Answer:
Explanation:
A) False.
Glucosidase (not calnexin nor calreticulin) helps to remove glucose residue.
Both calnexin and calreticulin rather have an affinity for last glucose residue of misfolded protein (Only misfolded proteins are marked by glycosyltransferase by attaching glucose residue). They attach with misfolded protein and with the help of other proteins like ERp57 (a type of protein disulfide isomerase) and try to fold it properly. If protein is properly folded then glucosidase removes the glucose residue thereby releasing the properly folded protein from calnexin or calreticulin. and now protein is transported to the Golgi body. If folding is still not proper then the same cycle of glycosylation -binding of calnexin/calreticulin and effort to fold it properly is repeated.
B) True.
Transketolase is a key enzyme of the pentose phosphate pathway. It contains thiamine diphosphate (TPP) as a cofactor. it does transfer 2 carbon residue from a ketose to aldose. So, effectively it converts one ketose sugar to aldose with 2 carbonless and aldose to ketose with 2 carbon more.
C) True.
Theoretically, for the evolution of one molecule of oxygen, only 8 photons are required. But in practice, it is known that there are many variants like wavelength and the energy of the photon. The larger the wavelength, like the one which is used in PS1 (more than 700nM), the lesser the energy. Secondly, the energy of the photon is also wasted as heat energy. Because of these factors, more than 8 photons are needed in reality.
D) Wrong.
Fructose 2,6 bisphosphate is a key substrate and affects both the enzymes- phosphofructokinase and fructose bisphosphatase allosterically during gluconeogenesis. It strongly favors the breakdown of glucose during glycolysis by activating phosphofructokinase but it inhibits fructose bisphosphatase. Hence it activates the kinase enzyme while inhibiting the phosphatase and maintains a huge supply of glucose in the system.
E) Wrong.
The Calvin cycle shares similarity with the pentose phosphate pathway as both are involved in the synthesis of sugar (Triose and Ribose). However, it does not share similarity with enzymes of glycolysis (which is primarily focused on the breakdown of glucose) and gluconeogenesis.
Calculate the rate of emission of SO2 in g/s that results in a centerline (y = 0) concentration at ground level of 1.786 x 10-3 g/m3 one kilometer from the stack. The time of measurements was 1 P.M. on a clear summer afternoon. The wind speed was 2.0 m/s measured at a height of 10 m. The effective stack height is 96 m. No inversion is present.
Answer:
790 g/s
Explanation:
The first thing to do is to make sure to calculate the standard deviation of the y-axis and the z-axis and the values for both axis are; 215 m and 450 m for y-axis and z-axis respectively.
The next thing to do now is to find the rate of emission of SO2 in units of g/s using the relation by using the formula below;
Emission rate of pollutant, E = (x × π × by × bz × g) ÷ [ e^(-1/2) × (y/by)^2] × [e^(-1/2)× (h/bz)^2].
Where g = wind direction, y and h are the distance in metres.
Therefore, slotting in the values into the Emission rate of pollutant equation above, we have;
Emission rate of pollutant,E =[ 1.412 × 10^-3] × π × 215 × 450 × 1.8] ÷ [e^(-1/2)(0/215)^2] × e^ (-1/2)(94/450)^2.
Emission rate of pollutant,E = 790 g/s.
A tank contains helium gas at 191 mm HG nitrogen gas at 0.261 ATM and neon at 522 torr what is the total pressure in MM Hg
Answer:
911.36mmhg
Explanation:
1 torr is almost equivalent to 1mmhg.but to convert from atm to mmhg,multiply by 760
Final answer:
The total pressure in the tank, after converting all pressures to the same unit (mm Hg), is the sum of the partial pressures of helium, nitrogen, and neon, resulting in a value of 911.36 mm Hg.
Explanation:
To calculate the total pressure in a mixture of gases, you can use Dalton's Law of partial pressures which states that the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.
Since the pressure values are given in different units, we first need to convert them all to millimeters of mercury (mm Hg). The pressure of helium gas is already given in mm Hg. The nitrogen gas pressure needs to be converted from atmospheres (ATM) to mm Hg by multiplying by 760 mm Hg/ATM, and the neon gas pressure in torr simply needs to be directly converted to mm Hg because 1 torr equals 1 mm Hg.
Now we can sum these values to find the total pressure in the tank.
The conversion and summation are as follows:
Helium: 191 mm Hg
Nitrogen: 0.261 ATM x 760 mm Hg/ATM = 198.36 mm Hg
Neon: 522 torr = 522 mm Hg
Adding these together, the total pressure in the tank is:
191 mm Hg + 198.36 mm Hg + 522 mm Hg = 911.36 mm Hg
Therefore, the total pressure in the tank in mm Hg is 911.36 mm Hg.
What information about water is needed to calculate the enthalpy change for converting 1 mol H2O(g) at 100 °C to H2O(l) at 80 °C? (a) Heat of fusion, (b) heat of vaporization, (c) heat of vaporization and specific heat of H2O(g), (d) heat of vaporization and specific heat of H2O(l), (e) heat of fusion and specific heat of H2O(l).
The information needed includes the heat of vaporization and specific heat of H₂O(l).
Explanation:The information needed to calculate the enthalpy change for converting 1 mol H₂O(g) at 100 °C to H₂O(l) at 80 °C is the (d) heat of vaporization and specific heat of H₂O(l).
The heat of vaporization is required because it accounts for the energy needed to convert water from a liquid to vapor state. The specific heat of H₂O(l) is needed to determine the amount of heat required to change the temperature of liquid water from 100 °C to 80 °C.
Consider an electrolytic cell with a platinum anode and a silver cathode in a 1.0 M AgNO3(aq) solution.
a) (3 pts) What species can be reduced in this solution? Which species is preferentially reduced? Write the reduction half- reaction. (Note that oxyanions like nitrate are not commonly reduced in aqueous electrolysis due to kinetic reasons.)
b) (2 pts) Which species is oxidized during the electrolysis? Write the oxidation half-reaction. Note that acid will form as a byproduct of the oxidation.
c) (2 pts) Write the overall electrolysis reaction in net ionic and molecular forms.
d) (2 pts) Determine the external electric potential needed for the electrolysis under standard conditions.
e) (5 pts) If the electrolysis is carried out for 1.00 hour using 1.00 A current, how many grams of metal will be deposited at the cathode and how many liters of gas will form at the anode at 1.00 atm pressure and 25°C?
Answer:
Check the attached image below
Explanation:
The step by step solution to the question above can be seen in the attached image below.
What happens if you break a magnet in half?
a) Each half will be a new magnet, with both a north and south pole.
b) One half will have a north pole only and one half will have a south pole only.
c) Neither half will have a pole.
d) Neither half will be able to attract or repel
Answer:
neither half will have a pole
14 H+ + 6 Fe2+ + Cr2O72- . 2 Cr3+ + 6 Fe3+ + 7 H2O This reaction is used in the
titration of an iron solution. What is the concentration of the iron solution if it takes 45.20 mL
of 0.1000 M Cr2O72- solution to titrate 50.00 mL of an acidified iron solution?
a.
0.5424 M
b.
0.1000 M
c.
1.085 M
d.
0.4520 M
e.
0.2712 M
Answer:
Option A 0.5424 M
Explanation:
To do this, let's write the reaction again and see the mole ratio:
14H⁺ + 6Fe²⁺ + Cr₂O₇⁻ ---------> 2Cr³⁺ + 6Fe³⁺ + 7H₂O
According to this reaction, we have a mole ratio between the dicromate and iron of 1:6. In other words, 6 moles of dicromate reacts with only 1 mole of iron, therefore we can use the following expression:
nFe = 6 * nCr₂O₇
This is a REDOX reaction, where Iron is being oxidized while cromium is being reduced. Now, we have the concentrations and volume, let's express the moles in function of concentration and volume:
M₁V₁ = 6M₂V₂
Replacing the given data above and solving for concentration of iron (M₁)
M₁ = 6 * 45.20 * 0.1 / 50
M₁ = 0.5424 M
This is the concentration of the iron solution
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer. 2. A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in
To calculate the concentration of hydronium ion and the moles in a buffer solution, use the formula [H3O+] = 10^(-pH) and Moles = Concentration x Volume.
Explanation:The concentration of hydronium ion can be calculated using the formula:
[H3O+] = 10^(-pH)
For example, if the starting pH is 4.74, then the concentration of hydronium ion is 10^(-4.74) M
The moles of hydronium ion present in 20.0 mL of the buffer can be calculated using the formula:
Moles = Concentration x Volume
Assuming the concentration of hydronium ion is 10^(-4.74) M, then the moles of hydronium ion in 20.0 mL is (10^(-4.74) M) x (20.0 mL / 1000) L.
How do neutrons stars generate the energy they need to survive
The neutron stars are the smallest but heaviest density supergiant star. The surfaces of the neutron stars undergo thermonuclear fusion to produce energy to survive.
What is thermonuclear fusion?Thermonuclear fusion or just simply nuclear fusion is a nuclear reaction that involves the combining (fusion) of the atomic nuclei together to produce different nuclei in the presence of high temperatures.
Unlike nuclear fission, it needs high temperature and energy to produce atomic nuclei. It is a process generally seen in the universe including the Sun, stars, galaxies, etc.
Neutron stars also use their energy to undergo thermonuclear fusion and produce heavy nuclei from lighter nuclei. The hydrogen atom combines fuses to form helium atoms resulting in high energy production.
Therefore, neutron stars generate energy by thermonuclear fusion.
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At constant pressure, 50.0 mL of 0.100 M KOH and 45.0 mL of 0.100 M HNO3 are mixed in a styrofoam cup.The two solutions are initially at 25.00 °C and the final temperature is 25.65 °C. Calculate the ΔHneutralization (kJ/mol) in term of moles of HNO3 since it’s the limiting reagent. Assume a specific heat capacity of 4.184 J/(°C•g) and that the density of the solution is the same as water (0.997 g/mL).
Answer:
≈ -57.2 kJ/mol
Explanation:
Total volume of the solution = (50.0 + 45.0) mL = 95.0 mL.
Density of the solution = 0.997 g/mL.
Mass of the solution = (volume of the solution)*(density of the solution)
= (95.0 mL)*(0.997 g/mL)
= 94.715 g
Heat gained by the solution, qsoln = (mass of the solution)*(specific heat capacity)*(change in temperature)
= (94.715 g)*(4.184 J/ºC.g)*(25.65 – 25.00)ºC
= 257.5869 JAccording to the principle of thermochemistry,
qsoln + qrxn = 0
where qrxn denotes the heat change during the neutralization reaction.
Therefore,
(257.5869 J) + qrxn = 0
======> qrxn = -257.5869 J
HNO3 is the limiting reactant.
Moles of limiting reactant = (volume of the limiting reactant in L)*(molarity of limiting reactant)
= (45.0 mL)*(0.100 M)
= (45.0 mL)*(1 L)/(1000 mL)*(0.100 M)
= 0.0045 mole.
ΔHneutralization = qrxn/(moles of HNO3)
= (-257.5869 J)/(0.0045 mol)
= -57241.5 J/mol
= (-57241.5 J/mol)*(1 kJ)/(1000 J)
= -57.2415 kJ/mol
≈ -57.2 kJ/mol
Given a balanced chemical equation below: 3Cu(s) + 2H3PO4 --- > Cu3(PO4)2 + 3H2 How many moles of copper are needed to react with 5 moles of phosphoric acid?
Answer:
7.5 moles
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
3Cu + 2H3PO4 —> Cu3(PO4)2 + 3H2
From the balanced equation above,
3 moles of Cu reacted with 2 moles of H3PO4.
Therefore, Xmol of Cu will react with 5 moles of H3PO4 i.e
Xmol of Cu = (3 x 5)/2
Xmol of Cu = 7.5 moles
Therefore, 7.5 moles of Cu are needed to react with 5 moles of H3PO4.
Determine the number of bonding electrons and the number of nonbonding electrons in the structure of SI2. Enter the number of bonding electrons followed by the number of nonbonding electrons, separated by a comma, in the dot structure of this molecule.
The molecule SI2 does not exist but if the student meant SiI2, it would have 8 bonding electrons and 14 nonbonding electrons.
Explanation:The molecule SI2 does not exist. It may have been a typo. If the student meant Si2, this molecule also doesn't exist because silicon, a member of group 14 on the periodic table, needs 4 electrons to achieve a full outer shell, and thus generally forms 4 bonds. If we are talking about SiI2, or silicon diiodide, it would have 8 bonding electrons and 14 nonbonding electrons. This is because all four of the bonding electrons from silicon and four from iodine atoms make up the bonding electrons while the remaining 7 electrons from each iodine atom make up the nonbonding electrons.
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Which of the following is used to represent each element?
formula
equation
symbol
O
coefficient
What is the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 ⋅ 10-8. What is the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 x 10-8.
a. 7.7
b. 0
c. 7.14
d. 14.28
e. 9.05
f. 6.86
Answer:
c. 7.14
Explanation:
The buffer solution is formed by a weak acid ( hypochlorous acid, HClO) and its conjugate base (hypochlorite ClO⁻, coming from sodium hypochlorite NaClO). We can calculate the pH using the Henderson-Hasselbach equation.
pH = pKa + log [base]/[acid]
pH = -log 3.8 × 10⁻⁸ + log 0.111/0.211
pH = 7.14
A biochemist carefully measures the molarity of glycerol in of photobacterium cell growth medium to be . Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to . Calculate the new molarity of glycerol in the photobacterium cell growth medium. Round each of your answers to significant digits.
Note:
The Complete Question
A biochemist carefully measures the molarity of glycerol in 913 mL of photobacterium cell growth medium to be 81.3 μM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 11.1 mL. Calculate the new molarity of glycerol in the photobacterium cell growth medium. Round each of your answers to 3 significant digits.
Answer:
669 μM
Solution:
[Find the attachment]
Which pair of elements can be used to determine the age of a fossil that is over one billion years old?
Answer:
Potassium and Argon (⁴⁰K-⁴⁰Ar)
Explanation:
Potassium-Argon dating is a method used in geology and archeology to date rocks or volcanic ash, generally the oldest.
It is based on the principle of radioactive decay, it is a process that has a half-life of 1248 million years, during which time the gas is concentrated in the rock crystals. Taking advantage of this known rhythm and half-life, the method lends itself to dating samples ranging from 100.000 years to several billion years.
Answer:
Its D to sum it up.
Explanation:
Some people must eat a low-sodium diet with no more than 2,000 mg of sodium per day. By eating 1 cracker, 1 pretzel, and 1 cookie, a person would ingest 149 mg of sodium. If a person ate 8 pretzels and 8 cookies, he or she would ingest 936 mg of sodium. By eating 6 crackers and 7 pretzels, the person would take in 535 mg of sodium. Find the amount of sodium in each.
Answer:
The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
Explanation:
Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.
So, the following three equations can be written as per given information:
x+y+z = 149 ........(1)
8y+8z = 936 ........(2)
6x+7y = 535 .........(3)
From equation- (2), y+z = [tex]\frac{936}{8}[/tex] = 117
By substituting the value of (y+z) in equation- (1) we get,
x = 149-(y+z) = 149-117 = 32
By substituting the value of x into equation- (3) we get,
y = [tex]\frac{535-(6\times 32)}{7}[/tex] = 49
By substituting the value of y into equation- (2) we get,
z = (117-49) = 68
So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
Final answer:
To find the amount of sodium in each item, set up a system of equations and solve to get x = 9, y = 55, and z = 85.
Explanation:
To find the amount of sodium in each item, we can set up a system of equations.
Let x represent the amount of sodium in 1 cracker, y represent the amount of sodium in 1 pretzel, and z represent the amount of sodium in 1 cookie.
From the given information, we can create the following equations:
x + y + z = 1498y + 8z = 9366x + 7y = 535Solving this system of equations, we can find the values of x, y, and z. The amount of sodium in 1 cracker is 9 mg, 1 pretzel is 55 mg, and 1 cookie is 85 mg.
9) 20 g of calcium chloride per 100 g of water at 90°C will be...
10 points
Сас,
KNO,
K₂ Cigar
Solubility (g of salt in 100 g H,0)
Pb(NO3)2
KCR
20
A
KCIO,
0
10
Ce (80)
20 30 40 50 60 70 80 90 100
Temperature (°C)
O
saturated
O
unsaturated
O
supersaturated
According to solubility, 20 g of calcium chloride per 100 g of water at 90°C will be saturated solution.
What is solubility?Solubility is defined as the ability of a substance which is basically solute to form a solution with another substance. There is an extent to which a substance is soluble in a particular solvent. This is generally measured as the concentration of a solute present in a saturated solution.
The solubility mainly depends on the composition of solute and solvent ,its pH and presence of other dissolved substance in it . It is also dependent on temperature and pressure which is maintained through the process.Concept of solubility is not valid for chemical reactions which are irreversible. The dependency of solubility on various factors is due to interactions between the particles, molecule or ions.
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On a hot summer day in Florida, Sarah was having a picnic with her family. She took a glass, which had been sitting in the Sun, and filled it with ice cubes from a cooler. Then she added cold lemonade from a thermos jug. Which of the following describes one direction in which heat will flow?
A.From the ice cube to the glass.
B. From the lemonade to the glass.
C. From the lemonade to the ice cube.
D. From the ice cube to the lemonade.
The objects will exchange thermal energy, until thermal equilibrium is reached. This means until their temperatures are equal. Here the direction in which heat will flow is from the lemonade to the glass. The correct option is B.
What is heat flow?The transfer of heat energy from an object or substance to another because of the difference in the temperature between them is defined as the heat flow. The flow of heat occurs from an object of higher temperature to an object of lower temperature.
The heat flows from hot objects to cold objects by several processes such as radiation, conduction and convection. However evaporation allows an object to lose the heat even if the surrounding medium in which it is immersed is hotter than the object itself.
Here lemonade has higher temperature than the glass. So heat flows from lemonade to the glass.
Thus the correct option is B.
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Consider the titration of a 20.0 mL sample of 0.500 M HCN (Ka =6.17x10-10) with 0.250 M KOH. a. (6pt) What is the initial pH? b. (7pt) What is the pH at 6.00 mL base added? c. (8pt) What is the pH when 40.00 mL of KOH is added to reach the equivalence point? d. (8pt) What is the pH if 42.00 mL of KOH is titrated with the solution?
Answer:
a. pH = 4.75
b. pH = 9.20
c. pH = 8.42
d. pH = 13.53
Explanation:
This is a titration between a strong base, the KOH and a weak acid, HCN.
The initial pH is the pH, when you did not add the base yet, so it is the pH of the HCN
HCN + H2O ⇄ H₃O⁺ + CN⁻
Initial 0.5 - -
Eq. 0.5-x x x
Ka = x² / (0.5-x) = 6.17ₓ10⁻¹⁰
Ka is really small, so we can say that 0.5-x = 0.5. Then,
x² = 6.17ₓ10⁻¹⁰ . 0.5
x = √(6.17ₓ10⁻¹⁰ . 0.5) = 1.75×10⁻⁵ → [H₃O⁺]
pH = - log [H₃O⁺] → - log 1.75×10⁻⁵ = 4.75
b. First of all, we determine the moles of base, we are adding.
0.250 mol/L . 0.006 L = 0.0015 moles
In conclussion we have 0.0015 moles of OH⁻
Now, we determine the moles of our acid.
0.500 mol/L . 0.020L = 0.01 moles
The 0.0015 moles of OH⁻ will be neutralized with the acid, so:
HCN + OH⁻ → H₂O + CN⁻
0.01 0.0015 0.0085
The hydroxides are neutralized with the proton from the weak acid, so we have 0.0085 moles of cyanide and 0.0085 moles of HCN. (0.01-0.0015)
Our new volume is 20 mL and 6mL that we added, so, 26mL
This is a buffer with the weak acid, and its conjugate base.
Our concentrations are 0.0085 moles / 0.026 L = 0.327 M
We apply Henderson-Hasselbach
pH = pKa + log (base/acid) → pH = 9.20 + log (0.327/0.327)
pH = pKa
c. When we add 40 mL, our volume is 20mL +40mL = 60 mL
These are the moles, we add:
0.040 L . 0.250 mol/L = 0.01 moles of KOH (moles of OH⁻)
HCN + OH⁻ → H₂O + CN⁻
0.01 0.01 0.01
All the hydroxides have neutralized all the moles from the HCN, so we only have in solution, cyanhide. This is the equivalence point.
0.01 moles / 0.060 L = 0.16 M → [CN⁻]
pH at this point will be
CN⁻ + H₂O ⇄ HCN + OH⁻ Kb = 1.62ₓ10⁻⁵ (Kw/Ka)
In. 0.16 - -
Eq. 0.16-x x x
Kb = x² / (0.16-x)
We can also assume that 0.16-x = 0.16. Then:
[OH⁻] = √(Kb . 0.16) → √(1.62ₓ10⁻⁵ . 0.16) = 2.59×10⁻⁶
- log [OH⁻] = pOH → - log 2.59×10⁻⁶ = 5.58
pH = 14 - pOH → 14 - 5.58 = 8.42
This is a basic pH, because the titration is between a weak acid and a strong base.
d. When we add 42 mL of base, our volume is 20mL + 42 mL = 62 mL
We add 0.5 mol/L . 0.062L = 0.031 moles
These are the moles of OH⁻ , so as we have neutralized all the acid with 40 mL, with 42 mL of base, we only have base in solution.
0.031 moles - 0.01 moles = 0.021 moles of OH⁻
[OH⁻] = 0.021 moles / 0.062L = 0.34M
- log [OH⁻] = pOH → - log 0.34 = 0.47
pH = 14-pH → 14 - 0.47 = 13.53
This answer details the titration of HCN with KOH, explaining ways to calculate the pH values at different stages of the titration: the initial state, partway through, at the equivalence point when all acid has been neutralized, and after the equivalence point.
Explanation:We are considering the titration of a solution of HCN with KOH. First, we calculate the initial pH using the formula pH=-log10[H+], where the H+ concentration can be determined from the Ka expression for HCN.
For question b, when 6.00 mL base KOH is added, we need to determine which ion is in excess, and then use its concentration to calculate the pH. In this case, there are still excess HCN ions leading to a buffer solution. The Henderson-Hasselbalch equation can be used to determine the pH.
For question c, at the equivalence point, 40.00 mL of base KOH has been added. All HCN has been titrated to form CN-, so the pH is determined by the OH- ions hydrolyzed from CN-. We use the Kb of CN- (which can be calculated using Kw/Ka) and the total volume to calculate [OH-], and then pH = 14 - pOH.
Finally, for question d, after 42.00 mL of KOH, OH- is in excess. We can calculate the [OH-] and then find the pH using pH = 14 - pOH.
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2. Carbon monoxide is a poisonous carbon oxygen compound (as opposed to carbon
dioxide). Provide the complete MO diagram including the atomic orbitals for both
carbon and oxygen as well as the MOs for CO assuming the bonding is the same as for
C2 in the reference. Also indicate the bond order for CO from the MO diagram. (14
points)
Answer:
Carbon monoxide has a bond order of 3
Explanation:
A molecular orbital diagram ( MO diagram ), is a qualitative descriptive tool tha explains chemical bonding in molecules in terms of the molecular orbital theory in general and particularly adopts the linear combination of atomic orbitals (LCAO) method in its approach. A fundamental principle of these theories is that as atoms bond to each other to form molecules, a certain number of atomic orbitals combine to form an equal number of molecular orbitals , although the electrons involved may be redistributed among the orbitals. This tool is very well suited for simple diatomic molecules such as hydrogen molecule , oxygen molecule, and carbon monoxide but becomes more complex when discussing even comparatively simple polyatomic molecules, such as methane. MO diagrams usually explains why some molecules exist and others do not exist. It is used to predict bond strength of molecules, as well as the electronic transitions that can take place.
The MO diagram for carbon monoxide is shown in the image attached. It can clearly be seen that carbon monoxide has a bond order of three. The combination of carbon and oxygen atoms to form molecular orbitals of appropriate energy is also seen in the image.
19.24 The mechanism for acetal hydrolysis has been heavily investigated. In one study, which explored rates as well as stereochemical aspects,[5] compound 1 was treated with aqueous acid to afford compound 2. Draw the structure of 2, clearly showing the configuration of any chiral center(s). c19s124
Answer:
Explanation:
check the attached file below for the diagram and better expalnation.
A student adds solid KCl to water in a flask. The flask is sealed with a stopper and thoroughly shaken until no more solid KCl dissolves. Some solid KCl is still visible in the flask. The solution in the flask is A) saturated and is at equilibrium with the solid KCl B) saturated and is not at equilibrium with the solid KCl C) unsaturated and is at equilibrium with the solid KCl D) unsaturated and is not at equilibrium with the solid KCl
Answer:
Option (A) saturated and is at equilibrium with the solid KCl
Explanation:
A saturated solution is a solution which can not dissolve more solute in the solution.
From the question given above, we can see that the solution is saturated as it can not further dissolve any more KCl as some KCl is still visible in the flask.
Equilibrium is attained in a chemical reaction when there is no observable change in the reaction system with time. Now, observing the question given we can see that there is no change in flask as some KCl is still visible even after thorough shaking. This simply implies that the solution is in equilibrium with the KCl solid as no further dissolution occurs.
Which isoelectronic series is correctly arranged in order of increasing radius?
A) K+ < Ca2+ < Ar < Cl
B) < Ar < K+ < Ca2+
C) Ca2+ < Ar < K+ < Cl
D) Ca2+ < K+ < Ar < Cl
E) Ca2+ < K+ < Cl- < Ar
Answer:
D) Ca2+ < K+ < Ar < Cl-
Explanation:
The isoelectronic series is correctly arranged in order of increasing radius is option D. Ca2+ < K+ < Ar < Cl.
Arranging of isoelectronic series:Since k+, Ca2+, Cl- and Ar represent the same kind of e- (18) due to this the positive charge increased. The size should be decreased due to the net attraction on outer e- increased and a negative charge should increase the size because the electronic repulsion should be increased.
Therefore, the option d is correct.
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A 475 ml sample of a gas was collected at room temperature of 23.5 °C and a pressure of
756 mm Hg. Calculate the volume of the gas if the conditions were altered to 10.0 °C and a
pressure of 722 mm Hg.
Answer:
The volume when the conditions were altered is 0.5109 L or 510.9 mL
Explanation:
Using the general gas equation,
P1 V1 / T1 = P2 V2 / T2
where;
P1 = 756 mmHg
V1 = 475 ml = 0.475 L
T1 = 23.5°C = 23.5 + 273K = 275.5 K
P2 = 722 mm Hg
T2 = 10°C = 10 + 273 K = 283 K
V2 = ?
Rearranging to make V2 the subject of the formula, we obtain:
V2 = P1 V1 T2 / P2 T1
V2 = 756 * 0.475 * 283 / 722 * 275.5
V2 = 101, 625.3 / 198911
V2 = 0.5109 L or 510.9 mL
What volume would 32.0 g of NO2 gas occupy at 3.12 atm and 18.0 Celsius
Answer:
Volume of 32.0 g [tex]NO_{2}[/tex] at 3.12 atm and 18.0 [tex]^{0}\textrm{C}[/tex] is 5.33 L.
Explanation:
Molar mass of [tex]NO_{2}[/tex] = 46.0055 g/mol
So, 32.0 g of [tex]NO_{2}[/tex] = [tex]\frac{32.0}{46.0055}[/tex] mol of [tex]NO_{2}[/tex] = 0.696 mol of [tex]NO_{2}[/tex]
Let's assume [tex]NO_{2}[/tex] behaves ideally.
Then [tex]P_{NO_{2}}V_{NO_{2}}=n_{NO_{2}}RT[/tex] , where P, V, n, R and T represents pressure, volume, number of moles, gas constant and temperature (in kelvin scale) respectively.
Here [tex]P_{NO_{2}}[/tex] = 3.12 atm, [tex]n_{NO_{2}}[/tex] = 0.696 mol, R = 0.0821 L.atm/(mol.K) and T = (273+18.0)K = 291 K
So, [tex]V_{NO_{2}}=\frac{n_{NO_{2}}RT}{P_{NO_{2}}}[/tex]
or, [tex]V_{NO_{2}}=\frac{(0.696mol)\times (0.0821\frac{L.atm}{mol.K})\times (291K)}{3.12atm}[/tex]
or, [tex]V_{NO_{2}}=5.33L[/tex]
So, volume of 32.0 g [tex]NO_{2}[/tex] at 3.12 atm and 18.0 [tex]^{0}\textrm{C}[/tex] is 5.33 L.
The Volume of NO₂ is 542.8L
What is Ideal Gas Law ?Ideal Gas Law relates the macroscopic property of gas , It states that the product of pressure and volume of 1 mole of a gas is equal to the product of absolute temperature and universal gas constant.
For n mole of gas,
PV = nRT
In the question it is given that mass of NO₂ is 32 g
P = 3.12 atm
T = 18°Celsius = 291 K
Volume = ?
n = m/M = 32/46 = 0.7
3.12 * V = 0.7*0.0821*291
V = 5.36 L
Therefore the Volume of NO₂ is 5.36L
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Antifreeze is used in car engines. It is produced by a local manufacturer, but they’ve run out of an ingredient: ethane (C2H6). As a junior scientist, you know you can produce ethane by reacting sulfuric acid with diethyl zinc gas. The local manufacturer will pay $1.25 per gram of ethane you produce. If you have 45.00 mL of 8.00 M H2SO4 and 30.00 mL of Zn(CH3CH2)2, what is the maximum amount of money you can make?
Answer:
Maximum amount of money = $21.95
Explanation:
Required reaction:
H2SO4 + Zn(CH3CH2)2 -------------> 2 C2H6 + ZnSO4
No.of moles of H2SO4= (45 x 10-3 L) x (8 M) = 0.360 mol
No.of moles of Zn(CH3CH2)2 = (30 x 1.205) / (123.50 g/mol) = 0.2927 mol
Here H2SO4 is in excess amount and Zn(CH3CH2)2is the limiting reagent.
Hence, Maximum no.of moles of ethane that can be produced = 2 * 0.2927 mol = 0.484 mol
Hence, Mass of ethane that can be produced = 0.484 mol x 30 g/mol = 17.56 g
Therefore,Maximum amount of money that you can make :
Maximum amount of money = (17.56 g) x ($1.25 /g)
Maximum amount of money = $21.95
Give the formula for the alkyne containing 40 hydrogens
Answer:
Molecular formula is C21H40
Explanation:
Use the formula CnH2n-2
That is the alkyne formula
The formula for the alkyne containing 40 hydrogens is C21H40. This is further explained below.
What is Molecular formula?Generally, Molecular formula is simply defined as a chemical formula that shows how many atoms of each element there are in each molecule of a material.
In conclusion, Using the alkyne formula which is C_nH2n-2.
We have C21H40 as the alkyne containing 40 hydrogens.
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4.
What volume of 0.120 M HNO3(aq) is needed to
completely neutralize 150.0 milliliters of 0.100 M
NaOH(aq)?
A. 62.5 mL
B. 125 ml
C.
180. mL
D. 360. mL
Answer:
B) 125 mL
Explanation:
M1V1=M2V2
(0.120M)(x)=(150.0 mL)(0.100M)
x= 125 mL
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c. What predictions can you make about ΔH, ΔS, and ΔG for the general reaction A(s) + 2B(g) 3C(g) + D(g) + heat? What conclusions can you make about the spontaneity of the reaction? Explain your reasoning. (4 points)
d. The reaction 2NO2(g) N2(g) + 2O2(g) is spontaneous at all temperatures, yet it hardly happens under normal conditions. How can this be? (2 points)
ANSWER:
C.
1) Entropy (∆S) is spontaneous
2) Enthalpy (∆H) is not spontaneous
3) Gibbs free energy (∆G) is spontaneous
Therefore the reaction will be spontaneous at high temperature.
D.
Because Nitrogen is favourable to be produced under high temperature, and oxygen under low temperature. Which favours the product side of the equation. But when at a room temperature, which means the temperature is neither low nor high, the product side won't be favoured, and the reaction will not be spontaneous.
EXPLANATION:
C.
1) Entropy is the measure of disorderliness in a system, which increase more in gaseous substances, because the molecules of gases are not stable
Because almost all the reacting substances are in their gaseous state, the entropy of the reaction will be high. Therefore entropy will be positive, which makes the entropy of the system spontaneous.
2) Enthalpy is the measure of heat change in the system. Since their is an intake of heat in the system, therefore the reaction is endothermic and ∆H will be positive. Enthalpy of a system can only be spontaneous in an Exothermic reaction, where ∆H is negative. Therefore Enthalpy is not spontaneous.
3) Gibbs free energy is equal to the change in enthalpy minus the product of temperature and change in entropy. Since entropy and enthalpy are positive, the Gibbs free energy will be negative, which shows that the reaction can be spontaneous if some conditions are met. ∆G will be spontaneous because it is negative.
Therefore in summary, the reaction will favour the product side more, if the temperature of the system is increased, which will make the reaction to become more spontaneous.
D.
2NO2(g) --> N2(g) + 2O2(g)
This reaction is not spontaneous under atmospheric pressure and room temperature (normal conditions) because, Nitrogen can only be produced very fast at a high temperature, while oxygen production is preferable in a low temperature. For the reaction to favour the product side, the temperature should be increased or decreased. Because an increase or a decrease in the temperature will favour either nitrogen or oxygen, which are the product side of equation. This means that the reaction should not be spontaneous if you wish to achieve an equal proportion of the products.
Question 51 pts A breathalyzer is a device used to estimate the blood alcohol content of a suspected drunk driver by measuring the amount of alcohol in one's breath. The fuel cell breathalyzer employs the reaction below: CH3CH2OH(g)+O2(g)→HC2H3O2(g)+H2O(g) When a suspected drunk driver blows his or her breath through the fuel-cell breathalyzer, the device measures the current produced by the reaction and calculates the percent alcohol in the breath. How many moles of electrons are transferred per mole of ethanol, CH3CH2OH, in the reaction?
Answer:
Four moles of electrons
Explanation:
The reactions in a breathalyzer are redox reactions. Fuel cell breathalyzers consists of fuel cells with platinum electrodes. The current produced depends on the amount of alcohol in the breath. Detection of alcohol involves the oxidation of ethanol to ethanoic acid. The two half cells set in the process are;
Anode;
C2H5OH(aq) + 4OH^-(aq) ----------> CH3COOOH(aq) + 3H2O(l) + 4e
Cathode;
O2(g) + 2H2O(l) +4e--------> 4OH^-(aq)
Hence four electrons are transferred in the process.