The mass of AgI in the precipitate is approximately 38.65 grams (rounded to two significant figures) with the appropriate units.
To find the mass of AgI in the precipitate, we first need to determine the moles of AgI formed. We'll use the information provided and follow these steps:
1. Write the balanced chemical equation for the precipitation reaction:
[tex]\[ \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \][/tex]
[tex]\[ \text{Hg}_2^{2+} + 2\text{I}^- \rightarrow \text{HgI}_2 \][/tex]
2. Determine the limiting reactant:
- For AgI: [tex]\( \text{Ag}^+ + \text{I}^- \)[/tex] (1 mole of Ag per mole of I)
- For HgI2: [tex]\( \frac{1}{2}\text{Hg}_2^{2+} + \text{I}^- \)[/tex] (1 mole of Hg per 2 moles of I)
The limiting reactant is the one that produces the fewer moles of I^-, as it determines the amount of AgI formed.
[tex]\[ \text{moles of I}^- = 0.100 \, \text{L} \times 1.71 \, \text{mol/L} = 0.171 \, \text{mol} \][/tex]
The limiting reactant is AgI, as it requires 0.171 moles of I^-, while HgI2 would require 0.342 moles of I^-.
3. Calculate the moles of AgI formed:
[tex]\[ \text{moles of AgI} = \text{moles of I}^- \times \frac{1 \, \text{mol AgI}}{1 \, \text{mol I}^-} = 0.171 \, \text{mol} \][/tex]
4. Calculate the mass of AgI formed:
[tex]\[ \text{mass of AgI} = \text{moles of AgI} \times \text{molar mass of AgI} \][/tex]
The molar mass of AgI is the sum of the atomic masses of Ag (107.87 g/mol) and I (126.904 g/mol).
[tex]\[ \text{mass of AgI} = 0.171 \, \text{mol} \times (107.87 + 126.904) \, \text{g/mol} \][/tex]
[tex]\[ \text{mass of AgI} \approx 38.65 \, \text{g} \][/tex]
Therefore, the mass of AgI in the precipitate is approximately 38.65 grams (rounded to two significant figures) with the appropriate units.
Wavelength (nm)n2E (J) 404.7 435.8 546.1 579.0 Show your work (or send in a separate sheet with your work on it): Q1: A sodium vapor lamp is similar to a mercury vapor lamp. Sodium displays a single visible emission line at 589.3 nm. Why is it better to use a mercury vapor lamp for calibration purposes
Answer: Please see answer below
Explanation:
Mecury vapor lamp is better to use than Sodium vapor light, this is because because
---The Filaments of the lamp in sodium emit fast moving electrons, which causes valence electrons of the sodium atoms to excite to higher energy levels which when electrons after being excited, relax by emitting yellow light which concentrates on the the monochromatic bright yellow part of the visible spectrum which is about 580-590 or about (589nm) which will fall incident on the calibrations making it difficult to see
While
In Mercury vapor lamp, The emitted electrons from the filaments, after having been excited by high voltage, hit the mercury atoms but the excited electrons of mercury atoms relax and emits an ultraviolet uv invisible lights falling on the mecury vapour lamp to produce white light covering a wide range of (380-780 nm) which is visible that is why it is used for calibrations purposes in lightening applications.
Write a balanced half-reaction for the reduction of dichromate ion Cr2O−27 to chromium ion Cr+3 in acidic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
14 H⁺(aq) + Cr₂O₇²⁻(aq) + 6 e⁻ → 2 Cr⁺³ + 7 H₂O(l)
Explanation:
Let's consider the half-reaction for the reduction of dichromate ion Cr₂O₇²⁻ to chromium ion Cr⁺³ in acidic aqueous solution.
Cr₂O₇²⁻(aq) → Cr⁺³
Step 1: Perform the mass balance adding H₂O(l) and H⁺(aq) where appropriate.
14 H⁺(aq) + Cr₂O₇²⁻(aq) → 2 Cr⁺³ + 7 H₂O(l)
Step 2: Perform the charge balance adding electrons where appropriate.
14 H⁺(aq) + Cr₂O₇²⁻(aq) + 6 e⁻ → 2 Cr⁺³ + 7 H₂O(l)
when the iron(III) oxide reacts with hydrochloric acid, iron (III) chloride and water are formed. How many gras of iron (III) chloride are formed from 10.0g of iron (III) oxide
Answer:
10.1 g of FeCl₃ are formed by the reaction
Explanation:
First step is to determine the reaction where the reactants are Fe₂O₃ and HCl in order to produce FeCl₃ and H₂O.
Equation is: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O
We assume the acid is in excess, so the limiting reagent will be the oxide.
Let's work with mass:
1 mol of Fe₂O₃ is 159.7 g
2 mol of FeCl₃ is 162.2 g
So now we propose a rule of three:
159.7 g of oxide can produce 162.2 grams of chloride
Then, 10 g of oxide will produce (10 . 162.2) / 159.7 = 10.1 g of FeCl₃
Answer:
20.3 grams of FeCl3 will be formed
Explanation:
Step 1: Data given
iron(III) oxide = Fe2O3
hydrochloric acid = HCl
iron (III) chloride = FeCl3
water = H2O
Mass of Fe2O3 = 10.0 grams
Molar mass Fe2O3 = 159.69 g/mol
Step 2: The balanced equation
Fe2O3 +6HCl → 2FeCl3 + 3H2O
Step 3: Calculate moles Fe2O3
Moles Fe2O3 = mass Fe2O3 / molar mass Fe2O3
Moles Fe2O3 = 10.0 grams/ 159.69 g/mol
Moles Fe2O3 = 0.0626 moles
Step 4: Calculate moles of FeCl3
For 1 mol Fe2O3 we need 6 moles HCl to produce 2 moles FeCl3 and 3 moles H2O
For 0.0626 moles Fe2O3 we'll have 2*0.0626 = 0.1252 moles FeCl3
Step 5: Calculate mass FeCl3
Mass FeCl3 = moles FeCl3 * molar mass FeCl3
Mass FeCl3 = 0.1252 moles * 162.2 g/mol
Mass FeCl3 = 20.3 grams
20.3 grams of FeCl3 will be formed
A mixture of He
, N2
, and Ar
has a pressure of 13.6
atm at 28.0
°C. If the partial pressure of He
is 1831
torr and that of Ar
is 997
mm Hg, what is the partial pressure of N2
?
Gaseous ammonia (NH3) reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water. Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
The chemical equation is given as:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]
Explanation:
When gaseous ammonia reacts with gaseous oxygen it gives nitrogen monoxide gas and water vapors as product.
The chemical equation is given as:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]
According to reaction, 4 moles of ammonia reacts with 5 moles of oxygen gas to give 4 moles of nitrogen monoxide gas and 6 moles of water vapor.
To write the skeletal equation, begin by writing the chemical formula for each reactant and product.
Reactants:
1. Ammonia's chemical formula is given as NH3.
2. Gaseous oxygen exists as a diatomic molecule with the chemical formula O2.
Products:
1. Nitrogen monoxide has the chemical formula of NO.
2. The chemical formula for water is H2O.
Therefore, the skeletal equation is written as:
NH3(g)+O2(g)→NO(g)+H2O(g)
Now count the number of each atom on each side of the equation to determine if the equation is balanced.
Reactants
1N atom
3H atoms
2O atoms
Products
1N atom
2H atoms
2O atoms
Begin by balancing the number of hydrogen atoms by adding a coefficient of 2 to NH3 and a coefficient of 3 to H2O. Next, balance the number of nitrogen atoms by adding a coefficient of 2 to NO. Now there are two oxygen atoms on the reactant's side and five oxygen atoms on the product's side of the reaction. Since only whole number coefficients should be used, all coefficients need to be increased by a factor of two to balance the oxygen atoms. Thus the coefficient for NH3 is 4, the coefficient for H2O is 6, and the coefficient for NO is 4. Finally, balance the oxygen atoms by adding a coefficient of 5 to O2. The balanced equation is:
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
A variation of the acetamidomalonate synthesis can be used to synthesize serine. The process involves the following steps: Ethoxide ion deprotonates diethyl acetamidomalonate, forming enolate anion 1; Enolate anion 1 makes a nucleophilic attack on formaldehyde, forming tetrahedral intermediate 2; Protonation of the oxyanion forms alcohol 3; Acid hydrolysis yields dicarboxyamino alcohol 4; Decarboxylation leads to the final amino acid. Write out the mechanism on a separate sheet of paper, and then draw the structure of dicarboxyamino alcohol 4.
Answer:
See detailed mechanism in the image attached
Explanation:
The mechanism shown in detail below is the synthesis of serine in steps.
The first step is the attack of the ethoxide ion base on the diethyl acetamidomalonate substrate giving the enolate and formaldehyde.
The second step is the protonation of the oxyanion from (1) above to form an alcohol as shown.
Acid hydrolysis of the alcohol formed in (3) above yields a tetrahedral intermediate, a dicarboxyamino alcohol.
Decarboxylation of this dicarboxyamino alcohol yields serine, the final product as shown in the image attached.
A 1.31 mol sample of CO2 gas is confined in a 31.4 liter container at 19.0 �C.
If the volume of the gas sample is decreased to 15.7 L holding the temperature constant, the number of molecule-wall collisions per unit area per unit timewill
A. remain the same
B. not enough information to answer the question
C. increase
D. decrease
2. A 1.18 mol sample of CO2 gas is confined in a 27.8 liter container at 14.5 �C.
If the volume of the gas sample is increased to 55.7 L holding the temperature constant, the average kinetic energy will
A. remain the same
B. decrease
C. increase
D. not enough information to answer the question
3.A 0.855 mol sample of Xe gas is confined in a 20.5 liter container at 19.6 �C.
If the volume of the gas sample is decreased to 10.3 L, holding the temperature constant, the pressure will increase. Which of the following kinetic theory ideas apply?
Choose all that apply.
A. With less available volume, the molecules hit the walls of the container more often.
B. At lower volumes molecules have higher average speeds.
C. With higher average speeds, on average the molecules hit the walls of the container with more force.
D. For a given gas at constant temperature, the force per collision is constant. Some other factor must cause the pressure increase.
E. None of the Above
Ideal gas law is valid only for ideal gas not for vanderwaal gas. The equation used for ideal gas is PV=nRT. The number of molecule-wall collisions per unit area per unit time will remain the same.
What is ideal gas equation?Ideal gas equation is the mathematical expression that relates pressure volume and temperature.
Mathematically the relation between Pressure, volume and temperature can be given as
PV=nRT
where,
P = pressure of gas
V= volume of gas
n =number of moles of gas
T =temperature of gas
R = Gas constant = 0.0821 L.atm/K.mol
If the volume of the gas sample is decreased to 15.7 L holding the temperature constant, the number of molecule-wall collisions per unit area per unit time will remain the same.
Therefore, the number of molecule-wall collisions per unit area per unit time will remain the same.
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Balance the following unbalanced redox reaction (assume acidic solution if necessary): Cr2O72- + Cl- → Cl2 + Cr3+ Indicate the coefficient that will be used for Cl2 (g) in this reaction
Answer:
14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O
The coefficient that will be used for Cl₂ in this reaction is 3
Explanation:
We use the method of electron-ion to the balance.
We assume that the redox reaction is happening at acidic medium.
Cr₂O₇²⁻ + Cl⁻ → Cl₂ + Cr³⁺
Chloride is raising the oxidation state from -1 in the chloride, to 0 in the chloride dyatomic. This is the half reaction of oxidation
2Cl⁻ → Cl₂ + 2e⁻ Oxidation
In the dichromate anion, chromium acts with +6 in oxidation state, and we have 2 Cr, so the global charge of the element is +12. To change to Cr³⁺ it has release 3 electrons, but we have 2 Cr, so it finally released 6 e-. The oxidation state was decreased, so this is the reduction half reaction.
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O Reduction
As we have 7 O in the product side, we add 7 water, to the opposite place. In order to balance the H (protons) we, add the amount of them, in the opposite side, again.
(2Cl⁻ → Cl₂ + 2e⁻) ₓ3
(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) ₓ1
We multiply the half reactions, in order to remove the electrons and we sum, the equations:
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6Cl⁻ → 3Cl₂ + 6e⁻ + 2Cr³⁺ + 7H₂O
Now, that we have the same amount of electrons, they can cancelled, so the balanced redox reaction is:
14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O
To balance the given redox reaction, follow the steps: assign oxidation states, balance the elements undergoing oxidation or reduction, balance the elements not involved in redox, balance the charge, balance oxygen atoms, combine H2O molecules, and finally, balance the equation. The coefficient for Cl2 (g) is 3.
Explanation:To balance the redox reaction Cr2O72- + Cl- → Cl2 + Cr3+ in an acidic solution, we need to follow these steps:
Assign oxidation states to each element in the reaction. Chromium (Cr) in Cr2O72- is assigned an oxidation state of +6 on the left side and +3 on the right side. Chlorine (Cl) in Cl- has an oxidation state of -1 on the left side and 0 in Cl2 on the right side.Balance the elements that are undergoing oxidation or reduction first. In this case, Cr is being reduced from +6 to +3, so we need to add 3 electrons to the left side of the equation.Next, balance the elements that are not involved in redox reactions. In this case, Cl- stays the same on both sides.Finally, balance the charge by adding H+ ions to the side that needs it. In this case, we need to add 14 H+ ions to the left side.Balancing the oxygen atoms by adding H2O molecules. There are 7 oxygen atoms on the left side, so we add 7 H2O molecules to the right side.Combine the H2O molecules on the right side to form OH- ions, then cancel out the same ions present on both sides.Now, the equation is balanced in terms of atoms and charges. The coefficient for Cl2 (g) is 3.Learn more about Balancing redox reaction here:https://brainly.com/question/32431557
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A tank at is filled with of dinitrogen difluoride gas and of carbon dioxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits. dinitrogen difluoride molar fraction: partial pressure: carbon dioxide mole fraction: partial pressure: Total pressure in tank:
Answer:
For N₂F₂:
Molar fraction = 0.84
Partial pressure = 1.12 atm
For SF₄:
Molar fraction = 0.16
Partial pressure = 0.208 atm
Explanation:
It seems your question is missing the values required to solve the problem. However, an internet search showed me the following values for your question. If the values in your problem are different, your answer will be different as well, however the solving method will remain the same:
" A 5.00L tank at 0.7°C is filled with 16.5g of dinitrogen difluoride gas and 5.00g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. "
First we calculate the moles of each gas, using their molar mass:
16.5 g N₂F₂ ÷ 66 g/mol = 0.25 mol N₂F₂5.00 g SF₄ ÷ 108 g/mol = 0.0463 mol SF₄Total mol number = 0.25 + 0.0463 = 0.2963 mol
Mole Fraction N₂F₂ = 0.25/0.2963 = 0.84Mole Fraction SF₄ = 0.0463/0.2963 = 0.16Now we use PV=nRT to calculate the partial pressure of each gas:
P = ?
V = 5.00 L
T = 0.7 °C ⇒ 0.7 + 273.16 = 273.86 K
For N₂F₂:
P * 5.00 L = 0.25 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.86 KP = 1.12 atmFor SF₄:
P * 5.00 L = 0.0463 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.86 KP = 0.208 atm
Final answer:
To calculate the mole fraction, you divide the number of moles of a gas by the sum of the moles of all gases in the mixture. The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. The total pressure in the tank is the sum of the partial pressures of each gas.
Explanation:
The mole fraction of a gas is the ratio of the moles of that gas to the total moles of all gases in the mixture. To calculate the mole fraction of dinitrogen difluoride (N2F2), divide the moles of N2F2 by the sum of the moles of N2F2 and CO2:
Mole fraction of N2F2 = moles of N2F2 / (moles of N2F2 + moles of CO2)
To calculate the mole fraction of carbon dioxide (CO2), divide the moles of CO2 by the sum of the moles of N2F2 and CO2:
Mole fraction of CO2 = moles of CO2 / (moles of N2F2 + moles of CO2)
The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. To calculate the partial pressure of N2F2, multiply its mole fraction by the total pressure in the tank:
Partial pressure of N2F2 = mole fraction of N2F2 * total pressure
To calculate the partial pressure of CO2, multiply its mole fraction by the total pressure in the tank:
Partial pressure of CO2 = mole fraction of CO2 * total pressure
The total pressure in the tank is the sum of the partial pressures of N2F2 and CO2:
Total pressure = partial pressure of N2F2 + partial pressure of CO2
N2 +3H2 → 2NH3 What volume of ammonia at STP is produced if 30.0 g of nitrogen gas is reacted with an excess of hydrogen gas?
Answer:
V= 48L
Explanation:
Moles of N2 = m/M = 30/28= 1.07moles
From the equation of reaction
N2 +3H2 → 2NH3
1mole of N2 produces 2mole of NH3
Hence
1.07moles will produce 1.07×2= 2.14 moles
At STP, 1mole occupy 22.4L
Hence volume of N2 produced = 2.14×22.4= 48 L
The price of gold on April 15,2000 was $282/t.oz. How much did 100.0cm^3 of gold cost that day if 1.00 t.oz=28.4 grams?
Answer:
price ($) Au = $ 19183.94
Explanation:
april 15,2000:
∴ price Au = $ 282/t.oz
∴ 1.00 t.oz = 28.4 g
∴ V Au = 100.0 cm³ ⇒ price ($) = ?
∴ δ Au = 19.32 g/cm³
⇒ mass Au = (100.0 cm³)*(19.32 g/cm³)
⇒ m Au = 1932 g
⇒ price ($) = (1932 g Au)*(1.00 t.oz/28.4 g Au)*( $ 282/t.oz)
⇒ price ($) = $ 19183.94
To calculate the cost of 100.0cm3 of gold on April 15, 2000, convert the volume to grams, then to troy ounces, and multiply by the price per troy ounce.
Explanation:To calculate the cost of 100.0cm3 of gold on April 15, 2000, we need to convert the volume of gold to grams and then to troy ounces. First, convert 100.0cm3 to grams by multiplying it by the density of gold (19.3 g/cm3). This gives us 1930 grams. Next, convert grams to troy ounces by dividing by the conversion factor of 28.4 grams per troy ounce. This gives us approximately 67.96 troy ounces. Finally, multiply the number of troy ounces by the price per troy ounce to find the cost. Therefore, 100.0cm3 of gold on April 15, 2000 would have cost $19,191.12.
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When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of potassium bromide are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for potassium bromide in . Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.
Answer:
The factor is 2
Explanation:
Van't Hoff factor is defined as the ratio between the species of a solute before the addition to the solvent and particles produced when the substance is dissolved. It is used, principally, in colligative properties.
Before solution, potassium bromide, KBr, has just one specie, that is, KBr. When KBr is dissolved (As a salt):
KBr(aq) → K⁺(aq) + Br⁻(aq)
There are produced two species, K⁺ and Br⁻. By definition of Van't Hoff factor, for this salt, the factor is 2.
You are given mixture made of 290 grams of water and 14.2 grams of salt. Determine the % by mass of salt in the salt solution.
Answer:
Solution is 4.67% by mass of salt
Explanation:
% by mass is the concentration that defines the mass of solute in 100g of solution.
In this case we have to find out the mass of solution with the data given:
Mass of solution = Mass of solute + Mass of solvent
Solute: Salt → 14.2 g
Solvent: Water → 290 g
Solution's mass = 14.2 g + 290g = 304.2 g
% by mass = (mass of solute / mass of solution) . 100
(14.2 g / 304.2g) . 100 = 4.67 %
(h) The student made observations related to the contents of the Erlenmeyer flask during the titration. Identify an observation that could have led the student to conclude that a chemical change took place during the titration.
Answer:
Explanation:
During titration indicators are often used to identify chemical changes between reacting species.
For colorless solutions in which no noticeable changes can easily be seen, indicators are the best bet. Most titration processes involves a combination of acids and bases to an end point.
Indicators are substances whose color changes to signal the end of an acid-base reaction. Examples are methyl orange, methyl red, phenolphthalein, litmus, cresol red, cresol green, alizarin R3, bromothymol blue and congo red.
Most of these indicators have various colors when chemical changes occur.
Also, there are heat changes that accompanies most of these reactions. These are also indicators of chemical changes.
Final answer:
An observation indicative of a chemical change during titration could include temperature change, light emission, unexpected color change, or the formation of bubbles signaling the production of gas, any of which suggest that a new substance has been formed.
Explanation:
An observation that could have led the student to conclude that a chemical change took place during the titration includes the following:A temperature change such as the solution in the Erlenmeyer flask becoming warmer or cooler, indicating an exothermic or endothermic reaction.The emission of light which could suggest an energy release during a chemical reaction.An unexpected color change indicating the formation of a new substance.The formation of bubbles not caused by boiling, suggesting the production of a gas during the reaction.For instance, if the student added an indicator to the acid and after titrating with a base, noticed a color change, this would be evidence of a chemical change. Similarly, if the student observed the solution fizzing but it was not reaching boiling temperature, it could indicate the formation of a gaseous product.
Problem PageQuestion Suppose an iron atom in the oxidation state formed a complex with three hydroxide anions and three water molecules. Write the chemical formula of this complex.
The Question is incomplete here is the complete question " Suppose an iron atom in the oxidation state +3 formed a complex with three hydroxide anions and three water molecules. Write the chemical formula of this complex.
Answer:
{Fe(OH)3(H20)3}
Explanation:
Oxidation state is the electron gained or lost by and atom, So if iron is in +3 state in formula it must have lost three electron.
We know that OH posses the oxidation state of -1 and water have zero oxidation state. SO, Let's take iron equal to y and find its oxidation state in the formula
y + 3 ( - 1 ) + 3 ( 0 ) = 0
y - 3 + 0 = 0
y-3=0
y= + 3
Hence it's proved that iron has +3 oxygen state.
Calculate the number of moles of sugar in 800 ml of a 2.6 M solution
Answer:
2.08moles
Explanation:
Given parameters:
Volume of solution = 800mL = 0.8L
Molarity = 2.6M
Unknown:
Number of moles of sugar in the solution = ?
Solution:
We need to understand the relationship between number of moles, volume and molarity of a solution.
The molarity of a solution is the number of moles of solute of a substance dissolved in a given solvent.
Molarity = [tex]\frac{number of moles }{volume}[/tex]
Number of moles of solute = molarity x volume of solution
Now input the parameters and solve;
Number of mole of solute = 2.6 x 0.8 = 2.08moles
The number of moles of sugar is 2.08moles
Time and concentration data were collected for the reaction A ⟶ products A⟶products t (s) [A] (M) 0 0.52 0.52 20 0.43 0.43 40 0.35 0.35 60 0.29 0.29 80 0.23 0.23 100 0.19 0.19 The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at t = 40 s. t=40 s. Approximate the instantaneous rate of this reaction at time t = 40 s.
Answer:
instantaneous rate at 40 s= 0.0035 M /s.
Explanation:
Instantaneous rate at 40 s is the slope of the line (tangent to the curve)
= Δp/Δt
From, the straight orange line
ΔP = (0.48 - 0.16) M.
Δt = (92 -0) s
Now, instantaneous rate at 40 s
= 0.48 - 0.16/92 - 0
instantaneous rate at 40 s= 0.0035 M /s.
Answer:
0.0035
Explanation:
check the picture
Which of the following aqueous solutions are good buffer systems? . 0.37 M hydrocyanic acid + 0.24 M sodium cyanide 0.14 M sodium nitrite + 0.27 M nitrous acid 0.21 M hypochlorous acid + 0.13 M potassium hypochlorite 0.20 M nitric acid + 0.22 M potassium nitrate 0.32 M ammonium bromide + 0.31 M ammonia
Answer : The good buffer systems are, (a), (b), (c) and (e)
Explanation :
Buffer : It is a solution that prevent any changes in the pH of the solution on the addition of an acidic and basic components.
Or, it is a solution that maintain the pH of the solution by adding the small amount of acid or a base.
There are two types of buffer which are acidic buffer and basic buffer.
Acidic buffer : It is the solution that have the pH less than 7 and it contains weak acid and its salt. For example : Acetic acid (weak acid) and sodium acetate (salt).
Basic buffer : It is the solution that have the pH more than 7 and it contains weak base and its salt. For example : Ammonia (weak base) and ammonium chloride (salt).
The conditions for a good buffer system is:
(1) a weak acid and its conjugate base.
(2) a weak base and its conjugate acid.
(a) 0.37 M hydrocyanic acid + 0.24 M sodium cyanide
It is a combination of weak acid and its conjugate base. So, it is a good buffer.
(b) 0.14 M sodium nitrite + 0.27 M nitrous acid
It is a combination of weak acid and its conjugate base. So, it is a good buffer.
(c) 0.21 M hypochlorous acid + 0.13 M potassium hypochlorite
It is a combination of weak acid and its conjugate base. So, it is a good buffer.
(d) 0.20 M nitric acid + 0.22 M potassium nitrate
It is a combination of strong acid. So, it will not form buffer solution.
(e) 0.32 M ammonium bromide + 0.31 M ammonia
It is a combination of weak base and its conjugate acid. So, it is a good buffer.
Hence, the good buffer systems are, (a), (b), (c) and (e)
Good buffer systems are those that consist of a weak acid and its conjugate base or a weak base and its conjugate acid. The given combinations that are good buffer systems include hydrocyanic acid and sodium cyanide, sodium nitrite and nitrous acid, hypochlorous acid and potassium hypochlorite, and ammonium bromide and ammonia.
Explanation:The aqueous solutions that are good buffer systems from the provided options are:
0.37 M hydrocyanic acid + 0.24 M sodium cyanide0.14 M sodium nitrite + 0.27 M nitrous acid0.21 M hypochlorous acid + 0.13 M potassium hypochlorite0.32 M ammonium bromide + 0.31 M ammoniaA good buffer system consists of a weak acid and its conjugate base or a weak base and its conjugate acid. These pairs are capable of neutralizing small amounts of added acid or base, thus maintaining a relatively stable pH in the solution. In contrast, solutions like 0.20 M nitric acid + 0.22 M potassium nitrate do not form a buffer because nitric acid is a strong acid and does not create a conjugate weak acid-base pair needed for buffering.
Recall from Chapter 12.6 that chlorofluorocarbons (CFCs) catalyze the decomposition of ozone in the stratosphere. Infrared light is absorbed by CFCs while visible light is transmitted. Ultraviolet light can break the bonds in these molecules. Answer the following questions about CFCs. A) Draw the three-dimensional structure of trichlorofluoromethane, CC13F. B) How will CFCs interact with solar radiation in the troposphere? C) How will CFCs interact with solar radiation in the stratosphere?
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Magnesium metal reacts with gaseous oxygen in a combination reaction. Write a balanced equation to describe this reaction. Include states of matter in your answer. Click in the answer box to open the symbol palette.
Answer: 2Mg (s) + O2 (g) ----> 2MgO (s)
Explanation:
Lets write the equation of the reaction.
Mg(s) + O2 (g) ----> MgO (s)
Counting the number of atoms for each element we have:
Left hand side: Mg =1 O = 2
Right hand side : Mg = 1, O =2
To balance this equation input "2" as coefficient for "MgO" on the right hand side and "2", as coefficient for "Mg" on the left hand side of the equation. Hence our balanced equation will be
2Mg (s) + O2 (g) -----> 2MgO (s)
Final answer:
The balanced chemical equation for the combination reaction between magnesium metal and gaseous oxygen to form magnesium oxide is: 2Mg (s) + O₂(g) → 2MgO (s). This reaction follows the conservation of mass.
Explanation:
The reaction between magnesium metal and gaseous oxygen is a combination reaction in which magnesium is oxidized. When magnesium (Mg) combines with oxygen (O₂), it forms magnesium oxide (MgO). The equation representing this exothermic reaction is written as:
2Mg (s) + O₂(g) → 2MgO (s)
This reaction adheres to the law of conservation of mass, meaning the total mass of the reactants equals the total mass of the products. During the reaction, magnesium atoms lose electrons (are oxidized) and the oxygen molecule gains electrons.
1) Draw the structure of only the Zaitsev product that would be obtained from the dehydration of 1-methylcyclohexanol. 2) Write out a detailed mechanism of the acid catalyzed dehydration of cyclohexanol. Use only hydronium as your acid source. 3) With respect to GC, what effect
Question:
Question 3 is not complete. here is the complete question;
3. With respect to GC, what effect would raising the column temperature have on the retention time?
Answer:
See the attached file for the structures
Explanation:
Find attached of question 1 and 2.
3. Raising the column temperature enhances the internal energy of the analyse. Therefore, it will escape from the liquid phase faster and predominantly be present in the vapour phase. On increasing temperature, retention time will be DECREASED, because analyte would not be able to interact with stationary phase efficiently.
The Zaitsev product from the dehydration of 1-methylcyclohexanol would have the double bond between the carbon connected to the OH group and the adjacent more substituted carbon. The mechanism of acid-catalyzed dehydration of cyclohexanol involves protonation, departure of a water molecule forming a carbocation, and subsequent loss of a proton which forms a pi bond. Changes in temperature and pressure in Gas Chromatography (GC) can impact analysis outcomes.
Explanation:The basis of your question involves the Zaitsev product from the dehydration of 1-methylcyclohexanol, the mechanism of the acid-catalyzed dehydration of cyclohexanol using hydronium, and effects in relation to Gas Chromatography (GC).
1) When dehydrating 1-methylcyclohexanol, the Zaitsev rule tells us that the most substituted alkene will be the major product. Because you are dehydrating 1-methylcyclohexanol, the Zaitsev product will have the double bond between the carbon connected to the OH group and the adjacent carbon which is more substituted.
2) The second part of the question relates to the acid-catalyzed dehydration of cyclohexanol.
Firstly, in the presence of acid, the hydroxyl group on cyclohexanol is protonated forming a good leaving group. The water molecule leaves creating a carbocation. A proton is then lost resulting in the pi bond of the cyclohexene. Water acts as a base to receive this proton.
3) With respect to Gas Chromatography (GC), changes in temperature and pressure can affect the results. Higher temperatures can increase resolution but at the risk of over-separation.
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Calculate the final concentration of ONPG (in mM) if you add 1.42 mL of 3.3 mM ONPG and dilute to a final volume of 10 mL with PBS buffer. Report your final answer to two places after the decimal.
Answer : The final concentration of ONGP is, 0.47 mM
Explanation :
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume
[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume
We are given:
[tex]M_1=3.3mM\\V_1=1.42mL\\M_2=?\\V_2=10mL[/tex]
Now put all the given values in above equation, we get:
[tex]3.3mM\times 1.42mL=M_2\times 10mL\\\\M_2=0.47mM[/tex]
Hence, the final concentration of ONGP is, 0.47 mM
How many mL of 0.50 M NaOH solution are required to completely titrate 15.0 mL of 0.20 M HNO3 solution?
Answer:6.0 ML
Explanation:
A 27g pancake (61 Calories) is burned in a calorimeter that contains 2.500 kg of water. How much does the temperature of the water increase if all of the energy from the pancake is transferred into the water?
If all of the energy from the pancake is transferred into the water, the temperature of the water will increase by 0.02 °C
Data obtained from the questionFrom the question given above, the following data were untainted:
Mass of pancake = 27 gEnergy of package released (Q) = 61 caloriesMass of water (M) = 2.5 Kg = 2.5 × 1000 = 250 g Specific heat capacity of water (C) = 1 cal/gºC Change in temperature of water (ΔT) =? How to determine the increase in temperatureThe increase in temperature can be obtained as follow:
Q = MCΔT
Divide both side by MC
ΔT = Q / MC
ΔT = 61 / (2500 × 1)
ΔT = 61 / 2500
ΔT = 0.02 °C
Thus, the temperature of the water will increase by 0.02 °C
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What is the molarity of a solution containing 56 grams of solute in 959 mL of solution? (molar mass of solute is 26 g/mol)
Answer:
Molarity = 2.25M
Explanation:
n= m/M= 56/26=2.15mol, V= 959ml= 0.959L
n=C×V
2.15= C× 0.959
Simplify
C= 2.25M
Answer:
2.2M
Explanation:
Suppose the surface-catalyzed hydrogenation reaction of an unsaturated hydrocarbon has a rate constant of 0.725 M/min. The reaction is observed to follow zero-order kinetics. If the initial concentration of the hydrocarbon is 5.90 M, what is the half-life of the reaction in seconds? *Please report 3 significant figures. Numbers only, no unit. No scientific notation.
Answer : The half-life of the reaction in seconds is, 244
Explanation :
The expression used for zero order reaction is:
[tex]t_{1/2}=\frac{[A_o]}{2k}[/tex]
where,
[tex]t_{1/2}[/tex] = half-life of the reaction = ?
[tex][A_o][/tex] = initial concentration = 5.90 M
k = rate constant = 0.725 M/min
Now put all the given values in the above formula, we get:
[tex]t_{1/2}=\frac{5.90}{2\times 0.725}[/tex]
[tex]t_{1/2}=4.069min=244.14s\approx 244s[/tex]
conversion used : (1 min = 60 s)
Thus, the half-life of the reaction in seconds is, 244
Consider the volumes of benzaldehyde and acetone that you used for your scaled-down version of the lab (as described on the Aldol Condensation page and in the Aldol Lab quiz), and consider how these reactants are added to the reaction mixture. There is a potential problem associated with the preparation and addition of the benzaldehyde/acetone mixture, which would be exacerbated by the scaling down of the reaction. What is this problem, and why would this become a bigger problem at smaller scale
Answer:
Aldol condensation is possible only when their is alpha Hydrogen atom is present. It ia present only in the acetophenone and not in benzaldehyde.
Explanation:
Consider the following half-reactions: Half-reaction E° (V) I2(s) + 2e- 2I-(aq) 0.535V 2H+(aq) + 2e- H2(g) 0.000V Zn2+(aq) + 2e- Zn(s) -0.763V (1) The weakest oxidizing agent is: Zn2+(aq) enter formula (2) The strongest reducing agent is: Zn(s) (3) The strongest oxidizing agent is: I2(s) (4) The weakest reducing agent is: I-(aq) (5) Will I-(aq) reduce Zn2+(aq) to Zn(s)? (6) Which species can be reduced by H2(g)? If none, leave box blank.
Answer:
The weakest oxidizing agent is Zn^2+(aq)
The strongest reducing agent is Zn(s)
The strongest oxidizing agent is I2(s)
The weakest reducing agent is I^-(aq)
I^- cannot reduce Zn^2+ to Zn(s)
I2(s) can be reduced by hydrogen gas
Explanation:
In looking at oxidizing and reducing agents, our primary guide is the reduction potentials of each specie. The more negative the reduction potential of a specie, the better its function as a reducing agent. Zn has a very negative reduction potential hence it a very good reducing agent. Similarly, iodine has a very positive reduction potential hence it is a good oxidizing agent.
Only a specie having a more negative reduction potential than zinc can reduce it in aqueous solution. Similarly, the reaction potential of hydrogen is less than that of iodine hence hydrogen gas can reduce iodine.
Final answer:
In electrochemistry, the strongest reducing agent is Zn(s) and the weakest oxidizing agent is Zn2+(aq). I2(s) is the strongest oxidizing agent, while I-(aq) is the weakest reducing agent. I-(aq) cannot reduce Zn2+(aq) to Zn(s), and none of the species provided can be reduced by H2(g).
Explanation:
To determine the oxidizing and reducing agents, we refer to their standard electrode potentials (E°). A strong oxidizing agent has a higher positive E° value, indicating a greater tendency to gain electrons and be reduced. Conversely, a strong reducing agent has a lower (more negative) E° value, reflecting a higher tendency to lose electrons and be oxidized.
The weakest oxidizing agent is Zn²+(aq) because it has the most negative E° value (-0.763V), meaning it is the least likely to gain electrons. The strongest reducing agent is Zn(s) because zinc in its solid state is more willing to be oxidized (lose electrons) as indicated by its half-reaction (Zn(s) → Zn²+(aq) + 2e⁻).
Similarly, the strongest oxidizing agent is I₂(s) due to its higher E° value (0.535V), which demonstrates its greater ability to take up electrons. The weakest reducing agent in this set is I⁻(aq) because it is derived from I₂(s), which is a strong oxidizer, thus its conjugate, I⁻, would be the weakest reducer.
Regarding whether I⁻(aq) can reduce Zn²+(aq) to Zn(s), the answer is no, because I⁻(aq) is the weakest reducing agent and Zn²+(aq) is not a strong oxidizer. Lastly, H₂(g) has an E° value of 0.000V, making it neutral in this context, and therefore it will not reduce any of the species provided in the question.
For Kinetic Trial 2, Alicia was distracted when the color change occurred but decided to record the time lapse read from her watch. Will this distraction cause an increase or decrease in he slop of the log (rate) vs log [I-]0? Explain.
The distraction during Kinetic Trial 2 will not affect the slope of the log (rate) vs log [I-]0. The relationship between the log of the rate and the log of the initial concentration of I- is determined by the reaction kinetics and is not influenced by external factors like distractions.
Explanation:The distraction that Alicia experienced when the color change occurred during Kinetic Trial 2 will not affect the slope of the log (rate) vs log [I-]0.
This is because the distraction only affected the timing of the color change, not the actual reaction rate.
The relationship between the log of the rate and the log of the initial concentration of I- is determined by the reaction kinetics and is not influenced by external factors like distractions.
Thermal decomposition of 5.0 metric tons of limestone to lime and carbon dioxide requires 9.0 x 106 kJ of heat. Convert this energy to joules A. 9.0 x 108 J B. 9.0 x 104 J C. 9.0 x 103 J D. 9.0 x 109 J E. None of these is within 5% of the correct answer
Answer : The correct option is, (D) [tex]9.0\times 10^9J[/tex]
Explanation :
As we are given that the energy require for decomposition is, [tex]9.0\times 10^6kJ[/tex].
Now we have to calculate the energy in joules.
Conversion used :
1 kJ = 1000 J
As, 1 kJ of energy = 1000 J
So, [tex]9.0\times 10^6kJ[/tex] of energy = [tex]\frac{9.0\times 10^6kJ}{1kJ}\times 1000J[/tex]
= [tex]9.0\times 10^9J[/tex]
Therefore, the energy in joules is, [tex]9.0\times 10^9J[/tex]
Thermal decomposition of 5.0 metric tons of limestone to lime and carbon dioxide requires 9.0 × 10⁹ Joules of heat.
What is energy?Enegy is the quantitative property which is used by any system to perform any work.
Chemical reactions generally involves energy in the form of heat energy and given amount of energy is 9.0 × 10⁶ kJ.
We know that:
1 kJ = 1000 J
So, 9.0 × 10⁶ kJ = 9.0 × 10⁶ kJ × 1000
9.0 × 10⁶ kJ = 9.0 × 10⁹ J
Hence, option (D) is correct i.e. 9.0 × 10⁹ J.
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