The standard entropy change of a reaction has a positive value. This reaction results in: Select the correct answer below: a decrease in entropy. an increase in entropy. no entropy change. neither an entropy increase nor decrease.
Explanation:
The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.
Positive entropy means the system has increased its degree of disorderness.
Which of the following ligands is not capable of exhibiting linkage isomerism?
a. NCO-
b. -OH
c. -CN
d. -SCN
Answer:
a
...
........
...........
In practice, the second law of thermodynamics means that:
a. Systems move from ordered behavior to more random behavior.
b. Systems move from random behavior to more ordered behavior.
c. Systems move between ordered and random behavior patterns based on temperature.
d. Systems are constantly striving to reach equilibrium.
Answer:
Systems move from ordered behavior to more random behavior.
Explanation:
Entropy refers to the degree of disorderliness in a system. The second law of thermodynamics can be restated in terms of entropy as follows; “any spontaneous process in any isolated system always results in an increase in the entropy of that system.''(science direct)
According to this law, systems tend towards a more disorderly behaviour (increase in entropy) hence the answer given above.
For an atoms electrons, how many energy sublevels are present in the principal energy level n = 4?
A. 4
B. 9
C. 10
D. 16
E. 32
Answer:
by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32
Consider the reaction below. How much heat is absorbed if 5.00 moles of nitrogen react
with excess oxygen?
2 N2 (8) + O2(g) → 2 N20 (8) AHrxn- +163.2 kJ
Explanation:
The given chemical reaction is:
[tex]2 N_2 (g) + O_2(g) -> 2 N_20 (g) delta Hrxn= +163.2 kJ[/tex]
When two moles of nitrogen reacts with oxygen, it requires 163.2kJ of energy.
When 5.00 mol of nitrogen requires how much energy?
[tex]5.00 mol x \frac{163.2 kJ }{2 mol} \\=408 kJ[/tex]
Hence, the answer is 408 kJ of heat energy is required.
Why is bromine more electronegative than iodine?
Answer
Accordingly the order of electronegativity of the given elements would be: Fluorine > Chlorine > Bromine > Iodine. ( Fluorine has the highest electronegativity.)
A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is ________ g/mol.
Answer:
160 g/mol
Explanation:
Step 1: Calculate the molarity of the solution
We will use the following expression.
π = M × R × T
where,
π: osmotic pressure of a nonelectrolyteM: molarityR: ideal gas constantT: absolute temperature (25 °C = 298 K)M = π / R × T
M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L
Step 2: Calculate the moles of solute in 550 mL (0.550 L)
0.550 L × 0.0752 mol/L = 0.0413 mol
Step 3: Calculate the molecular weight of the nonelectrolyte
0.0413 moles weigh 6.60 g.
6.60 g/0.0413 mol = 160 g/mol
how is the molecule of substance formed
Answer:
When atoms approach one another closely, the electron clouds interact with each other and with the nuclei. If this interaction is such that the total energy of the system is lowered, then the atoms bond together to form a molecule.
Explanation:
Consider the reaction: A(aq) + 2B (aq) === C (aq). Initially 1.00 mol A and 1.80 mol B
were placed in a 5.00-liter container. The mole of B at equilibrium was determined to
be 1.00 mol. Calculate K value.
0.060
5.1
25
17
Ugh
Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
Many important analgesic compounds are derived from simple aromatic starting materials.
a. True
b. False
The Lewis dot model of a molecule is shown.
Based on the model, which of the following is true?
Each carbon has three lone pairs of electrons on it.
The octet of carbon atom remains incomplete in the molecule.
The two carbon atoms share a total of six electrons with each other.
The difference between the electronegativity of carbon and hydrogen is greater than 1.7.
Answer:
The two carbon atoms share a total of six electrons with each other.
Explanation:
Looking at the structure of the molecule H-C≡C-H as shown in the question, we will notice that there exists a triple bond between the two carbon atoms.
Each bond between the two carbon atoms represents two electrons shared. Since there are three bonds between the two carbon atoms, then a total of six electrons were shared between the two carbon atoms hence the answer chosen above.
balance equation of potassium sulphate+ water
Answer:
2KHCO
3
+H
2
SO
4
→K
2
SO
4
+2CO
2
+2H
2
O
A substance is made up of slow-moving particles that have very little space between them. Based on this information, what can most likely be concluded about this substance? O It is not a gas because its particles do not move continuously. It is a gas because its particles move continuously in a straight line. 0 It is not a gas because its particles do not have large spaces between them. It is a gas because its particles move in many different directions.
Answer:
o
Explanation:
it is not a gas because the particles do not move freely it may be a liquid or a solid partly and mostly liquidized.
Consider a hypothetical metal that has a density of 10.6 g/cm3, an atomic weight of 176.8 g/mol, and an atomic radius of 0.130 nm. Compute the atomic packing factor if the unit cell has tetragonal symmetry, values for the a and c lattice parameters are 0.570 and 0.341, respectively.
Answer:
0.3323
Explanation:
GIven that:
Density of the metal = 10.6 g/cm^3
atomic weight = 176.8 g/mol
atmic radius = 0.130 nm
values of a and c = 0.570 nm and 0.341 nm respectively
For us to determine the atomic packing factor, we need to first determine the volume of all spheres (Vs) and the volume of unit cell (Vc).
However, the number of atoms in the unit cell (n) can be computed as:
[tex]n = \dfrac{\rho * V_c *N_A}{A} \\ \\ n = \dfrac{(10.6) * (5.7)^2 (3.41)*(10^{-24}) *(6.022*10^{23})}{176.8}[/tex]
n = 4.0
Thus, the number of atoms in the unit cell is 4
∴
The atomic paking factor (APF) is calculated by using the formula:
[tex]\dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *R^3 }{a^2 *c} \\ \\ \\ \dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *(1.30*10^{-8})^3 }{(5.70*10^{-8})^2 *(3.41*10^{-8})}[/tex]
= 0.3323
2. XC12 is the chloride of metal X. The formulae of its sulphate is
Answer:
XSO₄
Explanation:
XCl₂ is the chloride of metal X. The sum of the charges of the cation and the anion must be zero because the salt is electrically neutral. The charge of the cation of X is:
1 × X + 2 × Cl = 0
1 × X + 2 × (-1) = 0
X = +2
X has a charge +2 and sulphate (SO₄²⁻) a charge -2. The neutral salt they form is XSO₄.
For the following acids of varying concentrations, which are titrated with 0.50 M NaOH, rank the acids in order of least to most volume of base needed to completely neutralize the acid.
a. 0.2M H2C6H5O7
b. 0.2M H2C2O4
Answer:
0.2M H2C6H5O7 < 0.2M H2C2O4
Explanation:
A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.
More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.
Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.
H₂C₆H₅O₇ is a weaker acid than H₂C₂O₄, and will require the least volume of 0.50 M NaOH to be neutralized.
H₂C₆H₅O₇ < H₂C₂O₄
The strength of an acid is related to the value of its dissociation constant, Ka or its pKa (negative logarithm of Ka)
Strong acids have high Ka values or low pKa value, whereas weak acids have low Ka values and high pKa values.
Between two acids, the acid with a higher Ka or lower pKa values is the stronger acid.
Acids are classified as either strong or weak depending on how well it ionizes in solution to produce hydrogen ions.
Strong acids ionizes completely to produce hydrogen ions.
Weak acid ionizes partially to a varying degrees in water to produce hydrogen ions.
In neutralization reactions between acids and bases, stronger acids will require the most volume of base or alkali in order to be neutralized.
H₂C₂O₄ has a Ka value of 5.9 x 10⁻² and a pKa value of 1.23
H₂C₆H₅O₇ has a Ka value of 8.4 x 10⁻⁴ and a pKa value of 3.08
Hence H₂C₂O₄ is a stronger acid than H₂C₆H₅O₇
For equal molar concentrations of the two acids, H₂C₂O₄ will produce more hydrogen ions than H₂C₆H₅O₇, and thus, will require more volume base (0.50 M NaOH) to be neutralized.
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The reversible reaction 2H2 CO <------> CH3OH heat is carried out by mixing carbon monoxide and hydrogen gases is a closed vessel under high pressure with a suitable catalyst . After equilibrium is established at high temperature and pressure, all three substances are present. If the pressure on the system is lower, with the temperature kept constant, what will be the result
Answer:
The amount of CH3OH present in the mixture would decrease
Explanation:
According to Le Cha-telier's principle, when a reaction is in equilibrium and one of the constraints that influence the rate of reactions is applied, the equilibrium would shift so as to neutralize the effects of the constraint.
In this case, looking at the equation of the reaction:
2H2 + CO <------> CH3OH + heat
the total number of moles on the reactant's (left hand) side is 3 (2+1) while on the product's (right hand) side, it is 1. If the pressure on the system is increased, more CH3OH (and less of H2 and CO) will be produced because its side has the lower number of moles out of the two sides.
If the pressure on the system is otherwise lowered, more of H2 and CO would be produced while the amount of CH3OH present would gradually decrease.
The fact that a beam of particles was deflected in the presence of an electric
or magnetic force led J.J. Thomson to conclude that the particles had a(n)
O A. large mass
B. electric charge
O C. negligible mass
O D. neutral charge
Answer:
electric charge
Explanation:
Charged particles are deflected in an electric or a magnetic field. The particles discovered by J.J. Thomson were charged particles.
When these charged particles are passed through electric and magnetic fields, deflection occurs depending on the nature of the charge.
A positive charge is deflected towards the negative part of an electric field or the south pole of a magnetic field.
A negative charge is selected towards the positive end of an electric field or the north pole of a magnetic field.
A person slips over banana pills. Give reason
Answer:
We slip when we step on a banana peel because the inner side of banana peel being smooth and slippery reduces the friction between the sole of our shoe and the surface of road.
Determine the effect each given mutation would have on the rate of glycolysis in muscle cells.
a. loss of binding site for fructose 1 ,6-bisphophate in pyruvate kinase.
b. loss of allosteric binding site for ATP in pyruvate kinase.
c. loss of allosteric binding site for AMP in phosphofructokinase.
d. loss of regulatory binding site for ATP in phosphofructokinase.
1. Increase
2. decrease
3. No effect
Answer:
a. Decrease
b. Increase
c. Increase
d. No effect
Explanation:
Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.
a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease
b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect
c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase
d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase
A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.
B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.
C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.
D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.
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What is the correct ratio of carbon to hydrogen to oxygen in glucose (CH1206)?
•12:12:6
•2:1:1
•1:2:1
•6:6:12
Answer:
1:2:1 is the correct ratio of carbon hydrogen to oxygen in glucose.
If Sterling silver is 90.0% silver and 10.0% copper, what is the maximum amount of Sterling silver that can be made if you have 48.3 grams of silver metal and an unlimited amount of copper
Answer:
[tex]x=54g[/tex]
Explanation:
From the question we are told that:
Content of Sterling silver:
Let x be sterling silver
Silver [tex]S=0.9x[/tex]
Copper [tex]C=0.1x[/tex]
Total silver available [tex]T=48.3[/tex]
Generally the equation for Total amount to be made is mathematically given by
[tex]T=\frac{x*90}{100}[/tex]
[tex]x=\frac{48.3*100}{90}[/tex]
[tex]x=54g[/tex]
If a buffer is composed of 23.34 mL of 0.147 M acetic acid and 33.66 mL of 0.185 M sodium acetate, how many mL of 0.100 M NaOH can be added before the buffer capacity is reached
Answer:
25.5mL of 0.100M NaOH are needed to reach buffer capacity.
Explanation:
The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:
Moles sodium acetate / Moles Acetic acid = 10
The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:
HA + NaOH → H2O + NaA
That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.
The initial moles of each species is:
Acetic acid:
23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid
Sodium Acetate:
33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate
We can write the moles of each species when NaOH is added as:
Moles sodium acetate / Moles Acetic acid = 10
0.00623 moles + X / 0.00343 moles - X = 10
Where X are moles of NaOH added
Solving for X:
0.00623 moles + X = 0.0343 moles - 10X
11X = 0.0281
X = 0.00255 moles of NaOH are needed
In Liters:
0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =
25.5mL of 0.100M NaOH are needed to reach buffer capacity
Spell out the full name of the compound.
Answer:
it is propane
C3H8 it is propane
The Full name of the given compound is Propane.
What is Propane ?Propane is a three-carbon alkane with the molecular formula C₃H₈.
It is a gas at standard temperature and pressure, but compressible to a transportable liquid.
In the figure given ;
Black balls represents Carbon atomsWhite balls represents Hydrogen atomsIn the given figure, there is single bond present between Carbon and Hydrogen. Hence, The Full name of the given compound is Propane.
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During the reaction of 2-methyl-2-butanol with the nucleophile-solvent mixture two layers are formed after shaking the reaction for 5 minutes. After removing the aqueous layer with a Pasteur pipette the organic layer is diluted with 1 mL dichloromethane. The organic phase is washed with 1 mL water. Two layers are obtained.
a. Top layer is Aqueous (H20/ H2SO4/NH4CI)
b. Top layer is Organic (CH2Cl2 and product)
c. Bottom layer is Organic (CH2Cl and product)
d. Top layer is Aqueous (H20)
Answer:
Top layer is Organic (CH2Cl2 and product)
Explanation:
In a solvent mixture, there are usually two phases, the organic phase and the aqueous phase.
It is usual that the organic phase is almost always less dense than the aqueous phase hence the organic phase tend to remain on top of the aqueous phase.
Hence, the top layer is expected to be the organic CH2Cl2 and product.
A sample of oxygen occupied 621 mL when the pressure increased to 1095.93mm Hg. At constant temperature, what volume did the gas initially occupy when the pressure was 774.29mm Hg?
a.) 879.0
b.) 438.7
c.) 890.2
d.) 1366
Answer:
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Explanation:
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You need to make an aqueous solution of 0.182 M aluminum sulfate for an experiment in lab, using a 250 mL volumetric flask. How much solid aluminum sulfate should you add
Answer:
Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.
Explanation:
To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask
The molar mass of Aluminum Sulfate = 342.15 g/mol
Using the molarity formula:-
Molarity = Number of moles/Volume of solution in a liter
Number of moles = Given weight/ molar mass
Molarity = (Given weight/ molar mass)/Volume of solution in liter
0.187 M = (Given weight/342.15 g/mol)/0.250 L
Given weight = 15.99 g
A hot pot of water is set on the counter to cool. After a few minutes it has lost 495 J of heat energy. How much heat energy has the surrounding air gained?
_____unit_____
Answer:
495 J
Explanation:
When the hot pot was set on the counter to cool, heat energy was lost from the pot. Note that according to the first law of thermodynamics, heat is neither created nor destroyed.
This implies that, the heat energy lost from the pot must be gained by the surrounding air. Therefore, if 495 J of energy is lost from the pot, then 495 J of energy is gained by the surrounding air.
Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answer:
21.4g of HBr is the minimum mass that could be left over.
Explanation:
Based on the reaction:
HBr + NaOH → NaBr + H2O
1 mole of HBr reacts per mole of NaOH
To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:
Moles NaOH -40.0g/mol-
17g * (1mol/40.0g) = 0.425 moles NaOH
Moles HBr -Molar mass: 80.91g/mol-
55.8g * (1mol/80.91g) = 0.690 moles HBr
The difference in moles is:
0.690 moles - 0.425 moles =
0.265 moles of HBr could be left over
The mass is:
0.265 moles * (80.91g/mol) =
21.4g of HBr is the minimum mass that could be left over.During electrophilic aromatic substitution, a resonance-stabilized cation intermediate is formed. Groups, already present on the benzene ring, that direct ortho/para further stabilize this intermediate by participating in the resonance delocalization of the positive charge. Assume that the following group is present on a benzene ring at position 1 and that you are brominating the ring at positon 4. Draw the structure of the resonance contributor that shows this group actively participating in the charge delocalization.
-----OCH3
Answer:
See explanation and image attached
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.
Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.
Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.
The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.
