex]
Now balance the charge.
Oxidation : [tex]S^{2-}\rightarrow S+2e^-[/tex]
Reduction : [tex]Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O[/tex]
In order to balance the charge, we multiply oxidation reaction by 3. Thus, added both equation, we get the balanced redox reaction.
The balanced chemical equation in acidic medium will be,
[tex]3S^{2-}+Cr_2O_7^{2-}+14H^+\rightarrow 3S+2Cr^{3+}+7H_2O[/tex]
From the balanced chemical equation we conclude that, the number of moles of [tex]H^+[/tex] consumed from every mole of [tex]Cr_2O_7^{2-}[/tex] are 14 moles.
Hence. the correct option is, (E) 14
A 17.55 mL solution of potassium nitrate (KNO3) was diluted to 125.0 mL, and 25.00 mL of this solution was then diluted to 1.000 × 103 mL. The concentration of the final solution is 0.00383 M. Calculate the concentration of the original solution.
Answer:
[tex]M_2 = 1.094 M[/tex]
Explanation:
Given data:
V1 = 25.0 ml
M2 = 0.00383 M
V2 = 1000 ml
we knwo that [tex]M_1 V_1 = M_2 V_2[/tex]
∴ [tex]M_1 = \frac{M_2 V_ 2}{V_1}[/tex]
[tex] = \frac{0.00383 M \times 1000}{25.0}[/tex]
= 0.1532 M
0.1532 M is concentration in 25 ml but taken from 125 ml solution.
∴ The concentration in 125.0 ml solution is = 0.1532 M
M1 = 0.1532 M
V1 = 125.0ml and V2 = 17.5.0ml
[tex]M_2 = \frac{M_1 V_1}{V_2}[/tex]
[tex] = \frac{0.1532 M \times 125.0}{17.5}[/tex]
[tex]M_2 = 1.094 M[/tex]
Which process is a chemical reaction? Which process is a chemical reaction? Limescale is formed on the kettle walls Dew condensing on grass during the night Gasoline evaporating from a gasoline tank Lake freezes in winter
Answer:
Limescale formed on kettle walls
Explanation:
A chemical reaction is one which is associated with a chemical change. While the other two examples are mere change in physical state, the formation of limescale on kettle is a chemical change. It is called the furring of kettles.
These limescales are formed when Calcium bicarbonate decomposes into calcium carbonate. It is this calcium carbonate that causes the furring of kettles.
It is one of the consequence of using temporary hard water. Temporarily hard water contains soluble magnesium bicarbonate and calcium bicarbonate. Now the heating of this water causes the decomposition of the calcium bicarbonate into calcium carbonate which forms these scales on the body of the kettle.
Calcium bicarbonate decomposes into calcium carbonate according to the following equation;
CaH(CO3)2 (aq) ---------> CaCO3 (s) + H2O (l) + CO2 (g)
According to your textbook, how many of the following are properties of covalent compounds:
a. *Can be gas, liquid or solid at room temperature
b. *Does not conduct electricity
c. *Exist as an extended array of atoms
d. *Involves the sharing of electrons
Answer:
a, b and d are correct.
Explanation:
There are two types of attractive forces in covalent compounds:
The force that holds the atoms together in a molecule (the covalent bond).The force that operates between molecules, called an intermolecular force.Because intermolecular forces are usually quite weak compared with the forces holding atoms together within a molecule, molecules of a covalent compound are not held together tightly. Consequently, covalent compounds are usually gases, liquids, or low-melting solids.
Most covalent compounds aqueous solutions generally do not conduct electricity, because the compounds are nonelectrolytes.
How many moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl?
2 Al(s) + 6 HCl(aq) → 2 AlCl3 (aq) + 3 H2(g)
Answer:
0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl
Explanation:
This is the reaction:
2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)
To make 3 moles of H₂, we need 2 moles of Al.
By conditions given, we will find out how many moles of H₂ do we have.
Let's use the Ideal Gas Law
P. V = n . R . T
1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K
(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n
0.182 mol = n
So the rule of three will be:
If 3 moles of H₂ came from 2 moles of Al
0.182 moles of H₂ will come from x
(0.182 .2) / 3 = 0.121 moles
Entry to the citric acid cycle involves the generation of acetyl CoA. The enzyme that catalyzes the production of acetyl CoA is _____. This enzyme is negatively regulated by _____ in a high-energy charge situation.
Answer:
Pyruvate dehydrogenase / Feedback on pyruvate decarboxylase production
Explanation:
In the Krebs cycle, pyruvic acid from glycolysis undergoes an oxidative decarboxylation process through the action of the pyruvate dehydrogenase enzyme found within the mitochondria of eukaryotes, it reacts with coenzyme A (CoA).
The result of this reaction is the production of acetylcoenzyme A (acetylCoA) and a carbon dioxide (CO₂) molecule.
When acetyl CoA and ATP are at high concentrations (in addition to the increased NADH / NAD⁺ ratio), pyruvate carboxylase production is stimulated.
This process will eventually generate oxalacetic acid for gluconeogenesis (conversion of pyruvate to glucose).
By increasing oxalacetic acid concentrations, pyruvate dehydrogenase is eventually inhibited by a negative feedback mechanism.
In other words, as energy levels increase, the higher the pyruvate carboxylase production, and therefore the greater the inhibition of pyruvate dehydrogenase.
Calculate the percent ionization of 0.120 M lactic acid in a solution containing 8.5×10−3 M sodium lactate.
Express the percent ionization to two significant digits.
To calculate the percent ionization in this scenario, information regarding the ionization constant of lactic acid and the pH is required. The calculation would typically use the Henderson-Hasselbalch equation, but without additional data, we cannot provide a specific answer.
Explanation:The question involves the calculation of the percent ionization of a weak acid in the presence of its conjugate base. To calculate this, one would need the initial concentration of the acid, the acid ionization constant (Ka), and the concentration of the conjugate base. Generally, percent ionization is given by the ratio of the concentration of ionized acid to the initial concentration of the acid, multiplied by 100%. However, since the pH or hydronium ion concentration is not provided and the solution is a buffer system (acid with its conjugate base), the Henderson-Hasselbalch equation would typically be used to find the pH first, then the percent ionization could be calculated.
As specific values are not provided for the ionization constant of lactic acid or the pH of the solution, an accurate calculation cannot be completed without them. If these values were provided, the calculation would involve using the Henderson-Hasselbalch equation to solve for pH, determining the hydronium ion concentration, and subsequently finding the percent ionization.
HCl gas is introduced at one end of the tube, and simultaneously NH3 gas is introduced at the other end. When the two gases diffuse through the cotton plugs down the tube and meet, a white ring appears due to the formation of NH4Cl(s). Why is this?
Answer:
When the two gases are mixed, the ammonium chloride precipitates in the tube walls.
Explanation:
This is the reaction:
HCl (g) + NH₃(g) → NH₄Cl (s) ↓
As the product formed is solid at room temperature, a suspension is first formed in the internal air of the tube that appears as a cloud. Afterwards it finally precipitates into the walls forming a white layer
A sample of benzene was vaporized at 25◦C. When 37.5 kJ of heat was supplied, 95.0 g of the liquid benzene vaporized. What is the enthalpy of vaporization of benzene at 25◦C?
1. 30.8 kJ/mol
2. 13.4 kJ/mol
3. 43.0 kJ/mol
4. 6.09 kJ/mol
5. 25.6 kJ/mol
Answer:
Enthalpy of vaporization = 30.8 kj/mol
Explanation:
Given data:
Mass of benzene = 95.0 g
Heat evolved = 37.5 KJ
Enthalpy of vaporization = ?
Solution:
Molar mass of benzene = 78 g/mol
Number of moles = mass/ molar mass
Number of moles = 95 g/ 78 g/mol
Number of moles = 1.218 mol
Enthalpy of vaporization = 37.5 KJ/1.218 mol
Enthalpy of vaporization = 30.8 kj/mol
Final answer:
The enthalpy of vaporization of benzene at 25°C is calculated by dividing the heat energy supplied by the number of moles vaporized, resulting in 30.8 kJ/mol.
Explanation:
The enthalpy of vaporization of benzene at 25°C can be calculated using the amount of heat supplied and the mass of benzene vaporized. Here's a step-by-step explanation:
First, convert the mass of benzene to moles using its molar mass (78.11 g/mol for C6H6).
Divide the given heat energy by the number of moles to find the enthalpy of vaporization per mole.
The calculation is as follows:
Moles of benzene = 95.0 g / 78.11 g/mol = 1.2169 mol
Enthalpy of vaporization (ΔHvap) = 37.5 kJ / 1.2169 mol = 30.8 kJ/mol
The correct answer is 30.8 kJ/mol, which is option 1.
Which of the following is true about halogens? Group of answer choices
they have seven valence electrons
they have two valence electrons
they have one valence electron
they have eight valence electrons
Answer:they have one valence electron
Explanation:halogens are in group seven of the periodi table and they have seven electrons in their outermost shell thus need one electron to complete their octet condition. Thus goes into combination to accept one electrons from another element and that is their combining power which is their valency.
The halogens have seven valence electrons.
• In the periodic table, the elements present in group 7A are known as halogens.
• These are fluorine, chlorine, bromine, iodine, and astatine.
• The halogen term means salt former.
• The major characteristic of halogens is that they have seven valence electrons in their higher energy orbitals.
• They are one electron behind in forming a full octet, so these elements would seem to produce anions exhibiting -1 charges and are known as halides.
Thus, halogens having seven valence electrons is the right statement.
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Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and the other at Tc, are brought into contact and allowed to reach equilibrium. Evaluate the change for the two blocks of copper, each of mass 500 grams with Cpcm= 24.4 J KT-1 mol-1, taking Th = 500 K and Tc = 250 K.
Explanation:
Relation between entropy change and specific heat is as follows.
[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
The given data is as follows.
mass = 500 g, [tex]C_{p}[/tex] = 24.4 J/mol K
[tex]T_{h}[/tex] = 500 K, [tex]T_{c}[/tex] = 250 K
Mass number of copper = 63.54 g /mol
Number of moles = [tex]\frac{mass}{/text{\molar mass}}[/tex]
= [tex]\frac{500}{63.54}[/tex]
= 7.86 moles
Now, equating the entropy change for both the substances as follows.
[tex]7.86 \times 24.4 \times [T_{f} - 250][/tex] = [tex]7.86 \times 24.4 \times [500 -T_{f}][/tex]
[tex]T_{f} - 250 = 500 - T_{f}[/tex]
[tex]2T_{f}[/tex] = 750
So, [tex]T_{f}[/tex] = [tex]375^{o}C[/tex]
For the metal block A, change in entropy is as follows.[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
= [tex]24.4 log [\frac{375}{500}][/tex]
= -3.04 J/ K mol
For the block B, change in entropy is as follows.[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
= [tex]24.4 log [\frac{375}{250}][/tex]
= 4.296 J/Kmol
And, total entropy change will be as follows.
= 4.296 + (-3.04)
= 1.256 J/Kmol
Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.
If a polypeptide synthesized by a ribosome contains 99 amino acids in the chain, how many codons (including the start and stop codons) were required on the mRNA to make the polypeptide chain?
Explanation:
-101 mRNA codons
In the genetic code, an amino acid is encoded by 3 nucleotides, while there are just 4 bases . Each amino acid is specifically encoded by a codon- a triplet sequence of nucleotides...
Thus 99 amino acids= 99 codon sequences.
...along with a start and stop codon, this would require 101 mRNA codons
Further Explanation:
The nucleic acids are comprised of smaller units called nucleotides and function as storage for the body’s genetic information. These monomers include ribonucleic acid (RNA) or deoxyribonucleic acid (DNA). They differ from other macromolecules since they don’t provide the body with energy. They exist solely to encode and protein synthesis.
Basic makeup: C, H, O, P; they contain phosphate group 5 carbon sugar does nitrogen bases which may contain single to double bond ring.
Codons are three nucleotide bases encoding an amino acid or signal at the beginning or end of protein synthesis.
RNA codons determine certain amino acids so the order in which the bases occur within in the codon sequence designates which amino acid is to be made bus with the four RNA nucleotides (Adenine, Cysteine and Uracil) Up to 64 codons (with 3 as stop codons) determine amino acid synthesis. The stop codons ( UAG UGA UAA) terminate amino acid/ protein synthesis while the start codon AUG begins protein synthesis.
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Select all of the following statements that are incorrect. A. Solids, liquids, and gases are all able to act as solutes in a solution. B. Solids, liquids, and gases are all able to act as solvents in a solution. C. Solids are not able to act as a solvent if a gas is the solute. D. Gases are not able to act as a solvent if a solid is the solute.
Answer:
B
Explanation:
Explanation:
Adsorption is defined as a process in which gas molecules tend to adsorb on the surface of a solid solvent.
Therefore, the statement solids are not able to act as a solvent if a gas is the solute, is false.
On the other hand, gases can also act as a solvent if a solid is the solute.
For example, air present in the surrounding acts as a solvent and solid substances present in the surrounding acts as a solute.
Hence, the statement gases are not able to act as a solvent if a solid is the solute, is false.
Thus, we can conclude that following are the statements which are incorrect.
Solids are not able to act as a solvent if a gas is the solute. Gases are not able to act as a solvent if a solid is the solute.Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system, based on the following descriptions:
A. Surroundings get colder and the system decreases in volume.
B. Surroundings get hotter and the system expands in volume.
C.Surroundings get hotter and the system decreases in volume.
D. Surroundings get hotter and the system does not change in volume.
Also assume that the magnitude of the volume and temperature changes are similar among the reactions.
Rank from most energy released to most energy absorbed. To rank items as equivalent, overlap them.
Answer:
B > D > C > A
Explanation:
For the first law of the thermodynamics, the total energy variation in a process is:
ΔU = Q - W
Where Q is the heat, and W the work. If the system loses heat, Q < 0, if it absorbs heat, Q>0. If work is done in the system (volume decreases), W < 0, if the system does the work (volume increases), W > 0.
A. If the surroundings get colder, the system is absorbing heat, so Q>0, and the system decreases in volume so W < 0 :
ΔU = +Q - (-W) = +Q + W (absorbs a higher energy)
B. If the surroundings ger hotter, the system is losing heat, so Q<0, and the system expands, so W>0:
ΔU = -Q -W (loses higher energy)
C. Surroundings get hotter, Q<0, and the system decreases in volume, W<0
ΔU = - Q + W = 0 (magnitude of heat and work is similar)
D. Surroundings get hotter, Q<0, and the system is not changing in volume, W = 0.
ΔU = -Q (loses energy)
For the most released (more negative) for the most absorbed (most positive):
B > D > C > A
Final answer:
Reaction B is the most exothermic as it releases energy and the surroundings get hotter. Reaction D is less exothermic as the surroundings get warmer without volume change. Reaction A is most endothermic, absorbing energy, indicated by cooler surroundings.
Explanation:
Ranking the reactions from most energy released to most energy absorbed:
Exothermic reaction: Surroundings get hotter and the system expands in volume (B). This reaction releases the most energy, indicated by the heat given off to the surroundings.Surroundings get hotter but volume doesn't change (D). This is still an exothermic reaction but less energy is released compared to B due to no work done on expansion.Surroundings get hotter and the system decreases in volume (C). This describes an endothermic process where energy is absorbed, but because the surroundings get warmer, it's indicative of a net release of energy after accounting for the work done on the system (compression).Endothermic reaction: Surroundings get colder and the system decreases in volume (A). This reaction absorbs the most energy, indicated by the surroundings getting colder.In summary, the reactions involving the surroundings getting warmer are generally exothermic, while those involving the surroundings getting colder are endothermic. The volume change provides additional clues about energy changes; expansion suggests work is done by the system (releasing energy), while a decrease in volume suggests work is done on the system (absorbing energy).
How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.
Answer:
V H2O = 170.270 mL
Explanation:
QH2O ( heat gained) = Qcoffe ( heat ceded)⇒ Q = mCΔT
∴ m: mass (g)
∴ C: specific heat
assuming:
δ H2O = δ Coffe = 1.00 g/mLC H2O = C coffe = 4.186 J/°C.g....from literature⇒ Q coffe = (mcoffe)(C coffe)(60 - 95)
∴ m coffe = (180mL)(1.00 g/mL) = 180 g coffe
⇒ Q = (180g)(4.186 J/°C.g)(-35°C) = - 26371.8 J
⇒ Q H2O = 26371.8 J = (m)(4.186 J/°C.g)(60 - 23)
⇒ (26371.8 J)/(154.882 J/g) = m H2O
⇒ m H2O = 170.270 g
⇒ V H2O = (170.270 g)(mL/1.00g) = 170.270 mL
The beta oxidation pathway degrades activated fatty acids (acyl-CoA) to acetyl-CoA, which then enters the citric acid cycle. Additional enzymes are required to oxidize unsaturated, odd-chain, long-chain, and branched fatty acids.
Which of the following statements are true?
a.A 14-carbon monounsaturated fatty acid with cis configuration yields more ATP than a 14-carbon saturated fatty acid.
b.Enoyl-CoA isomerase, an enzyme that converts cis double bonds to trans double bonds in fatty acid metabolism, bypasses a step that reduces Q, resulting in a higher ATP yield.
c.Even-chain saturated fatty acids are oxidized to acetyl-CoA in the beta oxidation pathway.
d.Complete catabolism of the three-carbon remnant of a 15-carbon fatty acid requires some citric acid cycle enzymes.
e.Trans double bonds in unsaturated fatty acids are not recognized by beta oxidation enzymes.
f.The final round of beta oxidation for a 13-carbon saturated fatty acid yields acetyl-CoA and propionyl-CoA, a three-carbon fragment.
Final answer:
Enoyl-CoA isomerase bypasses a step that reduces Q, resulting in higher ATP yield. Even-chain saturated fatty acids are oxidized to acetyl-CoA in beta oxidation. Complete catabolism of a 15-carbon fatty acid requires some citric acid cycle enzymes.
Explanation:
Statement b is true. Enoyl-CoA isomerase, an enzyme in fatty acid metabolism, bypasses a step that reduces Q, resulting in higher ATP yield.
Statement c is true. Even-chain saturated fatty acids are oxidized to acetyl-CoA in the beta oxidation pathway.
Statement d is true. Complete catabolism of the three-carbon remnant of a 15-carbon fatty acid requires some citric acid cycle enzymes.
A hypothetical covalent molecule, X–Y, has a dipole moment of 1.93 1.93 D and a bond length of 109 pm. 109 pm. Calculate the partial charge on a pole of this molecule in terms of e , e, where e e is the charge on an electron.
Answer:
q= 110.5 ke
Explanation:
Dipole moment is the product of the separation of the ends of a dipole and the magnitude of the charges.
μ = q * d
μ= Dipole moment (1.93 D)
q= partial charge on each pole
d= separation between the poles(109 pm).
e= electronic charge ( 1.60217662 × 10⁻¹⁹ coulombs)
So,
q= [tex]\frac{1.93}{109 * 10^{-12} }[/tex] coulombs
q = [tex]\frac{1.93}{109 * 10^{-12} * 1.60217662 * 10^{-19} }[/tex] e
q = 1.105 * 10⁵ e
q= 110.5 ke
Enter your answer in the provided box.S(rhombic) + O2(g) → SO2(g) ΔHo rxn= −296.06 kJ/molS(monoclinic) + O2(g) → SO2(g) ΔHo rxn= −296.36 kJ/molcalculate the enthalpy change for the transformationS(rhombic) → S(monoclinic)(Monoclinic and rhombic are different allotropic forms of elemental sulfur.)_______kJ/mol
Answer: [tex]\Delta H^0=+0.3kJ/mol[/tex].
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
[tex]S_{rhombic}+O_2(g)\rightarrow SO_2(g)[/tex] [tex]\Delta H^0_1=-296.06kJ[/tex] (1)
[tex]S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ[/tex] (2)
The final reaction is:
[tex]S_{rhombic}\rightarrow S_{monoclinic}[/tex] [tex]\Delta H^0_3=?[/tex] (3)
By subtracting (1) and (2)
[tex]\Delta H^0_3=\Delta H^0_1-\Delta H^0_2=-296.06kJ-(-296.36kJ)=0.3kJ[/tex]
Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ
The enthalpy change for the transformation from rhombic sulfur to monoclinic sulfur is -0.30 kJ/mol.
The enthalpy change for the transformation S(rhombic) → S(monoclinic), we will use the enthalpy changes provided for the reactions involving sulfur and oxygen:
S(rhombic) + O₂(g) → SO₂(g) ΔHo rxn = −296.06 kJ/molS(monoclinic) + O₂(g) → SO₂(g) ΔHo rxn = −296.36 kJ/molGiven these values, the enthalpy change for the transformation from rhombic sulfur to monoclinic sulfur can be found as follows:
ΔH (transformation) = ΔHo(S(monoclinic) → SO₂) - ΔHo(S(rhombic) → SO₂)
Using the provided enthalpy changes:
ΔH (transformation) = -296.36 kJ/mol - (-296.06 kJ/mol)
ΔH (transformation) = -296.36 kJ/mol + 296.06 kJ/mol
ΔH (transformation) = -0.30 kJ/mol
The enthalpy change for the transformation S(rhombic) → S(monoclinic) is -0.30 kJ/mol.
For water ∆H°vap = 40.7 kJ/mol at 100.°C, its boiling point. Calculate ∆S° for the vaporization of 1.00 mol water at 100.°C and 1.00 atm pressure.
Answer : The value of change in entropy for vaporization of water is [tex]1.09\times 10^2J/mol[/tex]
Explanation :
Formula used :
[tex]\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}[/tex]
where,
[tex]\Delta S^o[/tex] = change in entropy of vaporization = ?
[tex]\Delta H^o_{vap}[/tex] = change in enthalpy of vaporization = 40.7 kJ/mol
[tex]T_b[/tex] = boiling point temperature of water = [tex]100^oC=273+100=373K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}[/tex]
[tex]\Delta S^o=\frac{40.7kJ/mol}{373K}[/tex]
[tex]\Delta S^o=1.09\times 10^2J/mol[/tex]
Therefore, the value of change in entropy for vaporization of water is [tex]1.09\times 10^2J/mol[/tex]
The entropy change ∆S° for the vaporization of 1.00 mol of water at its boiling point is calculated to be 109.09 J/K·mol using the given enthalpy of vaporization.
Explanation:To calculate ∆S° for the vaporization of 1.00 mol water at 100.°C and 1.00 atm pressure when ∆H°vap = 40.7 kJ/mol, we can use the thermodynamic relation ∆G° = ∆H° - T∆S° and the fact that at the boiling point, ∆G° for the phase change is zero. This would mean the ∆S° can be calculated as ∆S° = ∆H°vap/T. Plugging in the known values,
∆S° = 40.7 kJ/mol / 373.15 K
∆S° = 0.10909 kJ/K·mol
Since 1 kJ = 1000 J, we convert this to J:
∆S° = 109.09 J/K·mol
Therefore, the entropy change ∆S° for the vaporization of 1.00 mol of water at its boiling point is 109.09 J/K·mol.
Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and 791 mmHg. Calculate the percent by mass of calcium carbonate in the sample.
Final answer:
To calculate the percent by mass of calcium carbonate in the sample, we need to determine the mass of calcium carbonate present in the 1.14 L of carbon dioxide produced. We can use the ideal gas law to calculate the moles of carbon dioxide, and then use stoichiometry to convert from moles of CO2 to moles of CaCO3. Finally, we can calculate the percent by mass of CaCO3 in the sample by dividing the mass of CaCO3 by the mass of the sample and multiplying by 100.
Explanation:
To calculate the percent by mass of calcium carbonate in the sample, we need to determine the mass of calcium carbonate present in the 1.14 L of carbon dioxide produced. We can use the ideal gas law to calculate the moles of carbon dioxide, and then use stoichiometry to convert from moles of CO2 to moles of CaCO3. Finally, we can calculate the percent by mass of CaCO3 in the sample by dividing the mass of CaCO3 by the mass of the sample and multiplying by 100.
First, let's calculate the moles of CO2 using the ideal gas law:
mass of CO2 = (volume of CO2) x (molar mass of CO2) = (1.14 L) x (22.4 g/mol) = 25.536 g
Next, let's calculate the moles of CaCO3 using stoichiometry. From the balanced equation CaCO3 + 2HCl → CaCl2 + H2O + CO2, we know that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, moles of CaCO3 = moles of CO2 = 25.536 g / (molar mass of CaCO3) = 25.536 g / 100.09 g/mol = 0.255 mol
Finally, let's calculate the percent by mass of CaCO3:
percent by mass of CaCO3 = (mass of CaCO3 / mass of the sample) x 100 = (0.255 mol x (100.09 g/mol)) / 5.28 g x 100 = 4.85%
The principal source of the difference in the normal boiling points of ICI (97°C; molecular mass 162 amu) and Br2(59°C; molecular mass 160 amu) is:a. ICI has greater strength of hydrogen bonding than Br2 b. ICI has stronger dipole-dipole interactions than Br2c. ICI has stronger dispersion forces than Br2 d. The I-CI bond in ICl is stronger than the Br-Br bond in Br2 e. ICI has a greater molecular mass than Br2
Answer:
Option b
Explanation:
The difference in the boiling points of ICl and Br2 is mainly due to the dipole-dipole interactions present between iodine and chlorine in the ICl molecule because as there is difference in electronegativity between iodine and chlorine these type of forces arise
As the dipole-dipole interactions are stronger, the stronger will be the boiling point of that compound because the forces between the molecules increases and as a result the boiling point of the compound increases
In case of Br2 as both are bromine atoms there will be no difference in electronegativity and therefore these type of interactions are not present
Molecular mass can also be a explanation for difference in normal boiling points because more molecular mass means more will be the vander waals forces but as dipole interactions are stronger than vander waals forces the major factor will be due to dipole interactions
The principal source of difference in the normal boiling points of ICI and Br2 is that ICI has stronger dipole-dipole interactions. Although both have similar London dispersion forces due to similar molecular masses, ICI being a polar molecule exhibits stronger dipole-dipole attractions, requiring more energy to overcome and hence has a higher boiling point.
Explanation:The principal source of difference in the normal boiling points of ICI (97°C; molecular mass 162 amu) and Br2(59°C; molecular mass 160 amu) is that ICI has stronger dipole-dipole interactions than Br2. Both ICI and Br2 have similar masses and therefore experience similar London dispersion forces. However, compared to Br2 which is nonpolar, ICI is a polar molecule and thus also exhibits dipole-dipole attractions. These dipole-dipole attractions in ICI are stronger and require more energy to overcome, resulting in a higher boiling point for ICI.
Larger and heavier atoms and molecules like ICI exhibit stronger dispersion forces, leading to higher melting and boiling points. However, in the case of ICI and Br2, the effect of polar dipole-dipole attraction in ICI is more significant. This is apparent when we compare substances with similar molecular masses - the substance with the polar molecules has a higher boiling point, due to the stronger dipole-dipole attractions.
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A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Calculate the density, mole fraction, molarity, and molality of the solution.
Answer:
Density is: 1.05 g/ml
Mole fraction solute: 0.015
Mole fraction solvent: 0.095
Molarity: 0.80 M
Molality: 0.82 m
Explanation:
A typical excersise of solution.
It is more confortable to make a table for this.
| masss | volume | mol
solute | | |
solvent | | |
solution | | |
Let's complete, what we have.
| masss | volume | mol
solute | 10.8g | |
solvent | | 133 mL |
solution | | 137 mL |
We can first, know how many moles are 10.8 g
Molar Mass H3PO4 = 97.99 g/mol
Mass / Molar mass = mol
10.8 g / 97.99 g/m = 0.110 mol
Density of water is 1 g/ml (it is a very knowly value)
From this data, we can know water mass, solvent.
Density = mass / volume
1 g/ml = mass / 133 mL
Mass = 133 g
We can also have the moles, by the molar mass of water 18 g/m
133 g / 18 g/m = 7.39 mol
| masss | volume | mol
solute | 10.8g | | 0.110 mol
solvent | 133g | 133 mL | 7.39 mol
solution | 143.8g | 137 mL | 7.50 mol
Mass of solution will be solute mass + solvent mass
Moles of solution will be solute moles + solvent moles
Now we can calculate everything.
Molarity means mol of solute in 1 L of solution. (mol/L)
We have to convert 137 mL in L (/1000)
0.137L so → 0.110 m / 0.137L = 0.80 M
Molality means mol of solute in 1kg of solvent.
We have to convert 133g in kg (/1000)
0.133 kg so → 0.110 m/0.133 kg = 0.82 m
Density is mass / volume
Solution density will be solution mass / solution volume
143.8 g/137 mL = 1.05 g/m
Molar fraction is : solute moles / total moles or solvent moles/total moles.
You can also (x 100%) to have a percent of them.
Remember sum of molar fraction = 1
Molar fraction of solute = 0.110 mol / 7.50mol = 0.015
Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985
A 2.85-g sample of an unknown chlorofluorocarbon is decomposed and produces 564 mL of chlorine gas (Cl2) at a pressure of 752 mmHg and a temperature of 298 K. What is the percent chlorine (by mass) in the unknown chlorofluorocarbon
The percentage of chlorine by mass in the unknown chlorofluorocarbon is 56.8%
We'll begin by calculating the number of mole of Cl₂ produced using the ideal gas equation as illustrated below:
Volume (V) = 564 mL = 564 / 1000 = 0.564 LPressure (P) = 752 mmHg = 752 / 760 = 0.989 atmTemperature (T) = 298 KGas constant (R) = 0.0821 atm.L/KmolNumber of mole (n) =?PV = nRT
0.989 × 0.564 = n × 0.0821 × 298
0.557796 = n × 24.4658
Divide both side by 24.4658
n = 0.557796 / 24.4658
n = 0.0228 mole
Next, we shall determine the mass of 0.0228 mole of Cl₂
Mole = 0.0228 moleMolar mass of Cl₂ = 2 × 35.5 = 71 g/molMass of Cl₂ =?Mass = mole × molar mass
Mass of Cl₂ = 0.0228 × 71
Mass of Cl₂ = 1.62 g
Finally, we shall determine the percentage of chlorine in the unknown chlorofluorocarbon.
Mass of Cl₂ = 1.62 gMass of compound = 2.85 gPercentage of chlorine =?Percentage = (mass / total mass) × 100
Percentage of chlorine =(1.62 / 2.85) × 100
Percentage of chlorine = 56.8%
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The percent chlorine (by mass) in the unknown chlorofluorocarbon is 0%.
Explanation:In order to determine the percent chlorine (by mass) in the unknown chlorofluorocarbon, we need to calculate the mass of chlorine gas produced from the 2.85-g sample. From the equation 2.5g H and 7.5g C make up a 10.0g sample, we can calculate the percent composition of hydrogen and carbon to be 25% and 75%, respectively.
Knowing the percent composition of hydrogen and carbon, we can assume that the unknown chlorofluorocarbon consists of only hydrogen, carbon, and chlorine. Since chlorine gas is produced when the unknown chlorofluorocarbon is decomposed, the percent chlorine (by mass) can be calculated by subtracting the percent composition of hydrogen and carbon from 100%. Therefore, the percent chlorine is 100% - (25% + 75%) = 0%.
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For the reaction, calculate how many grams of the product form when 14.4 g of Br2 completely reacts. Assume that there is more than enough of the other reactant. 2K(s)+Br2(l)→2KBr(s)
Answer:
The answer to your question is 21.45 g of KBr
Explanation:
Chemical reaction
2K + Br₂ ⇒ 2KBr
14.4 ?
Process
1.- Calculate the molecular mass of bromine and potassium bromide
Bromine = 2 x 79.9 = 159.8g
Potassium bromide = 2(79.9 + 39.1) = 238 g
2.- Solve it using proportions
159.8 g of Bromine ------------ 238 g of potassium bromide
14.4 g of Bromine ------------ x
x = (14.4 x 238) / 159.8
x = 3427.2 / 159.8
x = 21.45g of KBr
The process where bacteria takes nitrogen out of the air and makes it usable for other organisms is called
Answer:
The process is called Nitrogen fixation
Explanation:
The nitrogen fixation is a process carried out by some prokaryotic microorganisms (bacteria), specifically those have the presence of the nitrogenase enzyme. The bacteria absorb the atmospheric nitrogen (N2) from the roots of plants, and the nitrogenase enzyme, with the help of two proteins that act as electron donors and acceptors (nitrogenase complex) reduce the nitrogen to ammonia (NH3), then the ammonia is ionized to NH4+ (ammonium). Followed by that, the ammonia is oxidated to nitrates and nitrites, which are finally absorbed again by plants.
Fireworks that contain metallic salts such as sodium, strontium, and barium can generate bright colors. A technician investigates what colors are produced by the metallic salts by performing flame tests. During a flame test, a metallic salt is heated in the flame of a gas burner. Each metallic salt emits a characteristic colored light in the flame. 3. State how bright-line spectra viewed through a spectroscope can be used to identify the metal ions in the salts used in the flame tests. 4. Explain, in terms of electrons, how a strontium salt emits colored light. 5. Explain why the electron configuration of 2-7-1-1 represents a sodium atom in an excited state.
Answer:
See explanation below
Explanation:
3. Each atom has unique electron energy level when compared to other, when electron relaxes from an excited energy level, it emits energy in the form of electromagnetic waves with energy equals to the difference between energy of ground and excited state. As electromagnetic wave wavelength obeys planck relation
[tex]c = \lambda f[/tex]
And the energy of each photon in electromangetic wave of specific frequency is
[tex]E=hf[/tex]
Where h is planck's constant
We can see that the wavelength, which determines color and position of bright line in spectroscope corresponds directly to the energy of electromagnetic wave, and thus, the characteristic energy of atom.
4. From the answer above, you now know that electronic transition can emit electron with specific wavelength. In strontium, emission of colored light occurs by the relaxation of electrons in excited state such that the difference between excited energy level and ground level falls in the energy of visible light spectral range.
5. In Na, ground state configuration is 2-8-1. 2-7-1-1 indicates that one of electron in ground state got excited by external energy to excited state, and thus, indicates that Na atom is in excited state
Final answer:
Bright-line spectra viewed through a spectroscope can be used to identify the metal ions in salts used in flame tests.
Explanation:
Bright-line spectra viewed through a spectroscope can be used to identify the metal ions in the salts used in flame tests. Each metal ion has a unique set of energy levels, and when these ions are heated in a flame, their electrons get excited and jump to higher energy levels. As the electrons return to their ground states, they release energy in the form of light. This emitted light can be observed through a spectroscope, which separates the different wavelengths of light, creating a bright-line spectrum. By comparing the bright-line spectrum of an unknown salt to the known spectra of different metal ions, the metal ions present in the unknown salt can be identified.