Answers
The formation of the phosphorylated intermediate (an anhydride) involves the formation of a penta-coordinate intermediate. This intermediate is formed by a nucleophilic attack of the sulfur on the phosphorus atom of the phosphate group.
In this mechanism, the sulfur atom of the sulfate group nucleophilically attacks the phosphorus atom of the phosphate group to form a penta-coordinate intermediate. This intermediate then rearranges to form a phosphorylated intermediate, which is an anhydride.
Mechanism showing the penta-coordinate intermediate and the formation of the phosphorylated intermediate are given as follows:
Step 1: Alkyl Phosphate Formation : The first step of the mechanism includes the formation of an alkyl phosphate. A proton is abstracted by OH− from the phosphate group to create the alkyl phosphate. The base catalyzes this step.
Step 2: Binding to Mg2+After the alkyl phosphate is created, the magnesium ion binds to it.
Step 3: Nucleophilic attack: Following that, the nucleophilic attack happens, with the nucleophile being the water molecule. It is coordinated with the magnesium ion. It occurs at phosphorus, causing it to be phosphorylated. It results in the creation of a pentacoordinate intermediate.
Step 4: Release of Orthophosphate: Orthophosphate is released as a result of the reaction between pentacoordinate intermediate and water. It results in the creation of a diester intermediate.
Step 5: Subsequent Hydrolysis: In the final step, the intermediate diester is hydrolyzed to form orthophosphate and the final product. This is accomplished via nucleophilic substitution.
The end result is a free phosphate group that is bound to the alcohol's oxygen. A phosphate anhydride is formed in the process.
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Related Questions
H2O+C2H5O→C2H5OH+OH NH4+OH→NH3+H20
Which of the two reactions is consistent with the Arrhenius theory of acids and bases
Give the Bronsted Lowry definition of an acid and base
List each acid and its conjugate base for each of the reactions above
Choose one of the above reactions and use Lewis structures to illustrate how it is consistent with the Lewis theory of acids and bases
Answers
The interaction between Arrhenius acid and base, which produces salt and water as a byproduct, is referred to as a neutralisation reaction. Strong acids include substances like HCl, HNO3, H2SO4, etc.
What does an Arrhenius reaction look like?The term "Arrhenius acid" refers to a material that contains a hydrogen atom that readily releases a hydrogen ion and proton when it is in contact with water. For instance, when hydrochloric acid dissolves in water, it produces the ions hydronium (H3O+) and chloride (Cl-).
That Arrhenius theory of bases and acids is which of the following?The hydrogen ion (H+) is one of the electrically charged molecules or atoms that are produced when an acid dissociates in water, according to the Arrhenius theory, which was first proposed by Swedish scientist Svante Berzelius in 1887. On the other hand, bases ionise in water to produce hydroxide ions (OH).
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The following chemical reaction takes place in aqueous solution: SnBr2 (aq) + K2S (aq) → SnS (s) + 2KBr (aq) Write the net ionic equation for this reaction.
Answers
To write the net ionic equation for this reaction, we need to first write the balanced molecular equation, and then break down all the soluble ionic compounds into their constituent ions:
Molecular equation:
SnBr2(aq) + K2S(aq) → SnS(s) + 2KBr(aq)
Complete ionic equation:
Sn2+(aq) + 2Br-(aq) + 2K+(aq) + S2-(aq) → SnS(s) + 2K+(aq) + 2Br-(aq)
Net ionic equation:
Sn2+(aq) + S2-(aq) → SnS(s)
The net ionic equation shows only the species that are involved in the actual reaction, which in this case are the tin cation (Sn2+) and the sulfide anion (S2-), which combine to form solid tin sulfide (SnS). The potassium cation (K+) and bromide anion (Br-) ions are spectator ions that do not participate in the reaction and are omitted from the net ionic equation.
What is an ionic ?Ionic refers to a type of chemical bond that occurs when one or more electrons are transferred from one atom to another, resulting in the formation of ions.
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Determine the
Cl
for NeMut1:Wt using the data presented in part 2 of the case study.
1×10 −10
1×10 10
1×10 −7
1×10 7
Cl
cannot be calculated from the data given 4. If the
LD s0
and/or
ID s 0
values of a Wt and mutant strain are similar in this type of experiment, does this automatically mean that the mutation does not affect a virulence factor? Why or why not? Part B. The researchers decided to determine the
Cl
of each of the mutants, again using the horse infection model. The results are summarized in the table below: 5. Determine the
CI
for NeMutl:Wt and NeMut2:Wt. 6. Interpret your results from question 5 above.
Answers
To determine the Cl for NeMutl:Wt, you need to use the data from part 2 of the case study. The data is given as 1x10-10 for the Wt strain and 1x10-7 for the mutant strain. To calculate the Cl, we use the following equation: Cl = 1/[(1/ID50) - (1/LD50)]. Using this equation, we can calculate the Cl to be 3x10-3.
To determine the Cl for NeMut2:Wt, we can use the same equation. Using the data from the table in part B, the Cl for NeMut2:Wt can be calculated to be 8x10-3.Interpreting these results, we can see that NeMut1:Wt has a Cl that is roughly 3 times lower than that of NeMut2:Wt. This suggests that the mutation of NeMut1 is significantly affecting a virulence factor, while NeMut2 may not be affecting a virulence factor as significantly.
It is important to note that similar LD50 and/or ID50 values of a Wt and mutant strain does not necessarily mean that the mutation does not affect a virulence factor. This is because the LD50 and ID50 values are used to measure how much of the pathogen is needed to produce a certain effect, but other aspects of the pathogen such as the speed or rate of infection or the amount of toxin produced can still be different and affect the virulence of the strain.
Cl for NeMut1:Wt cannot be calculated from the data presented in part 2 of the case study. The given results are:| Inoculum (LD50) | Mortality (LD50) | CFU/ml of blood | Wild-type | 6.5 × 10−7 | 6.5 × 10−7 | 7.0 × 103 | NeMut1 | 1.0 × 10−10 | 6.5 × 10−7 | 3.0 × 105 | NeMut2 | 2.0 × 10−7 | 2.0 × 10−7 | 2.2 × 103 |Since the Cl cannot be calculated from the data given, the correct option is (d) Cl cannot be calculated from the data given.If the LDs0 and/or IDs0 values of a Wt and mutant strain are similar in this type of experiment, it does not necessarily mean that the mutation does not affect a virulence factor.
This is because mutations can affect different aspects of virulence, and the specific virulence factor being measured may not be impacted by the mutation.In order to determine the CI for NeMut1:Wt and NeMut2:Wt, we need to use the following formula:CI = (output ratio of mutant) / (output ratio of wild-type)Output ratio = (CFU/ml of blood) / (inoculum)Using the data from the table, we get:
Output ratio of NeMut1:Wt = 3.0 × 105 / 1.0 × 10−10 = 3.0 × 1015Output ratio of wild-type = 7.0 × 103 / 6.5 × 10−7 = 1.1 × 1010CI of NeMut1:Wt = (3.0 × 1015) / (1.1 × 1010) = 2.7 × 105Output ratio of NeMut2:Wt = 2.2 × 103 / 2.0 × 10−7 = 1.1 × 1010CI of NeMut2:Wt = (1.1 × 1010) / (1.1 × 1010) = 1Interpretation of results from question 5 above: The CI of NeMut1:Wt is much greater than 1, indicating that NeMut1 is more virulent than the wild-type strain. The CI of NeMut2:Wt is equal to 1, indicating that NeMut2 does not exhibit any significant difference in virulence compared to the wild-type strain.
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How many calories are required to raise the temperature of a 35.0 g sample from 35 °C to 85 °C? The sample has a specific heat of 0.108 cal/g °C.
Answers
Answer:
First, we need to calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 85 °C - 35 °C
ΔT = 50 °C
Next, we can use the following formula to calculate the heat energy required:
Q = m·C·ΔT
where Q is the heat energy in calories, m is the mass of the sample in grams, C is the specific heat in cal/g °C, and ΔT is the change in temperature in °C.
Plugging in the given values, we get:
Q = 35.0 g · 0.108 cal/g °C · 50 °C
Q = 189.0 calories
Therefore, 189.0 calories are required to raise the temperature of the sample from 35 °C to 85 °C
a) The ionization constant for water (Kw) is 5.476 x 10-14 at 50 oC. Calculate [H3O+], [OH-], pH, and
pOH for pure water at 60oC.
b) Look at the following table with temperature, Kw, and pH values for pure water.
Temp. (oC)
Kw
pH
0
0.114 x 10-14
7.47
10
0.293 x 10-14
7.27
20
0.681 x 10-14
7.08
25
1.008 x 10-14
7.00
40
2.916 x 10-14
6.77
60
9.550 x 10-14
6.51
100
51.3 x 10-14
6.14
Notice that the pH falls as the temperature increases.
i) Does this mean that pure water becomes more acidic as the temperature rises? _______
ii) The reason for my answer is in pure water…(choose the correct letter from 1 - 5)._______
1. the pH becomes less than 7 as the temperature rises.
2. there are less H3O+ ions created as the temperature increases.
3. there are more H3O+ ions created as the temperature increases.
4. there are always the same # of hydronium and hydroxide ions even if the pH changes.
5. the OH- ions are evaporated as the temperature rises.
Answers
A) To solve this problem, we can use the formula for the ionization constant of water:
Kw = [H3O+][OH-]
At 50 oC, Kw = 5.476 x 10-14. We can assume that [H3O+] and [OH-] are equal since we are dealing with pure water.
Therefore,
[H3O+] = [OH-] = sqrt(Kw) = sqrt(5.476 x 10-14) = 7.40 x 10-8 mol/L
pH = -log[H3O+] = -log(7.40 x 10-8) = 7.13
pOH = -log[OH-] = -log(7.40 x 10-8) = 7.13
To find the values for pure water at 60oC, we can use the new value of Kw at that temperature:
Kw = 9.550 x 10-14
[H3O+] = [OH-] = sqrt(Kw) = sqrt(9.550 x 10-14) = 3.09 x 10-7 mol/L
pH = -log[H3O+] = -log(3.09 x 10-7) = 6.51
pOH = -log[OH-] = -log(3.09 x 10-7) = 6.51
b) i) No, it does not mean that pure water becomes more acidic as the temperature rises.
ii) The correct answer is 2. As the temperature increases, the ionization of water increases and more H3O+ and OH- ions are formed. However, since the concentration of H2O is also decreasing due to the increase in temperature, the increase in ionization does not result in an increase in [H3O+] and pH actually decreases.
Use the following equations to find the lattice energy of MgCl₂? (hint: first write the equation for
the lattice energy of MgCl2(s))
Mg(g) → Mg2+(g) + 2 e
CI(g) + eCl (g)
Mg(g) + 2Cl(g) → MgCl₂(s)
+2188 kJ/mol
-337 kJ/mol
AH = -642 kJ/mol
Answers
Answer:
Explanation:
The lattice energy (LE) of MgCl₂ can be calculated using the Born-Haber cycle, which relates the lattice energy to other thermodynamic quantities such as enthalpy of sublimation, ionization energy, electron affinity, and heat of formation.
The equation for the lattice energy of MgCl₂ is:
LE(MgCl₂) = H(sublimation of Mg) + IE(Mg) + EA(Cl) + 1/2 H₂(Cl₂) - DH(f)(MgCl₂)
where H(sublimation of Mg) is the enthalpy of sublimation of Mg, IE(Mg) is the ionization energy of Mg, EA(Cl) is the electron affinity of Cl, H₂(Cl₂) is the heat of formation of Cl₂, and DH(f)(MgCl₂) is the heat of formation of MgCl₂.
We are given the following equations:
Mg(g) → Mg2+(g) + 2 e ΔH₁ = +2188 kJ/mol
Cl(g) + e → Cl-(g) ΔH₂ = -337 kJ/mol
Mg(s) + Cl₂(g) → MgCl₂(s) ΔH₃ = -642 kJ/mol
Using these equations, we can calculate the values of H(sublimation of Mg), IE(Mg), EA(Cl), H₂(Cl₂), and DH(f)(MgCl₂) as follows:
H(sublimation of Mg) = ΔH₂(Mg(g)) + 1/2 ΔH₃(Cl₂(g)) - ΔH₁(Mg2+(g)) = -2220 kJ/mol
IE(Mg) = ΔH₁(Mg(g)) = +2188 kJ/mol
EA(Cl) = ΔH₂(Cl(g)) = -337 kJ/mol
H₂(Cl₂) = 0 (since Cl₂ is in the gas phase)
DH(f)(MgCl₂) = ΔH₃(Mg(s), Cl₂(g), MgCl₂(s)) = -642 kJ/mol
Substituting these values into the equation for the lattice energy, we get:
LE(MgCl₂) = -2220 + 2188 - 337 + 1/2(0) - (-642) = -3509 kJ/mol
Therefore, the lattice energy of MgCl₂ is approximately -3509 kJ/mol.
Which of the following molecules is drawn in a conformation that has a proton and a leaving group anti-periplanar? H₂C, Br Ph. H CH3 Br H H₂C Br H₂C Ph H₂C CH3 H Ph H₂C, Br H Ph Save for Later CH3 CH3 CH3 CH3 CH3 Sul
Answers
The molecule that is drawn in a conformation that has a proton and a leaving group anti-periplanar is H₂C, Br.
The A, B, C, and D bond angles of a molecule are referred to as anti-periplanar, or antiperiplanar, in organic chemistry. The dihedral angles of the A–B and C–D bonds in this conformer are larger than +150° or less than 150°. In textbooks, the term "anti-periplanar" is frequently used to refer to a strictly anti-coplanar structure with a 180° AB CD dihedral angle. The anti-periplanar functional groups will be 180° apart from one another and in a staggered configuration in a Newman projection of the molecule.
Conformation is an essential factor in predicting reactivity in organic molecules. The anti-periplanar conformation of a molecule is one that occurs when two atoms in a molecule are in the same plane and are separated by 180 degrees. In this case, the proton and leaving group are placed in a perpendicular plane to the atoms directly in between them. This is the most stable conformer of the molecule. A significant factor in predicting reactivity in organic molecules is conformation. In this case, the molecule H₂C, Br is drawn in a conformation that has a proton and a leaving group anti-periplanar.
Therefore, the correct option is H₂C, Br.
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Chemistry Question Help Please!
Use the phase diagram for water to answer the following questions.
35℃ and 85 kPa
-15℃ and 40 kPa
-15℃ and 0.1 kPa
60℃ and 50 kPa
Using the graph above, describe the phase changes that are in equilibrium along each black line. Describe what those equilibria mean.
The boiling point for nitrogen is -195.8℃. Imagine that you had a sealed container of nitrogen gas at room temperature. What are two ways that you could turn the gas into a liquid?
Answers
The two equilibria are the triple point, which is the point when water can exist as ice, liquid, or vapor, and the critical point, which is when water exists as a single phase that has properties of both a liquid and a gas.
What are the triple point and critical point of water?The line connecting the triple point and critical point on the phase diagram of water represents the conditions at which the solid, liquid, and gas phases can coexist in equilibrium.
At the triple point, which occurs at a temperature of 0.01℃ and a pressure of 0.006 atm, the solid, liquid, and gas phases of water exist in equilibrium. This means that at this point, water can exist as ice, liquid, or vapor, depending on the conditions.
At the critical point, which occurs at a temperature of 374℃ and a pressure of 218 atm, the distinction between the liquid and gas phases disappears, and the substance becomes a supercritical fluid. This means that at this point, water exists as a single phase, which exhibits the properties of both a liquid and a gas.
There are two ways to turn nitrogen gas into a liquid:
Cooling: Nitrogen can be cooled down to its boiling point (-195.8°C) using a cryogenic cooler or liquid nitrogen, which causes the gas to condense into a liquid.Increasing pressure: Nitrogen can also be compressed at room temperature to a pressure higher than its vapor pressure, causing it to condense into a liquid.Learn more about the phase diagram of water at: https://brainly.com/question/13518101
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What can you infer about these two parts of a microscope?
A They are lightweight
B They contain mirrors and lenses
C They are strong and sturdy
D They are made from glass or plastic
Answers
Answer:
the answer is for surely C
Explanation:
you can tell from the way is made and I also have one in my room
Consider the reaction between aqueous solutions of potassium hydroxide and chromium (III) chloride. Based on your balanced equation for this reaction, answer the following questions: 1) What are the spectator ions in this reaction? and 2) What is the formula for the precipitate formed in this reaction? 3) What is the sum of all the coefficients in the net ionic equation for this reaction? Net lonic Equati....pdf Hydrocarbon C....pdf
Answers
The net ionic equation for the given reaction is: 1 Cr3+(aq) + 3OH-(aq) → Cr(OH)3(s)
2. The formula for the precipitate formed in this reaction is Cr(OH)3.
3. The sum of all coefficients in the net ionic equation is 4.
Consider the reaction between aqueous solutions of potassium hydroxide and chromium (III) chloride. The balanced chemical equation for the given reaction is:KOH(aq) + CrCl3(aq) → KCl(aq) + Cr(OH)3(s)1) Spectator ionsThe ions that do not take part in the reaction are known as spectator ions.
These ions are present on both sides of the equation without undergoing any chemical changes.The ionic equation for the given reaction is:3K+(aq) + 3OH-(aq) + Cr3+(aq) + 3Cl-(aq) → 3K+(aq) + 3Cl-(aq) + Cr(OH)3(s)The spectator ions are K+ and Cl-.2)
PrecipitateThe precipitate is formed when the two reactants are combined together, and it can be identified from the ionic equation. In this reaction, the precipitate is formed when KOH is added to the aqueous solution of chromium(III) chloride.The formula for the precipitate formed in this reaction is Cr(OH)3.3) Sum of all coefficientsThe net ionic equation represents the actual chemical change occurring in the reaction.
The spectator ions are removed, and only the ions that participate in the reaction are shown. The net ionic equation for the given reaction is:Cr3+(aq) + 3OH-(aq) → Cr(OH)3(s)The sum of all coefficients in the net ionic equation is 4.
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ing The
Ionic bonds are made by electrons.
Answers
use the trendline equation in fig6.2 to determine the kelvin temperature at which the pressure equals .72 atm
Answers
When the pressure is 0.72 atm, the temperature in Kelvin is 156 K.
To determine the Kelvin temperature when the pressure is 0.72 atm, you will need to use the trendline equation given in Fig 6.2. First, find the equation of the trendline by using the graph's two points, (300 K, 1 atm) and (500 K, 2 atm).
The equation for the trendline is:
y = mx + b
Where y is pressure, x is the temperature in Kelvin, m is the slope, and b is the y-intercept. We can find the slope of the trendline by using the two points provided in the graph:
Slope (m) = (y2 - y1) / (x2 - x1)
Slope = (2 atm - 1 atm) / (500 K - 300 K)
Slope = 0.005 atm/K
The equation for the trendline can now be written: y = 0.005x + b. To find the y-intercept, b, we can use one of the two points: Solving for b:
1 atm = 0.005(300 K) + bb = 1.5 atm
Now we can use the equation for the trendline to find the temperature (x) at which the pressure (y) equals 0.72 atm:
0.72 atm = 0.005x + 1.5 atm
0.72 atm - 1.5 atm = 0.005x
-0.78 atm = 0.005xx
= -0.78 atm / 0.005x
= 156K
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exolain why the addition of 0.20 m naoh to 0.20 m ch3ch2cooh can result in the formation of a buffer solution.
Answers
The addition of 0.20 M NaOH to 0.20 M CH3CH2COOH can result in the formation of a buffer solution because NaOH is a strong base, and CH3CH2COOH is a weak acid. When these two react, they will produce a salt and water, and the salt will act as the buffer that helps keep the pH of the solution relatively constant.
The addition of 0.20 M NaOH to 0.20 M CH3CH2COOH results in the formation of a buffer solution because NaOH is a strong base, and CH3CH2COOH is a weak acid. When a strong base is mixed with a weak acid, a buffer solution is formed. A buffer solution is able to resist changes in pH when small amounts of an acid or base is added.
When NaOH is added to CH3CH2COOH, the weak acid (CH3CH2COOH) will react with the strong base (NaOH) to produce salt (CH3CH2COONa) and water (H2O). The CH3CH2COONa salt will then dissociate into CH3CH2COO- and Na+ ions, which will act as the buffer. The CH3CH2COO- ions will then accept H+ ions from an acid and donate H+ ions to a base. This helps to keep the pH of the solution relatively constant.
The Ka of the CH3CH2COOH is 1.8x10^-5 and its pKa is 4.75. This means that the buffer will work best when the pH is close to 4.75. The amount of NaOH added to the CH3CH2COOH will determine the pH of the buffer solution. If the amount of NaOH is too high, the pH will be above the pKa and the buffer will not work as efficiently.
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All of the following occur when the temperature of the preoptic area of the hypothalamus drops below its thermostat setting, except thata. blood flow to the skin increases.b. shivering thermogenesis occurs.c. nonshivering thermogenesis occurs.d. epinephrine levels rise.e. blood returning from limbs is shunted to deep veins.
Answers
The answer to this question is that blood flow to the skin does not increase; instead, it is directed inward to conserve heat.
The preoptic area of the hypothalamus serves as the body's thermostat, and when its temperature drops below the thermostat setting, a number of physiological processes are triggered. A) Shivering thermogenesis occurs, which is the production of heat by muscular contractions, B) Nonshivering thermogenesis occurs, which is the production of heat by increased metabolic activity in brown adipose tissue, C) Epinephrine levels rise to promote the mobilization of glucose from stored energy sources and D) Blood returning from the limbs is shunted to deep veins to conserve heat. However, the answer to this question is that blood flow to the skin does not increase; instead, it is directed inward to conserve heat.
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Please help!
What is the equilibrium concentration of C if the reaction begins with 0.200 M A and 0.250 M B?
Answers
The reaction starts with 0.200 M A and 0.250 M B, the equilibrium concentration of C.3.63x10⁻¹⁰ M.
Why are equilibrium concentrations important?Chemical equilibrium is the state in which both the reactants and the products of a reaction are at a concentration that does not change over time any longer. In this condition, the forward and backward reaction rates are equal.
According to the given information:2A(aq) + B(aq) <==> C(aq)
Equilibrium expression is
K = [C] / [A]2[B]
Prepare an ICE table:
2A(aq) + B(aq) <==> C(aq)
0.2..........0.15...............0........Initial
-2x...........-x.................+x......Change
0.2-2x....0.15-x............x........Equilibrium
Substitute in the equilibrium expression:
1.10x10⁻⁴ = (x) / (0.2-2x)2(0.15-x) ... b/c K is small, we can essentially avoid using the quadratic as follows..
1.1x10⁻⁴ = x/(0.2)2(0.15)
x = (2.2x10₋⁵) (1.65x10⁻⁵)
x = 3.63x10⁻¹⁰ M = [C]
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I WILL MARK BRAINLIEST
Question
Which of the following is an example of a valid experiment?
Placing balls of different masses at the top of a ramp and measuring the distance that they roll.
Asking your family members if they prefer ham or turkey,
Placing two plants in a dark closet
Which of the following should NOT be included in your conclusion?
Responses
A list of materials required for the experiment
A brief explanation of the purpose of your experiment
A scientific explanation for your conclusion,
A statement of whether your hypothesis was correct or not
Which of these is NOT a testable hypothesis?
Responses
Plants that receive more light will grow at a faster rate
Adding fertilizer to plants will cause them to grow at a faster rate
Plants that receive less water will grow at a slower rate,
Plants that receive compost will taste better
Answers
Answer:
1. placing balls
2. list of materials
Explanation:
Chemistry Help Please! It's worth a lot of points
1.Write the equilibrium expression for the following reactions
a. H2SO4(aq) + H2O(L) ⇆ HSO4-(aq) + H3O+(aq)
b. 4NH3(g) + 5O2(g) ⇆ 4NO(g) + 6H2O(g)
c. NH4Cl(s) ⇆ NH3(g) + HCl(g)
d. N2O4(g) ⇆ 2NO2(g)
2. The following reaction has a K value of 0.050. What does that mean about the concentrations of the reactants as compared to the products? Be specific in your answer.
N2(g) + 3H2(g) ⇆ 2NH3(g)
3. The following reaction has a K value of 6.8 x 103. What does that mean about the concentrations of the reactants as compared to the products? Be specific in your answer.
2SO3(g) ⇆ 2SO2(g) + O2(g)
4. When dissolving substances in water, the degree of solubility of a substance is often represented as the solubility product constant (Ksp). The solubility product constant is the same thing as the equilibrium constant for the dissolving reaction. Two substances that dissociate in water are shown below alone with the Ksp.
NaCl(s) ⇆ Na+(aq) + Cl-(aq) Ksp = 36
BaSO4(s) ⇆ Ba2+(aq) + SO42-(aq) Ksp = 1.1 x 10-16
5. Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
a. HNO3 + H2O ⟶ H3O+ + NO3−
b. CN− + H2O ⟶ HCN + OH−
c. H2SO4 + Cl− ⟶ HCl + HSO4−
d. HSO4− + OH− ⟶ SO42− + H2O
e. O2− + H2O ⟶2OH−
6. What is the conjugate acid of each of the following? What is the conjugate base of each of the following?
a. OH-
b. H2O
c. HCO3-
d. NH3
e. HSO4-
7. The following acids are shown with their equilibrium constants (also known as the acid dissociation constant). Rank these acids from strongest to weakest. Explain your ranking.
HCN(aq) + H2O(L) ⇆ H3O+(aq) + CN-(aq) K = 6.2 x 10-10
HC2H3O2(aq) + H2O(L) ⇆ H3O+(aq) + C2H3O-(aq) K = 1.75 x 10-5
H2CO3(aq) + H2O(L) ⇆ H3O+(aq) + HCO3-(aq) K = 4.5 x 10-7
HIO4(aq) + H2O(L) ⇆ H3O+(aq) + IO4-(aq) K = 2.3 x 10-2
8. Calculate the pH and the pOH of each of the following solutions.
a. 0.200 M HCl
b. 0.0143 M NaOH
c. 3.0 M HNO3
d. 0.0031 M Ca(OH)2
9. Wine has a pH of 3.6. What are the hydronium and hydroxide ion concentrations?
10. The hydroxide ion concentration in household ammonia is 3.2 x 10-3 M. What is the concentration of hydronium ions?
Answers
Answer:
1. Equilibrium expressions:
a. K = [HSO4-][H3O+]/[H2SO4][H2O]
b. K = [NO]^4[H2O]^6/[NH3]^4[O2]^5
c. K = [NH3][HCl]/[NH4Cl]
d. K = [NO2]^2/[N2O4]
2. Since K = 0.050, the concentrations of the reactants (N2 and H2) are larger than the concentrations of the products (NH3).
3. Since K = 6.8 x 10^3, the concentrations of the products (SO2 and O2) are larger than the concentrations of the reactant (SO3).
4. The Ksp expression for each of the reactions is:
a. Ksp = [Na+][Cl-]
b. Ksp = [Ba2+][SO42-]
5. Brønsted-Lowry acids and bases:
a. Acid: HNO3; Conjugate base: NO3-; Base: H2O; Conjugate acid: H3O+
b. Acid: HCN; Conjugate base: CN-; Base: H2O; Conjugate acid: HCN
c. Acid: H2SO4; Conjugate base: HSO4-; Base: Cl-; Conjugate acid: HCl
d. Acid: NH3; Conjugate base: NH2-; Base: H2O; Conjugate acid: NH4+
e. Acid: H2O; Conjugate base: OH-; Base: O2-; Conjugate acid: OH-
6. Conjugate acids and bases:
a. Acid: H2O; Conjugate base: OH-
b. Acid: H3O+; Conjugate base: H2O
c. Acid: H2CO3; Conjugate base: HCO3-
d. Acid: NH4+; Conjugate base: NH3
e. Acid: HSO4-; Conjugate base: SO42-
7. The strongest acid is HIO4 (highest K value), followed by HCN, HC2H3O2, and H2CO3 (lowest K value). The K values represent the degree to which the acids dissociate in solution. HIO4 is a strong acid, meaning it dissociates almost completely in solution, while H2CO3 is a weak acid, meaning it only dissociates partially.
8. pH and pOH calculations:
a. pH = -log[H3O+] = -log(0.200) = 0.699; pOH = -log[OH-] = -log(1.0 x 10^-14/0.200) = 12.301
b. pOH = -log[OH-] = -log(0.0143) = 1.844; pH = 14.000 - pOH = 12.156
c. pH = -log[H3O+] = -log(3.0) = 0.522; pOH = 13.478
d. pOH = -log[OH-] = -log(0.0062) = 2.206; pH = 14.000 - pOH = 11.794
9. Hydronium and hydroxide ion concentrations:
pH = 3.6; hydronium ion concentration = 10^-pH = 3.98 x 10^-4 M; hydro
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1.Explain why the incomplete combustion of alkanes is dangerous.
2.Outline the environmental impacts of burning fossil fuels such as the alkanes. Include an
explanation of how these impacts can be mitigated.(max 300 words)
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Answer- Incomplete combustions of alkanes are dangerous as it leads to toxic gases such as carbon monoxide man odorless,colorless and highly poisonous gas to be released in the air
Environmental impacts of burning fossil fuels such as alkanes can have a massive impact of global warming in which heat is prevented from the leaving the atmosphere due to a build-up of carbon dioxide ad other compounds, collectively known as greenhouse gases. In order to make these impacts less severe scientists have found alternative fuels, such as biofuels and biodiesels, this can reduce the risk of harm to the world as alternative fuels produce less CO2 and air pollutants then others that run on petrol and diesel
One set among the following sets of quantum numbers, {n, l, m_t, m_s} is erroneous. Which one and why? Identify the sets of four quantum numbers, {n, I, ml, m,}, that are forbidden for an electron in an atom and explain why they are in valid. Which combinations of n and/represent real orbitals and which do not exist? What are the principal and orbital angular momentum quantum numbers for each of the following orbitals? Orbital the principal quantum number, n the orbital angular momentum quantum number How many electrons can occupy the 4p-orbitals? the 3d-orbitals? the ls-orbital? the 4f-orbitals?
Answers
The 4p-orbital can hold a maximum of 6 electrons, the 3d-orbital can hold a maximum of 10 electrons, the 1s-orbital can hold a maximum of 2 electrons, and the 4f-orbital can hold a maximum of 14 electrons.
The set of quantum numbers {n, I, ml, m} is incorrect because I is not a quantum number; instead, it should be l, which stands for the angular momentum quantum number. This set of quantum numbers must satisfy the following conditions: n ≥ l ≥ m_l ≥ |m_s|. For an electron in an atom, any combination of quantum numbers where n < l is forbidden, because n must be greater than or equal to l. Additionally, any combination of quantum numbers where |m_l| > l is forbidden because m_l must be less than or equal to l.
Real orbitals have the following combinations of n and l: n = 1 and l = 0 (1s), n = 2 and l = 0, 1 (2s, 2p), n = 3 and l = 0, 1, 2 (3s, 3p, 3d), n = 4 and l = 0, 1, 2, 3 (4s, 4p, 4d, 4f).
The principal quantum number and the orbital angular momentum quantum number for the 4p-orbital is n = 4 and l = 1; for the 3d-orbital is n = 3 and l = 2; for the 1s-orbital is n = 1 and l = 0; for the 4f-orbital is n = 4 and l = 3.
The 4p-orbital can hold a maximum of 6 electrons, the 3d-orbital can hold a maximum of 10 electrons, the 1s-orbital can hold a maximum of 2 electrons, and the 4f-orbital can hold a maximum of 14 electrons.
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I received the answer of 2.0 atm. How did you get 2.6 atm? for the problem of A sample of the inhalation anesthetic gas Halothane, in a 500-mL cylinder has a pressure of 2.3...
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If the sample of the inhalation anesthetic gas Halothane, in a 500-mL cylinder has a pressure of 2.3, the pressure is given as 2.6 atm
How to solve for the pressure of the gasWe have the following data to solve the problem with
p1 = 2.3 atm
T1 = 273 K
P2 is unknown
T2 = 273 + 37
= 310 K
To solve further we would have to use the Third gas law. The third gas law or the Gay-Lussacs law tells us that
P1 / T1 = P2 / T2
We would have to put in the values in the formula that we have above
[tex]\frac{2.3 atm}{273K} = \frac{p2}{310}[/tex]
from here would cross multiply
310 * 2.3 = p2 * 273
divide through by 273
713 / 273
P2 = 2.61
The pressure is given as 2.61
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write a balanced equation for the redox reaction between calcium metal and oxygen gas
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a balanced equation for the redox reaction: 2 Ca(s) + O2(g) → 2 CaO(s)
What is a redox reaction?A redox reaction is a type of chemical reaction that involves the transfer of electrons between species. One species undergoes oxidation (loses electrons) while another species undergoes reduction (gains electrons).
Which species is being oxidized and which species is being reduced in the reaction between calcium metal and oxygen gas?In the reaction between calcium metal and oxygen gas, the calcium metal is being oxidized (loses electrons) and the oxygen gas is being reduced (gains electrons). This can be seen in the balanced equation where the calcium atoms go from having an oxidation state of 0 to +2, while the oxygen atoms go from having an oxidation state of 0 to -2.
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FILL IN THE BLANK. The molecular ion produced when an organic molecule is bombarded by a high-energy stream of electrons _____.
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The molecular ion produced when an organic molecule is bombarded by a high-energy stream of electrons is known as a radical cation.
Radical cations are highly reactive and can react quickly with other compounds in their environment. The number of electrons removed to form the radical cation is determined by the type of organic molecule that is bombarded. If an alkene is bombarded, two electrons are typically removed, forming an allyl radical cation, while the addition of three electrons is necessary to form a benzyl radical cation. Radical cations are stabilized by the formation of new bonds with other molecules, thereby forming an adduct. The adduct can then be separated and characterized using chromatographic techniques. Additionally, radical cations can also react with nucleophiles, resulting in a variety of other reaction products. Thus, the molecular ion produced when an organic molecule is bombarded by a high-energy stream of electrons is a radical cation.
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A 0.48 molar solution of a monoprotic acid (HA) in water reaches equilibrium at a concentration of 0.36 M. What is Ka for this acid? Please enter your answer rounded to two significant figures. Step by step please <3
Answers
Answer:
The dissociation of a monoprotic acid HA can be represented as follows:
HA ⇌ H+ + A-
The equilibrium constant expression for this reaction is:
Ka = [H+][A-]/[HA]
We are given the initial concentration of the acid (HA) as 0.48 M and the equilibrium concentration as 0.36 M. At equilibrium, the concentration of H+ and A- will also be 0.36 M.
Substituting the values into the equilibrium constant expression, we get:
Ka = (0.36)^2 / 0.48 = 0.27
Therefore, the value of Ka for the acid is 0.27, rounded to two significant figures.
List the following compounds from most reactive to least reactive toward electrophilic aromatic substitution. Rank the compounds from most to least reactive. To rank items as equivalent, overlap them. Reset Help toluene benzene ne phenol bromobenzene nitrobenzene
Answers
Phenol, toluene, benzene, bromo benzene, and nitrobenzene are the chemicals in this order.
Which aromatic substance reacts least favourably to electrophilic substitution?Due to the M effect, benzosulphonic acid is least reactive in an electrophilic aromatic substitution. Because nitrogen is more electronegative than carbon and functions as an electron withdrawing group, pyridine is less reactive than benzene towards electrophilic aromatic substitution. The meta hydrogen is thus replaced.
Which five electrophilic aromatic substitution reactions occur most frequently?Most beginning organic chemistry courses cover six fundamental electrophilic aromatic substitution reactions: chlorination, bromination, nitration, sulfonation, Friedel-Crafts alkylation, and Friedel-Crafts acylation.
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What is the pH of a 0.021 molar solution of NaOH? Enter your answer rounded to two decimal places. step by step
Answers
pH = 12.32
Step by step explanation:
To find the pH of a 0.021 molar solution of NaOH, we first need to calculate the concentration of hydroxide ions (OH-) in the solution using the following equation:
OH- concentration = NaOH concentration
OH- concentration = 0.021 M
Next, we can use the equation for the dissociation of water to calculate the concentration of hydrogen ions (H+) in the solution:
Kw = [H+][OH-]
where Kw is the ion product constant for water, which has a value of 1.0 x 10^-14 at 25°C.
Rearranging the equation, we get:
[H+] = Kw / [OH-]
[H+] = (1.0 x 10^-14) / (0.021)
[H+] = 4.76 x 10^-13 M
Now, we can use the equation for pH to calculate the pH of the solution:
pH = -log[H+]
pH = -log(4.76 x 10^-13)
pH = 12.32
Therefore, the pH of a 0.021 molar solution of NaOH is 12.32 (rounded to two decimal places).
Sulfur reacts with oxygen according to the following equation: 2S + 3O2 = 2SO3.
if 4 moles of sulfur react with 9.5 moles of oxygen, how many moles of oxygen would remain after the reaction?
Answers
Explanation:
From the balanced equation you can see that for every TWO moles of S , THREE moles of O2 are needed
so if you have four moles of S you will need SIX moles of O2 ....meaning
you will have ( 9.5 - 6 ) = 3.5 moles of O2 left over
Assuming that all is working properly, which of the following is at a higher energy level?
Answers
If all is working correctly, the electron in a higher orbital or shell will have more energy than the electron in a lower orbital or shell. The following statement is correct: The electron that is farther away from the nucleus is at a higher energy level.
In order for the electron to escape from the atom, it must be excited, meaning that it must absorb energy. When this occurs, the electron moves to a higher energy level, which is farther from the nucleus. Because the electron is now in an excited state, it is more vulnerable to being released from the atom if additional energy is provided to it. According to Bohr's model of the atom, electrons revolve around the nucleus in circular orbits with varying energy levels. As the distance between the nucleus and the electron increases, so does the energy level of the electron. The energy of electrons in the first energy level is the lowest, and as the energy level increases, so does the energy of electrons. As a result, electrons in the outermost shell have the highest energy levels.
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Component involves in simple cell
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Answer: two solid electrodes (must be two different metals in solution)
what does 2NaOH equal
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Calculate the wavelength in meters of electromagnetic radiation that has a frequency of 640.0 kHz. (c = 3.00 X 108 m/s)
Answers
Answer: 468.8m
Explanation:
640kHz x 1000Hz/1kHz = 6.4x10^5 Hz
λ=c/v
3.00x10^8m/s / 6.4x10^5s = 468.75m