Answer:
a) pH = 9.8
b) 2.2 x 10⁻⁸ = [ H₂C₃H₂O₄ ]
c) decrease
Explanation:
The equilibriums involved in this question are:
C₃H₂O₄²⁻ + H₂O ⇄ HC₃H₂O₄⁻ + OH⁻ (1) Kb₁ =[HC₃H₂O₄⁻][OH⁻]/[C₃H₂O₄²⁻ ]
HC₃H₂O₄⁻ + H₂O ⇄ H₂C₃H₂O₄ +OH⁻ (2) Kb₂=[ H₂C₃H₂O₄]/[OH⁻]/[HC₃H₂O₄⁻]
The kas for malonic acid, H₂C₃H₂O₄, from reference tables are:
Ka (H₂C₃H₂O₄ )= 1.4 x 10⁻³
Ka ( HC₃H₂O₄⁻ ) = 2.0 x 10⁻⁶
a) We can calculate the Kbs for the conjugate bases of the weak malonic acid from Kw = Ka x Kb
Kb (C₃H₂O₄²⁻) = 10⁻¹⁴ / 2.0 x 10⁻⁶ =5.0 x 10⁻⁹
Kb (HC₃H₂O₄⁻)= 10⁻¹⁴ / 1.0 x 10⁻³ = 7.1 x 10⁻¹²
Given the magnitudes of the Kbs ( Kb₂ is approximately 1000 times Kb1 ) , to calculate pOh we can neglect the contribution from (2). We then treat this problem as any equilibrium:
[K₂C₃H₂O₄] = 25 g/180.2 g/mol / 0.455 L = 0.30 M
Conc C₃H₂O₄²⁻ + H₂O ⇄ HC₃H₂O₄⁻ + 0H⁻
I 0.30 0 0 0
C -x +x +x
E 0.30 - x x x
Kb (C₃H₂O₄²⁻) = [ HC₃H₂O₄⁻ ] [OH⁻]/ [ C₃H₂O₄²⁻] = x² / 0.30 - x ≅ x² /0.30
x² /.030 = 5.0 x 10⁻⁹ ⇒ x = √(0.30 x 5.0 x 10⁻⁹ ) = 7.1 x 10⁻⁵ = [ÒH⁻]
(Verifying our approximation was good 7.1 x 10⁻⁵ / 0.30 = 2.4 x 10⁻⁴ so our approximation checks)
pOH = -log 7.1 x 10⁻⁵ = 4.2
pH = 14 -4.2 = 9.8
b) To answer this part we take equilibrium (2 ) and set up our usual ICE table to solve for the concentration of malonic acid:
Conc (M) HC₃H₂O₄⁻ + H₂O ⇄ H₂C₃H₂O₄ + OH⁻ (2)
I 7.1 x 10⁻⁵ 0 0
C -x +x +x
E 7.1 x 10⁻⁵ -x x x
7.1 x 10⁻⁵ - x ≅ 7.1 x 10⁻⁵
[ H₂C₃H₂O₄ ] [OH⁻] / [HC₃H₂O₄⁻] = Kb₂ = 7.1 x 10⁻¹² = x² / 7.1 x 10⁻⁵
x = √(7.1 x 10⁻¹² x 7.1 x 10⁻⁵) = 2.2 x 10⁻⁸ = [ H₂C₃H₂O₄ ]Again our approximation checks since [HC₃H₂O₄⁻] is almost 1000 times [ H₂C₃H₂O₄ ]
c) From eqn (1) :
C₃H₂O₄²⁻ + H₂O ⇄ HC₃H₂O₄⁻ + OH⁻
The salt K₂C₃H₂O₄ will react completely with the added acid, thereby decreasing the C₃H₂O₄²⁻ concentration, and according to Le Chateliers principle the system will shift to the left and the OH⁻ at equilibrium will decrease ( as also does [HC₃H₂O₄⁻] ) therefore the pOH will increase and the pH will decrease ( less OH⁻ higher pOH, smaller pH )
Final answer:
Calculate the pH of the solution, determine the concentration of malonic acid, and explain the impact of adding HCl on the concentration of malonic acid in the solution.
Explanation:
a) Calculate the pH of the solution:
Calculate the molarity of the potassium malonate solution.
Use the dissociation of K2C3H2O4 and the autoionization of water to find the concentration of OH- ions.
Convert the OH- concentration to pH using the formula pH = 14 - pOH.
b) Calculate the concentration of malonic acid:
Set up an equilibrium expression for the dissociation of malonic acid.
Use the pH calculated in part (a) to determine the concentration of malonic acid.
c) Impact of adding HCl:
The addition of HCl will shift the equilibrium towards the formation of malonic acid, leading to an increase in its concentration.
A solution is made by mixing 30.0 mL of 0.150 M compound A with 25.0 mL of 0.200 M compound B. At equilibrium, the concentration of C is 0.0454 M. Calculate the equilibrium constant, K, for this reaction.
Final answer:
To calculate the equilibrium constant, K, for the reaction A(aq) + 2B(aq) ⇒ 2C(aq), we can use the concentrations of A, B, and C at equilibrium. The equilibrium constant (K) for this reaction is approximately 5.07 x 10^-3.
Explanation:
To calculate the equilibrium constant, K, for the reaction A(aq) + 2B(aq) ⇒ 2C(aq), we need to use the concentrations of A, B, and C at equilibrium. Given that the equilibrium concentration of C is 0.0454 M, and the initial concentrations of A and B are 0.150 M and 0.200 M respectively, we can set up an expression for K.
The mathematical expression for the equilibrium constant, Kc, is given by:
Kc = [C]² / ([A] * [B]²)
Substituting the given equilibrium concentration and initial concentrations into the expression, we get:
Kc = (0.0454)² / (0.150 * (0.200)²) = 5.06666667 x 10^-3
Therefore, the equilibrium constant (K) for this reaction is approximately 5.07 x 10^-3.
Calculate the mass of butane needed to produce 59.8 g of carbon dioxide. Express your answer to three digits and include the appropriate units.
Answer:
We need 19.8 grams of butane
Explanation:
Step 1: Data given
Mass of CO2 produced = 59.8 grams
Molar mass CO2 = 44.01 g/mol
Molar mass butane = 58.12 g/mol
Step 2: The balanced equation
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
Step 3: Caclulate moles CO2
Moles CO2 = mass CO2 / Molar mass CO2
Moles CO2 = 59.8 grams / 44.01 g/mol
Moles CO2 = 1.36 moles
Step 4: Calculate moles butane
For 2 moles butane we need 13 moles O2 to produce 8 moles CO2 and 10 moles H2O
For 1.36 moles CO2 we need 1.36/4 = 0.34 moles butane
Step 5: Calculate mass butane
Mass butane = moles butane * molar mass butane
Mass butane = 0.34 moles * 58.12 g/mol
Mass butane = 19.76 grams ≈ 19.8 grams
We need 19.8 grams of butane
For a reaction at equilibrium, which of the following is the best way to think about it?
a. forward rate = reverse rate
b. The concentration of intermediates becomes negligible.
c. The concentration of reactants becomes negligible.
d. The concentration of reactants = concentration of products.
e. The concentration of reactants and products are not related.
Answer: Option A. forward rate = reverse rate
Explanation: for a system to be in dynamic equilibrium, it means that the rate of formation of product is the same as the rate of formation of reactant.
Collision theory states that the rate of a reaction depends on the number of collisions between molecules or ions, the average energy of the collisions, and their effectiveness. Does the general effect of concentration on reaction rate support the collision theory?
Explanation:
According to the collision theory the rate of chemical reaction directly proportional to number of reaction between the reactant molecules. More the number of collision between the reactant molecules, more would be the changes of reaction to occur that reactant converting into product.
Moreover, with an increase in concentration of reactant molecules, the chances of collision between the molecules increase and hence the rate of reaction also increase. Thus, we can say higher the concentration higher will be the rate of reaction.
The quantity of particles increases as that the concentration of those same reacting species increases. Whenever there are more particles, there seems to be a greater chance that they will collide, and therefore more collisions result in an improvement or enhancement throughout the response rate.
Thus the above response about the theory is correct.
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When 10.51 g CaCO3 reacts with excess hydrochloric acid, as below, 3.29 g of CO2 is produced.
CaCO3(s) + HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)
What is the percent yield of CO2?
Answer:
71.2 %
Explanation:
This is a problem we can solve by making use of the stoichiometry of the balanced ( very important !) chemical reaction:
CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)
Now determine the moles of CaCO3 and CO2 and perform the calculations
# mol CaCO3 = 10.51 g / 100.09 g/mol = 0.105 mol
From the stoichiometry of the reaction, we know 1 mol CO2 is produced per mol of CaCO3, thus whe should expect
0.105 mol CaCO3 x 1 mol CO2/ mol CaCO3 = 0.105 mol
lets compare this theoretical value with the one obtained:
mol CO2 obtained = 3.29 g / 44 g/mol = 0.075
Therefore our yield is
(0.075mol / 0.105 mol) x 100 = 71.2 %
A mass of 34.05 g of H2O(s) at 273 K is dropped into 185 g of H2O(l) at 310. K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equi- librium has been reached. Assume that CP, m for H2O is con- stant at its values for 298 K throughout the temperature range of interest.
Answer:
The temperature of the system once equilibrium is reached, is 292 Kelvin
Explanation:
Step 1: Data given
Mass of H2O = 34.05 grams
⇒ temperature = 273 K
Mass of H2O at 310 K = 185 grams
Pressure = 1 bar = 0.9869 atm
Step 2: Calculate the final temperature
n(ice)*ΔH(ice fusion) + n(ice)*CP(H2O)(Tfinal- Ti,ice) + n(H20)*CP(H2O)*(Tfinal-Ti,H2O) = 0
Tfinal = [n(ice)*CP(ice)*Ti(ice) + n(H2O)*CP(H2O)*Ti(H20) - n(ice)*ΔH(ice fusion)] / [n(ice)*CP(ice) +n(H2O)*CP(H2O)]
⇒ with n(ice) = moles of ice = 34.05 grams / 18.02 g/mol = 1.890 moles
⇒ with CP(ice) = 75.3 J/K*mol
⇒ with Ti(ice) = the initial temperature of ice = 273 K
⇒ with n(H2O) = the moles of water = 185.0 grams / 18.02 g/mol = 10.27 moles
⇒ with CP(H2O) = CP(ice) = 75.3 J/K*mol
⇒ with Ti(H2O) = the initial temperature of the water = 310 K
⇒ with ΔH(ice, fusion) = 6010 J/mol
Tfinal = [1.890 moles * 75.3 J/K*mol * 273 + 10.27 mol * 75.3 J/K*mol * 310 K - 1.890 moles * 6010 J/mol] / [1.890 moles *75.3J/k*mol + 10.27 mol * 75.3 J/K*mol]
38852.541 + 239732.61 - 11358.9 = 267226.251
Tfinal= 291.8 ≈ 292 Kelvin
The temperature of the system once equilibrium is reached, is 292 Kelvin
We must use the principle of conservation of energy to equate the heat gained by the ice to the heat lost by the water, in a given equation, to calculate the final equilibrium temperature.
Explanation:The topic here is thermodynamics, specifically calculating the final equilibrium temperature when two substances are mixed. Given the information, we can apply the principle of conservation of energy, which in this context is the heat gained by one substance is equal to the heat lost by the other. In this case, the heat gained by the ice (H2O(s)) as it melts and increases in temperature is equal to the heat lost by the water (H2O(l)). Therefore, we have the equation 34.05 g * 1 kcal/kg * K *(T - 273 K) + 34.05 g * 80 Cal/g = 185 g * 1 kcal/kg*K *(310 K - T) where T is the final temperature to be solved.
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The compound zoapatanol was isolated from the leaves of a Mexican plant.
Classify each oxygen in zoapatanol according to the functional group to which it belongs. If an oxygen is part of an alcohol, classify the alcohol as primary, secondary, or tertiary.
Answer:
The structure is shown below!
1. Ketone
2. Secondary alcohol
3. Ether
4. Primary alcohol
Explanation:
The molecule of zoapatanol is shown below, each oxygen is marked by a number. The functional group represents the organic function of the substance. Some substances can have more than one function.
The oxygen is presented in several organic compounds, such as alcohol, aldehyde, ketone, ether, ester, carboxyl acid, and others. The structure of each is different, so an alcohol has the oxygen bonded to a carbon and to a hydrogen (called hydroxyl); an aldehyde has the oxygen bonded to a terminal carbon; a ketone has the oxygen bonded to nonterminal carbon; the ether has an oxygen between carbons; the ester has an oxygen between carbons, and one of these carbons has a double bond with other oxygen, and a carboxyl acid has the terminal carbon bonded to an oxygen and to a hydroxyl.
If the hydroxyl is bonded to a primary carbon (which is bonded to only one carbon), then it's a primary alcohol, if it's bonded to a secondary carbon (which is bonded to 2 carbons), then it's a secondary alcohol, and if it's bonded to a tertiary carbon (which is bonded to 3 carbons), then it's a tertiary alcohol.
Analyzing the structure, the oxygens are:
1. Ketone
2. Secondary alcohol
3. Ether
4. Primary alcohol
6. What is the electric potential energy of a +3.0 μC charge placed at corner A?
The question is incomplete. The complete question is stated below:
Two point charges are held at the corners of a rectangle as shown in the figure. The lengths of sides of the rectangle are 0.050 m and 0.150 m. Assume that the electric potential is defined to be zero at infinity.
a. Determine the electric potential at corner A.
b. What is the electric potential energy of a +3 µC charge placed at corner A?
Answer / Explanation:
a )V(A) = 1 / 4πe° ( - 5 5x10∧6C / 0.150m + 2x10∧6C / 0.050m )
The answer to the equation above is : = +6.0x10∧4 j/c
b) U(A) = qV(A)= (3.0x10∧6C) (6.0x10∧4 . j/c) =
The answer to the equation above is : =0.18 J
Explanation:
Where V(A) is equivalent to the electric potential
U(A) is equivalent to the electric potential energy
Final answer:
Explanation of electric potential energy in physics.
Explanation:
Electric potential energy is the energy a charge possesses due to its position in an electric field. It is given by the formula U = k * Q1 * Q2 / r, where k is Coulomb's constant, Q1 and Q2 are the charges, and r is the distance between them.
)V(A) = 1 / 4πe° ( - 5 5x10∧6C / 0.150m + 2x10∧6C / 0.050m )
The answer to the equation above is : = +6.0x10∧4 j/c
b) U(A) = qV(A)= (3.0x10∧6C) (6.0x10∧4 . j/c) =
The answer to the equation above is : =0.18 J
Say, for example, that you had prepared a buffer c, in which you mixed 8.203 g of sodium acetate, nac2h3o2, with 100.0 ml of 1.0 m acetic acid. what would be the initial ph of buffer c? if you add 5.0 ml of 0.5 m naoh solution to 20.0 ml each of buffer b and buffer c, which buffer's ph would change less? explain
The initial pH of buffer C can be determined using the Henderson-Hasselbalch equation. Upon addition of NaOH, the buffer seeing less change in pH has greater buffer capacity, a measure of its resistance to pH changes upon addition of acid or base. Buffer solutions are most effective when the concentration of the weak acid and its conjugate base are equal.
Explanation:The topic you're dealing with pertains to buffer solutions in chemistry. In your given example with buffer C, you've mixed 8.203 g of sodium acetate with 100.0 ml of 1.0 M acetic acid. Here, the initial pH of the buffer can be calculated using the Henderson-Hasselbalch equation.
This equation states that pH = pKa + log([A-]/[HA]), where [A-] is the molar concentration of the base (sodium acetate) and [HA] is the molar concentration of the acid (acetic acid). You need to find the values of [A-] and [HA] and insert it in this formula to get the initial pH of buffer C.
Upon addition of NaOH solution to buffers B and C, the buffer whose pH changes less will be the one with the greater buffer capacity. This is because buffer capacity is a measure of the resistance of a buffer solution to pH change upon the addition of an acid or a base. Buffer capacity is maximized when [A-] = [HA], and a buffer solution loses its effectiveness when one component (either A- or HA) is less than about 10% of the other.
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2. At the molecular level, speculate on some ways that the environment might have an influence on DNA and its packaging
Answer: Histones, DNA methylation
Explanation:
The double stranded DNA are wrapped twice around a histone protein and depending on how loosely or tightly DNA are wounded around histion, DNA can be either read or not. The nature of histones have the ability to control which sections of DNA get copied and expressed. Histone synthesis stops when DNA synthesis ceases, histone are modified by acetylation, methylation ADP--ribosylation and phosphorylation. It has shown that environmental factors such as diet can have an effect on protein structure. Alteration of the histone protein structure can affect the expression of some genes.
DNA methylation can also inhibit the expression of some genes, which can be caused by environmental factit's.
Environmental factors can influence DNA and its packaging at the molecular level through epigenetic regulation and histone modification.
Explanation:Environmental factors can have an influence on DNA and its packaging at the molecular level. One way is through epigenetic regulation, where environmental factors can cause chemical modifications to DNA and histone proteins. For example, certain environmental factors can lead to the addition of methyl groups to specific cytosine nucleotides in DNA, which can affect gene expression. Another way is through the modification of histone proteins, such as acetylation and deacetylation, which can influence the packaging state of DNA and its accessibility for transcription.
The solubility of calcium chromate in water is 4.16 grams per liter. If a calcium chromate solution has a concentration of 4.16 grams per liter, is the solution saturated, unsaturated, or supersaturated?
Answer:
The concentration is 4.16 grams per liter, we call the solution saturated.
Explanation:
Step 1: Data given
The solubility of calcium chromate in water is 4.16 grams per liter.
Saturated Solution : A solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom.
Unsaturated Solution: A solution (with less solute than the saturated solution) that completely dissolves, leaving no remaining substances.
Supersaturated Solution: A solution (with more solute than the saturated solution) that contains more undissolved solute than the saturated solution because of its tendency to crystallize and precipitate.
Step 2:
If the calcium chromate solution has a concentration of 4.16 grams per liter.
If the concentration is less than 4.16 grams per liter, we call it unsaturated.
If the concentration is more than 4.16 grams per liter, we call it supersaturated.
The concentration is 4.16 grams per liter, we call the solution saturated.
Final answer:
A saturated solution contains the maximum amount of solute that can dissolve in a solvent at a specific temperature.
Explanation:
The solution is saturated because the concentration of calcium chromate in the solution is equal to its solubility in water, which is 4.16 grams per liter.
A saturated solution contains the maximum amount of solute that can dissolve in a given solvent at a specific temperature.
Saturated solutions are in a dynamic equilibrium where the rate of dissolution is equal to the rate of precipitation.
A chemist adds of a iron(III) bromide solution to a reaction flask. Calculate the mass in kilograms of iron(III) bromide the chemist has added to the flask. Round your answer to significant digits
Answer:
0.246 kg
Explanation:
There is some info missing. I think this is the original question.
A chemist adds 370.0mL of a 2.25 M iron(III) bromide (FeBr₃) solution to a reaction flask. Calculate the mass in kilograms of iron(III) bromide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
We have 370.0 mL of 2.25 M iron(III) bromide (FeBr₃) solution. The moles of FeBr₃ are:
0.3700 L × 2.25 mol/L = 0.833 mol
The molar mass of iron(III) bromide is 295.56 g/mol. The mass corresponding to 0.833 moles is:
0.833 mol × 295.56 g/mol = 246 g
1 kilogram is equal to 1000 grams. Then,
246 g × (1 kg/1000 g) = 0.246 kg
A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.) Fill in the missing part of this equation. pa * mm2
The question is incomplete, here is the complete question:
A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.) Fill in the missing part of this equation.
[tex]23.Pa.cm^3=?kPa.m^3[/tex]
Answer: The measurement after converting is [tex]23\times 10^{-9}kPa.m^3[/tex]
Explanation:
We are given:
A quantity having value [tex]23.Pa.cm^3[/tex]
To convert this into [tex]kPa.m^3[/tex], we need to use the conversion factors:
1 kPa = 1000 Pa
[tex]1m^3=10^6cm^3[/tex]
Converting the quantity into [tex]kPa.m^3[/tex], we get:
[tex]\Rightarrow 23.Pa.cm^3\times (\frac{1kPa}{1000Pa})\times (\frac{1m^3}{10^6cm^3})\\\\\Rightarrow 23\times 10^{-9}kPa.m^3[/tex]
Hence, the measurement after converting is [tex]23\times 10^{-9}kPa.m^3[/tex]
Which of the following represents a propagation step in the monochlorination of methylene chloride (CH2Cl2)?a. CHCl3 + Cl. Right arrow .CCl3 + HCl.b. CHCl2 + Cl2 right arrow CHCl3 + Cl..c. CH2Cl + Cl2 right arrow CH2Cl2 + Cl..d. CHCl2 + Cl. Right arrow CHCl3
Answer:
B = CHCl2 + Cl2 --> CHCl3 + Cl
Explanation:
Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.
Free radical chlorination is divided into 3 steps which are:
The initiation step
The propagation step
The termination step
So in reference to the question, propagation step involves two steps.
The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.
The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.
You would find in the attachment the 2 step mechanism.
The propagation steps in the monochlorination of methylene chloride (CH2Cl2) are option b and d. Propagation steps are characterized by the reaction of two radicals to form a non-radical species and a new radical, which will then enter the chain cycle. These steps occur after initiation and before termination in a radical chain reaction.
Explanation:The question asks which options represent a propagation step in the monochlorination of methylene chloride (CH2Cl2). The correct answer is option b and d. These steps depict the reaction of CHCl2 with Cl2 with chlorine radical (.) or molecular chlorine (Cl2) to generate either CHCl3 and Chlorine radical (.) in option b or CHCl3 in option d, which are required for the propagation process in a halogenation reaction. It is essential to distinguish between initiation, propagation and termination steps in radical chain reactions, like this one, where a hydrogen atom on the methylene chloride is replaced by a chlorine atom.
Propagation steps are the 'middle' steps of these reactions. They are characterized by two radicals reacting to form a non-radical species and a new radical, which re-enters the propagation cycle. Both initiation and termination steps involve either the creation or removal of radicals, respectively. Radical chain reactions, halogenation, and methylene chloride are key concepts for understanding this question.
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The following reaction takes place at high temperatures.
Solid chromium(III) oxide reacts with liquid aluminum to give liquid chromium and liquid aluminum oxide.
Write a balanced chemical equation (include the states of matter).
Answer:
[tex]Cr_2O_3 (s) + 2Al (l)[/tex] ⇒ [tex]2Cr (l) + Al_2O_3 (l)[/tex]
The balanced chemical equation (including the states of matter) is
Cr₂O₃(s) + 2Al(l) ⇒ 2Cr(l) + Al₂O₃(l)
Solid chromium(III) oxide reacts with liquid aluminum to give liquid
chromium and liquid aluminum oxide. This reaction occurs at high
temperature because the reactivity is increased under this condition.
TYhe liquid aluminium is oxidized to form liquid aluminum oxide and the
Solid chromium(III) oxide is reduced to form liquid chromium and can be
referred to as a redox reaction.
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Refer to the section in the Background titled, "Standard Tests Gaseous Products." Use a flaming splint and a glowing splint to identify the gas above the liquid in the cup. Write the name and formula of the gas in the space provided
To identify the gas above the liquid in the cup, you can use a flaming splint and a glowing splint. If the gas supports combustion, it is oxygen.
Explanation:The gas above the liquid in the cup can be identified using a flaming splint and a glowing splint. If the gas supports combustion and allows the flaming splint to burst into full flame, it is oxygen. If the gas extinguishes the glowing splint, it is a gas that does not contain oxygen. Oxygen is the only gas that will support combustion.
A real gas is in a regime in which it is accurately described by the ideal gas equation of state. 27.1 g of the gas occupies 50.0 L at a pressure of 0.500 atm and a temperature of 449 K. Identify the gas.
The gas is accurately described by the ideal gas equation. Calculations reveal that the given conditions match Argon (Ar) most closely. Thus, the correct option is B) Ar.
To identify the gas accurately, we can use the ideal gas equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's calculate the number of moles (n) using the given mass (m) of the gas and its molar mass (M). The molar mass of the gas is necessary for this step.
[tex]\[ n = \frac{m}{M} \][/tex]
For this calculation, we need to know the molar mass of each gas option. The molar masses are approximately:
- CO2: 44.01 g/mol
- Ar: 39.95 g/mol
- Ne: 20.18 g/mol
- N2: 28.02 g/mol
Let's go through the calculations step by step.
Given values:
- Mass (m) = 27.1 g
- Volume (V) = 50.0 L
- Pressure (P) = 0.500 atm
- Temperature (T) = 449 K
1. Calculate moles (n) using the given mass:
[tex]\[ n = \frac{m}{M} \][/tex]
For CO2: [tex]\( n_{\text{CO2}} = \frac{27.1 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.616 \, \text{mol} \)[/tex]
For Ar: [tex]\( n_{\text{Ar}} = \frac{27.1 \, \text{g}}{39.95 \, \text{g/mol}} \approx 0.679 \, \text{mol} \)[/tex]
For Ne: [tex]\( n_{\text{Ne}} = \frac{27.1 \, \text{g}}{20.18 \, \text{g/mol}} \approx 1.342 \, \text{mol} \)[/tex]
For N2: [tex]\( n_{\text{N2}} = \frac{27.1 \, \text{g}}{28.02 \, \text{g/mol}} \approx 0.968 \, \text{mol} \)[/tex]
2. Use the ideal gas equation to calculate moles (n):
[tex]\[ n = \frac{PV}{RT} \][/tex]
Substitute values:
[tex]\[ n = \frac{(0.500 \, \text{atm} \times 50.0 \, \text{L})}{(0.0821 \, \text{L}\cdot\text{atm/mol}\cdot\text{K} \times 449 \, \text{K})} \][/tex]
[tex]\[ n \approx \frac{25}{36.8029} \approx 0.681 \, \text{mol} \][/tex]
3. Compare the calculated moles:
- [tex]\( n_{\text{CO2}} \approx 0.616 \, \text{mol} \)[/tex]
- [tex]\( n_{\text{Ar}} \approx 0.679 \, \text{mol} \)[/tex]
- [tex]\( n_{\text{Ne}} \approx 1.342 \, \text{mol} \)[/tex]
- [tex]\( n_{\text{N2}} \approx 0.968 \, \text{mol} \)[/tex]
The closest match is [tex]\( n_{\text{Ar}} \approx 0.679 \, \text{mol} \)[/tex], so the gas is likely Argon (Ar). Therefore, option B) Ar is the correct answer.
Complete question :- A real gas is in a regime in which it is accurately described by the ideal gas equation of state. 27.1 g of the gas occupies 50.0 L at a pressure of 0.500 atm and a temperature of 449 K. Identify the gas.
A) CO2
B) Ar
C) Ne
D) N2
Many drugs are sold as their hydrochloric salts (R2NH2+Cl−), formed by reaction of an amine (R2NH) with HCl. Part 1 out of 4 Draw the major organic product formed from the formation of acebutolol with HCl. Acebutolol is a β blocker used to treat high blood pressure. Omit any inorganic counterions.Figure:a structure of Acebutolol is shown in the figure
Answer:
Hi
Acebutolol hydrochloride is the form of the hydrochloride salt of acebutolol, a synthetic derivative of butyranide with a hypotensive and antiarrhythmic activity. Acebutolol acts as a cardioselective beta-adrenergic antagonist with very little effect on bronchial receptors, having intrinsic sympathomimetic properties. Acebutolol is used in ventricular arrhythmias. Other indications may include hypertension, alone or in combination with other drugs. The salt scheme is found in the attached file.
Explanation:
Ammonium perchlorate is a powerful solid rocket fuel, used in the Space Shuttle boosters. It decomposes into nitrogen gas, chlorine gas, oxygen gas and water vapor, releasing a great deal of energy. Calculate the moles of ammonium perchlorate needed to produce of water. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Final answer:
To calculate the moles of ammonium perchlorate needed to produce a certain amount of water, use the balanced chemical equation and the mole ratio between ammonium perchlorate and water.
Explanation:
To calculate the moles of ammonium perchlorate needed to produce a certain amount of water, we need to use the balanced chemical equation. From the equation 10Al(s) + 6NH4C1O4(s) → 4Al2O3 (s) + 2AlCl3 (s) + 12H₂O(g) + 3N2 (8), we can see that for every mole of NH4C1O4, 12 moles of H₂O are produced. Therefore, the moles of ammonium perchlorate needed can be calculated by multiplying the given moles of water by the ratio of moles of NH4C1O4 to moles of water, which in this case is 6.
Which of the following would be expected to form hydrogen bonds with water? Choose all that apply.
(A) methylamine
(B) N-methylpropanamide
(C) acetaldehyde
(D) cyclopentane
(E) None of the Above
Answer: C
Explanation:
N-methylpropanamide and methylamine are the compounds that posses hydrogen bonds.
Hydrogen bonds are formed when hydrogen is covalently bonded to a highly electronegative atom. Hydrogen bonds are weaker than covalent bonds but they significantly impact on the chemistry of the molecules in which they occur.
Primary, secondary and tertiary amines all form hydrogen bonds. The compounds that has hydrogen bonds are;
methylamine N-methylpropanamideLearn more: https://brainly.com/question/7558603
By your calculations, you should administer 2.46 mL. What amount will you draw up in the syringe?
Answer:
The syringe should be at 2.50
Final answer:
To administer a 2.46 mL dose, draw up slightly more than this amount into the syringe and then adjust to the precise volume by expelling any excess. Ensure the plunger aligns with the 2.46 mL marking on the syringe for an accurate dose.
Explanation:
When administering medication in a clinical setting, it's important to measure the dosage accurately. If you've calculated that the correct dose is 2.46 mL, you should first draw up the liquid into the syringe to a volume slightly greater than needed, then carefully adjust down to the exact amount. For instance, you might withdraw 3-4 mL and then expel the excess until you have the precise 2.46 mL required for administration.
The technique of drawing up more than needed and then adjusting allows for removing air bubbles and ensuring an accurate dose. In practice, syringes often have volume marks in increments of 0.1 mL or 0.01 mL, depending on the syringe size. It is critical to use these markings and ensure that the top of the plunger aligns exactly with the marking for 2.46 mL when viewed at eye level.
write a net ionic equation for the reaction of aqueous acetic acid and aqueous potassium hydroxide
Answer:
[tex]CH_{3}CO_{2}H(aq.)+OH^{-}(aq.)\rightarrow CH_{3}CO_{2}^{-}(aq.)+H_{2}O(l)[/tex]
Explanation:
Acetic acid is an weak acid. Therefore it does not dissociate completely in aqueous solution.
Potassium hydroxide is a strong base. Therefore it dissociates completely in aqueous solution.
An acid-base reaction occurs between acetic acid and potassium hydroxide.
Molecular equation: [tex]CH_{3}CO_{2}H(aq.)+KOH(aq.)\rightarrow KCH_{3}CO_{2}(aq.)+H_{2}O(l)[/tex]
Total ionic: [tex]CH_{3}CO_{2}H(aq.)+K^{+}(aq.)+OH^{-}(aq.)\rightarrow K^{+}(aq.)+CH_{3}CO_{2}^{-}(aq.)+H_{2}O(l)[/tex]
Net ionic:[tex]CH_{3}CO_{2}H(aq.)+OH^{-}(aq.)\rightarrow CH_{3}CO_{2}^{-}(aq.)+H_{2}O(l)[/tex]
The net ionic equation for the reaction between aqueous acetic acid (CH3COOH) and aqueous potassium hydroxide (KOH) is CH3COOH + OH- --> CH3COO- + H2O. The K+ ions are spectator ions and thus removed from the net ionic equation.
Explanation:To write a net ionic equation, we need to first balance the chemical reaction equation and then omit the spectator ions to get the net ionic equation. The balanced chemical reaction between aqueous acetic acid (CH3COOH) and aqueous potassium hydroxide (KOH) is: CH3COOH + KOH --> CH3COOK + H2O. After balancing the equation, we split the compounds into their ions: CH3COOH + K+ + OH- --> K+ + CH3COO- + H2O. Because K+ ions are present at both sides of the equation, they can be considered as spectators and therefore removed from our net ionic equation. Thus, the net ionic equation for the reaction is: CH3COOH + OH- --> CH3COO- + H2O.
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. In chemistry, one often uses a unit of charge known as the Faraday, F, which has the magnitude of the charge of 1 mole of electrons. How many Coulombs are there in a Faraday
Answer: 96500 Coulombs are there in a Faraday.
Explanation:-
It is given that 1 mole of electrons has charge of 1 Faraday.
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]
Thus 96500 C= 1 Faraday
Thus 96500 Coulombs are there in a Faraday.
There are approximately 96,485.33289 Coulombs in one Faraday.
The Faraday constant, denoted by F, is the charge of one mole of electrons. It is a fundamental physical constant with a value that has been experimentally determined. The Faraday constant is defined as the amount of charge carried by one mole of electrons, and its value is approximately 96,485.33289 Coulombs per mole.
The relationship between the Faraday constant and the charge in Coulombs can be expressed as:
[tex]\[ 1 \text{ Faraday} = 1 \text{ mole of electrons} \times \frac{6.022 \times 10^{23} \text{ electrons}}{1 \text{ mole of electrons}} \times \frac{1.602 \times 10^{-19} \text{ Coulombs}}{1 \text{ electron}} \][/tex]
Here, [tex]\(6.022 \times 10^{23}\)[/tex] is Avogadro's number, which is the number of particles (in this case, electrons) per mole, and [tex]\(1.602 \times 10^{-19}\)[/tex] is the elementary charge, which is the charge of a single electron in Coulombs.
Multiplying these values together gives:
[tex]\[ 1 \text{ Faraday} = 6.022 \times 10^{23} \times 1.602 \times 10^{-19} \text{ Coulombs} \] \[ 1 \text{ Faraday} = 96,485.33289 \text{ Coulombs} \][/tex]
Therefore, one Faraday is equivalent to approximately 96,485.33289 Coulombs. This is the standard value used in chemistry and physics for calculations involving the charge of electrons on a molar scale.
All of the following statements concerning voltaic cells are true EXCEPT
1) oxidation occurs at the cathode.
2) a salt bridge allows cations and anions to move between the half-cells.
3) electrons flow from the anode to the cathode in the external circuit.
4) a voltaic cell can be used as a source of energy.
5) a voltaic cell consists of two-half cells.
Answer:
oxidation occurs at the cathode.
Explanation:
In a voltaic cell electrons move from anode to cathode. At the anode, species give up electrons. This is an oxidation reaction depicted by the oxidation half equation. At the cathode, species accept electrons and become reduced. This is depicted by the reduction half equation. In summary; in a Voltaic cell, oxidation occurs at the anode while reduction occurs at the cathode.
In a voltaic cell, oxidation does not occur at the cathode.
A voltaic cell is an electrochemical cell that converts chemical energy to electrical energy, hence it can be used as a source of energy. It consists of two separate half-cells. There is a salt bridge between the two half cells allowing for the free flow of ions from one cell to another.
There are two electrodes, the anode undergoes oxidation and the cathode undergoes reduction. Electrons flow from the anode to the cathode.
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Consider a compound that is 31.17% C, 6.54% H, and 62.29% O by mass. Assume that we have a 100 g sample of this compound. What are the subscripts in the empirical formula for this compound?
Answer: the empirical formula is CH3O and the subscripts are 1, 3, 1
Explanation:Please see attachment for explanation
Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Note: you are writing the molecular, and not the net ionic equation.
→ CaCl2(aq) + H20(l)
Answer:
Ca(OH)₂ and HCl(aq)
Explanation:
We have the products of a reaction, CaCl₂ and H₂O. Since the products are a salt and water, this is likely to be a neutralization reaction. In such a reaction, an acid and a base react to form a salt and water.
The acid provides the anion of the salt. Since the anion is Cl⁻, the acid is HCl(aq).The base provides the cation of the salt. Since the cation is Ca²⁺, the base is Ca(OH)₂(aq).The complete molecular equation is:
Ca(OH)₂ + 2 HCl(aq) → CaCl₂ + 2 H₂O
Rank the following solutes in order of increasing entropy when 0.0100 moles of each dissolve in 1.00 liter of water.(a) NaBr(b) Cr(NO3)3(c) CaCl2(d) C6H12O6
Answer:
The rank is: C₆H₁₂O₆ < NaBr < CaCl₂ < Cr(NO₃)₃
Explanation:
To take into account, the greater the number of ions that are produced in a solution, the entropy will be greater. In this case, we have that the number of ions of each compound is:
Cr(NO₃)₃ → Cr³⁺ + 3NO⁻₃ (it has 4 ions)
CaCl₂ → Ca²⁺ + 2Cl⁻ (it has 3 ions)
NaBr → Na⁺ + Br⁻ (it has 2 ions)
C₆H₁₂O₆ (does not ionize)
An igneous rock that contains quartz and potassium feldspar would have a mineralogic content placing it in the range of __________.
Answer:
The answer is granitic or felsic rocks
Explanation:
Felsic is a term used in geology applied to silicate minerals, magmas and rocks, rich in light elements such as silicon, oxygen, aluminum, sodium and potassium (describs igneous rocks that are relatively rich in elements that form feldspar and quartz). This term is a combination of the words "feldspar" and "silica". Felsic minerals are generally light in color and have a specific gravity of less than 3. The most common felsic minerals are quartz, muscovite, alkaline feldspars (eg orthoclase) and feldspars from the plagioclase series. The most common felsic rock is granite. At the opposite end of the rock spectrum are mafic (rich in iron) and ultramafic (rich in magnesium) rocks and minerals.
Final answer:
An igneous rock with quartz and potassium feldspar has a felsic mineralogic content, rich in silica and light-colored minerals, and typically has a lighter color due to low ferromagnesian content.
Explanation:
An igneous rock containing quartz and potassium feldspar would most likely be classified within the range of felsic igneous rocks. Felsic rocks are rich in silica and characteristically contain light-colored minerals such as quartz and feldspar. Specifically, the presence of quartz, a pure silica mineral, and potassium feldspar, which includes silica along with aluminum and potassium, indicates a felsic composition.
These rocks generally have high silica content (65-75%), and they also commonly include minor amounts of mafic minerals such as biotite mica and amphibole.
Felsic igneous rocks can be further understood by their mineral proportions which can include around 25% K-feldspar and 30% quartz, with additional albitic plagioclase and biotite or amphibole. Their color is generally lighter due to the predominance of light-colored minerals and their low content of ferromagnesian components like iron and magnesium.
Benzyl ethyl ether reacts with concentrated aqueous HI to form two initial organic products (A and B). Further reaction of product B with HI produces organic product C. Draw the structures of these three products.
Answer:
See answer for details
Explanation:
In this case, we have the following substract:
Ph-CH2 - O - CH2CH3
That's the benzyl ethyl ether.
When this compound reacts with a concentrated Solution of HI, as we have an acid medium, we will have an SN1 reaction, where the Hydrogen will attach to the oxygen and the iodine will go to the most stable carbon, in this case, the carbon next to the ring is the most stable (because of the double bond of the ring, charges can be stabilized better this way), so product A will be the benzyl with the iodine and product B, will be the alcohol:
A: Ph - CH2 - I
B: CH3CH2OH
When B reacts again with HI, it's promoting another SN1 reaction, where the OH substract the H from the HI, the OH2+ will go out the molecule, leaving a secondary carbocation (CH2+) and then, the Iodine (I-) can go there via SN1 and the final product would be:
C: CH3CH2I
See picture for mechanism:
Final answer:
The student's question pertains to the reaction of benzyl ethyl ether with concentrated aqueous HI, following an SN1 mechanism. The reaction generates a halogenated product (A) and an alcohol (B), with the alcohol further reacting to form a reduction product (C). Specific structures cannot be provided without more information on the starting compound.
Explanation:
The student is inquiring about the reaction of benzyl ethyl ether with concentrated aqueous HI and the subsequent products. The reaction is characteristic of the SN1 mechanism of ether cleavage, particularly with ethers that have benzylic, allylic, or tertiary alkyl groups. In such reactions, the benzyl or allylic group forms a relatively stable carbocation, leading to the formation of a halogen product (product A), while the ether's other alkyl substituent becomes an alcohol (product B).
Further reaction of product B with HI would lead to the formation of product C, which is typically the reduction of the alcohol to an alkane or a similar process depending on the specific structure of product B. Without the detailed structure of benzyl ethyl ether, we cannot provide the exact structures of the final products A, B, and C.
Define lower end and upper end of a melting point range. (Select all that apply)
O The lower end is the temp at which liquid first appears.
O The upper end is the temp at which liquid first appears.
O The upper end is the temp at which the last trace of crystals disppears.
O The lower end is the temp at which the last trace of crystals disppears.
Answer:
The lower end is the temp at which liquid first appearsThe upper end is the temp at which the last trace of crystals disappearExplanation:
Melting range is one of the properties of solid that can help to identify its purity. It is necessary to note the whole range of temperature through which the transition takes place. The change in this range of melting temperature is indicative of impurity in the sample.
The lower end is the starting temperature at which solid starts to melt while the upper end is the temperature at which the last trace of solid converts to liquid.