To calculate the concentration of a solution, moles of solute are divided by the final volume in liters. Without the specific mass of silver perchlorate, we cannot determine the concentration of the solution.
Explanation:To calculate the concentration of the chemist's silver perchlorate solution, we need to know the mass of silver perchlorate used and the final volume of the solution. The details of the mass have been omitted from the question, but typically, you would use the molar mass of silver perchlorate to determine the number of moles. To prepare a solution of desired concentration, the chemist would dissolve silver perchlorate in a volumetric flask and add water to the mark, ensuring the final volume accounts for the space occupied by the dissolved solute.
Once a known mass of silver perchlorate is dissolved to make a known final volume, the molarity (M) can be calculated by dividing the number of moles of silver perchlorate by the final volume of the solution in liters. For example, if we dissolved 50 grams of silver perchlorate and the molar mass is X g/mol, we would first calculate the moles as 50 g / X g/mol. Then if the final volume of the solution is 1 L, the concentration would be 50 g / X g/mol / 1 L = Y M.
If the question had provided specific values, we would use them in the calculation and round the answer to the appropriate number of significant digits. However, without the exact mass of silver perchlorate used, we cannot complete this calculation.
2HgCl2(aq) + C2O42–(aq) → 2Cl–(aq) +2CO2(g) + Hg2Cl2(s)
The above reaction was studied by the method of initial rates and the following results were obtained:
[HgCl2] (M)
[C2O42–] (M)
Initial Rate (M/min)
0.105
0.15
1.8x10–5
0.105
0.30
7.2x10–5
0.0525
0.30
3.6 x10–5
0.0525
0.15
9.0x10–6
How would you describe the kinetics of this reaction? What is the rate constant?
first-order wrt HgCl2, second-order wrt C2O42–, third-order overall
k = 7.6x10–3 M–2min–1
first-order with respect to (wrt) HgCl2, first-order wrt C2O42–, second-order overall
k = 1.1x10–3 M–1min–1
zero-order wrt HgCl2, second-order wrt C2O42–, second-order overall
k = 5.4x10–5 Mmin–1
first-order wrt HgCl2, second-order wrt C2O42–, second-order overall
k = 4.3x10–8 M–2min–1
Explanation:
Expression for rate of the given reaction is as follows.
Rate = k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y[/tex]
Therefore, the reaction equations by putting the given values will be as follows.
[tex]1.8 \times 10^{-5} = k[0.105]x [0.15]y[/tex] ............. (1)
[tex]7.2 \times 10^{-5} = k [0.105]x [0.30]y[/tex] ........... (2)
[tex]3.6 \times 10^{-5} = k [0.0525]x [0.30]y[/tex] ............ (3)
Now, solving equations (1) and (2) we get the value of y = 2. Therefore, by solving equation (2) and (3) we get the value of x = 1.
Therefore, expression for rate of the reaction is as follows.
Rate = [tex]k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y[/tex]
Rate = [tex]k [HgCl2]1 [C_{2}O^{-2}_{4}]2[/tex]
Hence, total order = 1 + 2 = 3
According to equation (1),
[tex]1.8 \times 10^{-5} = k[0.105]x [0.15]y[/tex]
[tex]1.8 \times 10^{-5} = k [0.105]1 [0.15]2[/tex]
k = [tex]7.6 \times 10^{-3} M^{-2} min^{-1}[/tex]
Thus, we can conclude that rate constant for the given reaction is [tex]7.6 \times 10^{-3} M^{-2} min^{-1}[/tex].
Final answer:
The kinetics of the reaction is first-order with respect to HgCl₂, second-order with respect to C₂O4²⁻, and third-order overall. The rate constant (k) is 7.6x10⁻³ M−²min−1, which can be calculated using the initial rates method from the given data.
Explanation:
The kinetics of the reaction 2HgCl₂(aq) + C₂O4²⁻(aq) → 2Cl–(aq) + 2CO₂(g) + HgCl₂(s) can be determined by using the method of initial rates. From the given data, we observe that when the concentration of HgCl₂ is halved (0.105 M to 0.0525 M), keeping the concentration of C₂O4²⁻ constant (0.30 M), the initial rate is also halved, indicating that the reaction is first-order with respect to HgCl₂. When the concentration of C₂O4²⁻ is doubled (0.15 M to 0.30 M), keeping the concentration of HgCl₂ constant (0.10⁵ M), the initial rate increases by a factor of four (1.8x10⁻⁵ to 7.2x10⁻⁵), suggesting the reaction is second-order with respect to C₂O4²⁻. Thus, the reaction is first-order with respect to HgCl₂ and second-order with respect to C₂O4²⁻, making it third-order overall.
To find the rate constant (k), we can choose any of the given rate data. For example, taking the data from the first experiment:
rate = k[HgCl₂][C₂O4²⁻]²
1.8x10⁻⁵ M/min = k(0.105 M)(0.15 M)²
k = 7.6x10⁻³ M−²min−1
Calculate the molecular weight when a gas at 25.0 ∘C and 752 mmHg has a density of 1.053 g/L . Express your answer using three significant figures.
Answer:
26.0 g/mol is the molar mass of the gas
Explanation:
We have to combine density data with the Ideal Gases Law equation to solve this:
P . V = n . R .T
Let's convert the pressure mmHg to atm by a rule of three:
760 mmHg ____ 1 atm
752 mmHg ____ (752 . 1)/760 = 0.989 atm
In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.
Moles = Mass / molar mass.
We can replace density data as this in the equation:
0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K
(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x
0.0405 mol = 1.053 g / x
x = 1.053 g / 0.0405 mol = 26 g/mol
Student measured the absorbance of substance X in aqueous solution depending on concentration. He used 0.5 cm cuvette for his measurements; concentration was expressed in moles/L and plotted on x-axis, while absorbance was on y-axis. His data were best fitted by the straight line with equation y = 2120x + 0.0112. The extinction coefficient for substance X in aqueous solution is:
Answer:
The extinction coefficient is 4240 M⁻¹cm⁻¹
Explanation:
The Beer-Lambert law for the absorbance of a monochromatic light by a molecule is given by the following mathematical expression:
[tex]A=EbC[/tex]----------------------(1)
Here, E is the extinction coefficient, b is the length of light pathway through the solution and C is the concentration in mol/L. This is a linear equation and fits well with its formula
y = mx + C-------------------(2)
Here, m is the slope and C is the y intercept. y and x are the two variables. Comparing equation 1 and 2, Absorbance (A) and molar concentration (C) are the two variables, while the product of extinction coefficient (E) and path length (b) makes the slope of the line. The y-intercept in the given equation is due to the solvent interference.
From this information, we can determine the extinction coefficient by the slope of the line with the following formula.
[tex]m=Eb[/tex]
[tex]E=\frac{2120 (M^{-1} )}{0.5(cm)}[/tex]
[tex]E= 4240 M^{-1}cm^{-1}[/tex]
What are the four important steps needed to obtain accurate meting points on a Mel-Temp? Select four options:
1. Grind the sample into a fine powder.
2. Put the smallest amount you can see in a capillary tube.
3. Fill the capillary tube with about 3mm of powder.
4. Set the voltage to rise quickly to 20 degrees below the expected temperature, then re-adjust so the temp rises at 2 degrees per minute.
5. Set the voltage to rise quickly to 20 degrees below the expected temperature, then re-adjust so the temp rises at 10 degrees per minute.
6. Record the temp at which liquid first appears and the temp at which the last crystal disappears.
7. Record the temp at which the sample completely melts and the temp when it boils.
To obtain accurate melting points on a Mel-Temp, grind the sample into a fine powder, put the smallest visible amount in a capillary tube, fill the tube with about 3mm of powder, set the voltage to rise quickly to 20 degrees below expected temperature and slow down to 2 degrees per minute, and record the temperature at which liquid first appears and the last crystal disappears.
Explanation:The four important steps needed to obtain accurate melting points on a Mel-Temp are as follows:
Grind the sample into a fine powder.Put the smallest amount you can see in a capillary tube. Fill the capillary tube with about 3mm of powder. Set the voltage to rise quickly to 20 degrees below the expected temperature, then re-adjust so the temperature rises at about 2 degrees per minute.Record the temperature at which liquid first appears and the temperature at which the last crystal disappears. This gives the range within which the melting point falls.By following these steps, you can ensure that you are determining the melting point accurately.
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Calculate the volume in liters of a silver(II) oxide solution that contains mg of silver(II) oxide . Round your answer to significant digits
The question is incomplete, here is the complete question:
Calculate the volume in liters of a [tex]8.75\times 10^{-5}M[/tex] silver (II) oxide solution that contains 200.g of silver (II) oxide . Round your answer to 3 significant digits.
Answer: The volume of solution is [tex]1.84\times 10^4L[/tex]
Explanation:
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
We are given:
Molarity of solution = [tex]8.75\times 10^{-5}M[/tex]
Given mass of silver (II) oxide = 200. g
Molar mass of silver (II) oxide = 124 g/mol
Putting values in above equation, we get:
[tex]8.75\times 10^{-5}M=\frac{200}{124\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{200}{124\times 8.75\times 10^{-5}}=1.84\times 10^4L[/tex]
Hence, the volume of solution is [tex]1.84\times 10^4L[/tex]
To calculate the volume in liters of a silver(II) oxide solution, you need to know the amount of the solution in milligrams and the density of the solution. Use the formula Volume (in liters) = x mg / (density in mg/L) and round the final answer to the appropriate number of significant digits.
Explanation:To calculate the volume in liters of a silver(II) oxide solution, we need to know the amount of the solution in milligrams. Let's say the solution contains x mg of silver(II) oxide.
To convert mg to liters, we need to know the density of the solution. Once we have the density, we can use the formula:
Volume (in liters) = x mg / (density in mg/L)
Round the final answer to the appropriate number of significant digits.
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Characterize the following alkene as having the E or Z configuration. Draw the product(s) of bromination of this compound, including all expected stereoisomers (if any).
Use wedge-and-dash bonds to designate the stereochemistry at any chirality centers, and make sure to draw an explicit hydrogen if a chirality center has one.
Answer:
The alkene has Z configuration as shown in the figure.
The products of bromination of this compound are shown in the figure too with they stereochemistry.
Explanation:
The cis/trans nomenclature system in alkenes is insufficient when there are three or more different substituents in the double bond. In these cases, the Z/E nomenclature system, adopted by IUPAC, is used for all alkenes. Z comes from the German word zusammen which means together and E from the German word entgegen which means opposite. They would be equivalent to the terms cis and trans respectively.
If a molecular configuration is Z or E is determined by the priority rules of Cahn, Ingold and Prelog. For each of the two carbon atoms of the double bond, it is determined individually which of the two substituents has the highest priority. If both substituents of higher priority are on the same side, the arrangement is Z. On the other hand, if they are on opposite sides, the arrangement is E.
Halogenation of alkenes
Takes place with the addition of halogen atoms to the double bond to give a neighborhood dihaloalkane.
Halogenation mechanism
Halogenations are carried out at room temperature and in inert solvents such as carbon tetrachloride. In the mechanism it is observed that the opening of the bromonium ion occurs on the opposite side to the positive bromine that is the leaving group, this causes the halogens to be anti in the final product.
3. The same well water contains dissolved calcium (Ca), magnesium (Mg), sodium (Na) and potassium (K) at 5.0, 1.0, 136 and 1.2 mg/L, respectively. Determine these concentrations in units of mol/L.
Answer: The concentration of calcium, magnesium, sodium and potassium are [tex]1.25\times 10^{-4}mol/L[/tex], [tex]4.20\times 10^{-5}mol/L[/tex], [tex]5.91\times 10^{-3}mol/L[/tex] and [tex]3.08\times 10^{-5}mol/L[/tex] respectively
Explanation:
To convert the mass from milligrams to grams, we use the conversion factor:
1 g = 1000 mg
To convert the given concentration fro grams to moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For Calcium:Concentration given = 5.0 mg/L = [tex]5.0\times 10^{-3}g/L[/tex]
We know that:
Molar mass of calcium element = 40 g/mol
Putting values in equation 1, we get:
[tex]\text{Concentration in mol/L}=\frac{5.0\times 10^{-3}g/L}{40g/mol}\\\\\text{Concentration in mol/L}=1.25\times 10^{-4}mol/L[/tex]
For Magnesium:Concentration given = 1.0 mg/L = [tex]1.0\times 10^{-3}g/L[/tex]
We know that:
Molar mass of magnesium element = 24 g/mol
Putting values in equation 1, we get:
[tex]\text{Concentration in mol/L}=\frac{1.0\times 10^{-3}g/L}{24g/mol}\\\\\text{Concentration in mol/L}=4.20\times 10^{-5}mol/L[/tex]
For Sodium:Concentration given = 136 mg/L = [tex]136\times 10^{-3}g/L[/tex]
We know that:
Molar mass of sodium element = 23 g/mol
Putting values in equation 1, we get:
[tex]\text{Concentration in mol/L}=\frac{136\times 10^{-3}g/L}{23g/mol}\\\\\text{Concentration in mol/L}=5.91\times 10^{-3}mol/L[/tex]
For Potassium:Concentration given = 1.2 mg/L = [tex]1.2\times 10^{-3}g/L[/tex]
We know that:
Molar mass of potassium element = 39 g/mol
Putting values in equation 1, we get:
[tex]\text{Concentration in mol/L}=\frac{1.2\times 10^{-3}g/L}{39g/mol}\\\\\text{Concentration in mol/L}=3.08\times 10^{-5}mol/L[/tex]
Hence, the concentration of calcium, magnesium, sodium and potassium are [tex]1.25\times 10^{-4}mol/L[/tex], [tex]4.20\times 10^{-5}mol/L[/tex], [tex]5.91\times 10^{-3}mol/L[/tex] and [tex]3.08\times 10^{-5}mol/L[/tex] respectively
Fracking involves the ______drilling to extract oil and natural gas from the rock. Select one: a. vertical b. lateral c. horizonta
Answer:
Fracking involves the ______drilling to extract oil and natural gas from the rock. Select one: a. vertical b. lateral c. horizontal
The answer is option C (horizontal)
Explanation:
Fracking commonly known as Hydraulic fracturing is an oil and gas production technique that has enabled the extraction of oil and gas from rock (shale rock).
A fracturing fluid typically water with additives is pumped under high pressure into the drilling pipe to widen fractures in the rock or to create new ones and and allow trapped gas or crude oil to flow through a pipe to a wellhead at the surface, deep holes are drilled down into the shale rock. These fractures are held open by an injected proppant, typically sand and release hydrocarbons. Fracking takes about three to five days. Since shale reserves are typically distributed horizontally rather than vertically, horizontal directional drilling method is employed.
Some of the impacts of Fracking on the environment are:
It consume large quantities of fresh waterThe injection of recovered fracking water into underground disposal wells and the high pressure used to extract oil and gas from rock may cause earthquakes.Fracking causes air and water pollution.Deforestation caused by fracking has deep effects on the environment.Fracking, or hydraulic fracturing, primarily involves horizontal drilling. Starting vertically, the drill transitions to a horizontal path to enhance contact with gas or oil-bearing rocks, thereby improving extraction efficiency.
Explanation:Fracking, also known as hydraulic fracturing, involves the horizontal drilling to extract oil and natural gas from the rock. The process begins with a vertical hole that reaches the desired depth and then transitions into lateral or horizontal drilling. This allows for increased contact with the oil or gas-bearing rock, improving the extraction process and its efficiency.
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Identify the molecularity of each elementary reaction below.
1. F(g)+H2(g)→HF(g)+H(g)
a. unimolecular
b. termolecular
c. bimolecular
d. quadrimolecular
e. Cannot be determined.
2. H2(g)+2I(g)→2HI(g)
a. unimolecular
b. termolecular
c. bimolecular
d. quadrimolecular
e. Cannot be determined.
3. NO2Cl(g)→NO2(g)+Cl(g)
a. unimolecular
b. termolecular
c. bimolecular
d. quadrimolecular
e. Cannot be determined.
The molecularity of each reaction is determined by the number of molecules or atoms participating as reactants. The first is bimolecular, the second is termolecular, and the third is unimolecular.
Certainly, here's a more detailed explanation for each reaction:
1. The reaction F(g) + H2(g) → HF(g) + H(g) is bimolecular in nature. Bimolecular reactions involve two molecules or atoms as reactants that come together to form products. In this case, one molecule of fluorine (F2) is reacting with one molecule of hydrogen gas (H2) to produce one molecule of hydrogen fluoride (HF) and one hydrogen atom (H). The balanced equation reflects this stoichiometry: 1 F2 + 1 H2 → 2 HF + 1 H.
2. The reaction H2(g) + 2I(g) → 2HI(g) is termolecular. Termolecular reactions are relatively rare and involve three molecules or atoms as reactants. In this reaction, one molecule of hydrogen gas (H2) is reacting with two molecules of iodine gas (I2) to form two molecules of hydrogen iodide (HI). The balanced equation represents this as 1 H2 + 2 I2 → 2 HI. The presence of three molecules as reactants in this equation makes it termolecular.
3. NO2Cl(g) → NO2(g) + Cl(g) is a unimolecular reaction. Unimolecular reactions involve only one molecule or species as a reactant that decomposes or rearranges to form products. In this reaction, a single molecule of nitrogen dioxide chloride (NO2Cl) is decomposing into one molecule of nitrogen dioxide (NO2) and one chlorine atom (Cl). The equation 1 NO2Cl → 1 NO2 + 1 Cl illustrates the unimolecular nature of this reaction.
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Consider a bathtub holding a system of water (where the boundaries of the tub are the boundaries of the system) under the following conditions:
(a) The bathtub has a closed drain and is already full of water at the start of analysis. No steam rises from the water. You want to examine heat transfer loss from the water contained in the tub over time.
(b) The bath tub drain is partially open and hot water is continually added to keep the tub filled and the temperature constant. You want to examine the flow rates required to maintain the temperature of the water in the tub.
For each system (a) and (b), determine if it is characterized as open or closed and state your reasoning.
Answer:
a) the system is closed
b) the system is open
Explanation:
a) the system can be characterised as closed system. If we define the boundaries of this system as an envelope that contains the bathtub, then there will not be any mass transfer to the environment through the boundary → then the system is closed ( the mass of the system remains constant)
b) on the other hand , taking the same system as in the first case, if the hot water is continually added to the tub , then there is water flow through the boundary (from the environment to the system) → then the system is open ( the mass of the system can be altered)
The carbon-carbon double bond in ethene is ________ and ________ than the carbon-carbon triple bond in ethyne.a. weaker; longerb. stronger; more polarc. weaker; shorterd. stronger; shortere. stronger; longer
Answer:
A
Explanation:
ethane weaker than ethyne because it is not as unsaturated as ethyne
Ethene is longer because it need to bond with two more hydrogen atom
Select the structural formula(s) for the product(s) obtained when 1-butene reacts with Hg(OAc)2 in THF and H2O, then NaBH4, OH-.
Answer:
CH3CH2CH(OH)CH3
Explanation:
This reaction follows the Markonikovs rule which states that the addition of an acid yo an assymetric alkene, the proton given off by the acid goes to the Carbon attached/with more hydrogen substituents and the rest of the acid ion goes to the Carbon with more alkyl substituents.
The reaction under goes 2 steps,
The first step called Oxymercuration involves the dissociation of Hg(OAc)2 into HgOAc+ and OAc-. The double bond in 1-butene (high electron site) bonds to an electron deficient HgOAc+ thereby forming an unstable complex, the lone pairs of oxygen in the water molecule (H2O) is attracted and a conplex is further formed with a proton on the now bonded H2O molecule.
To stabilise this, the proton(H+) is removed and then Deoxymercuration occurs. Deoxymercuration is simply removing the HgOAc+ molecule from the compound. This is done by the further reaction with the mixture of NaBH4 in alkaline medium; this is a string reducing reagent (reducting reagent - addition of excess Hydrogen) to form a secondary alcohol(Butan-2-ol)
Below us the mechanism in the attachment, i hope this was helpful.
Given the elements Cl, 784 kJ-mol"', match the atoms with their first ionization energies Ge, and K and three values of possible first ionization energies, 418, 1255, a. CI (418), Ge (784), and K (1255 kJ-mol) b. CI (1255), Ge (784), and K (418 kJ-mol) c. CI (784), Ge (1255), and K (418 kJ.mol) d. CI (1255), Ge (418), and K (784 kJ-mol"') e. CI (418), G (1255), and K (784 kJ-mol-)
Answer:
b. CI (1255 kJ/mol), Ge (784 kJ/mol), and K (418 kJ-mol)
Explanation:
In general, the first ionization energy for a given period increases as we go from left to right in the periodic table (there are some exeptions as with every rule), and the first ionization energy decreases a we go down in the periodic table. The reason for this are:
1. As we move from left to right in a group, the effective nuclear charge increases which makes it harder to remove the electron.
2. As we increase the period in going top to bottom in the periodic table we are adding another shell farther away from the nucleus, making it easier to remove the electron.
Given the three values for the first ionization energy in kJ/mol : 418, 784 and 1255, we expect the highest value to correspond to Cl which belongs to period 3 (K and Ge belong to period 4).
Now comparing K and Ge which belong to period 4, Ge will have a higher effective nuclear charge than K .
So the match will be Cl (1255 kJ/mol), Ge (784) and K(418)
Correct answer is b.
Note: there could be some confusion since the value of 784 was misplaced in the question statement, but we can deduce that in this question we are asked to match the values for the atoms.
A chemist adds 380.0mL of a ×9.7510−4mM magnesium fluoride MgF2 solution to a reaction flask. Calculate the micromoles of magnesium fluoride the chemist has added to the flask. Round your answer to 3 significant digits.
Answer:
0.370 micromoles of magnesium fluoride the chemist has added to the flask.
Explanation:
[tex]Molarity=\frac{moles}{\text{Volume of solution(L)}}[/tex]
Moles of magnesium fluoride = n
Volume of the solution = 380.0 mL = 0.380 L (1 mL = 0.001 L)
Molarity of the solution = [tex]9.75\times 10^{-4} mM=9.75\times 10^{-7} M[/tex]
(1 mM = 0.001 M)
[tex]9.75\times 10^{-7} M=\frac{n}{0.380 L}[/tex]
[tex]n=3.705\times 10^{-7} mol[/tex]
1 mole = [tex]10^6[/tex] micro mole
[tex]n=3.705\times 10^{-7} \times 10^6 \mu mol=0.3705 \mu mol[/tex]
0.370 micromoles of magnesium fluoride the chemist has added to the flask.
An electron charge e mass m and a positron charge e mass m revolve around their common center of mass under the influence of their attractive coulomb force Find the speed v of each particle in terms of e m k and their separation L
Answer: v = 2π2 Kme2 Z / nh
Explanation:
The formula for velocity of an electron in the nth orbit is given as,
v = 2π2 Kme2 Z / nh
v = velocity
K = 1/(4πε0)
m= mass of an electron
e = Charge on an electron
Z= atomic number
h= Planck’s constant
n is a positive integer.
Calculate the pH of a solution formed by dissolving 0.040 moles of solid sodium lactate, NaC3H5O3, in 50.0 mL of 1.0 M HC3H5O3. Assume that any volume change is negligible.
The pH of the solution is 3,76
Why?
To solve this problem we have to apply the Henderson-Hasselbach equation, which is used whenever we need to calculate the pH of a solution of an acid HA (Lactic Acid) and its conjugate base A⁻ (Sodium Lactate)
We can use either moles or concentrations for this equation. In this case, we are going to use the moles, and we are going to take the pKa of lactic acid as 3,86:
[tex]pH=pKa+log(\frac{nA^- }{nHA} )=3,86+log(\frac{(0,04moles)}{(0,05mL)*(1.0M)} )\\\\pH=3,76[/tex]
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Final answer:
To calculate the pH of a buffer solution containing lactic acid and its conjugate base sodium lactate, the Henderson-Hasselbalch equation is used. The calculated pH for the buffer solution is approximately 3.76.
Explanation:
The question involves calculating the pH of a solution containing a weak acid and its conjugate base, which is a typical buffer solution scenario. We recognize it as a buffer solution because it contains both the weak acid (lactic acid, HC3H5O3) and its conjugate base (sodium lactate, NaC3H5O3). The Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), can be used to calculate the pH of a buffer solution where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid.
Firstly, we need to determine the concentrations of the acid and the base. The concentration of the lactic acid remains at 1.0 M since the added volume is negligible, while the amount of sodium lactate added is given as 0.040 moles. Since the volume is 50.0 mL (or 0.050 L), the concentration of sodium lactate will be 0.040 moles/0.050 L = 0.80 M.
Using this information and the known pKa of lactic acid, which is approximately 3.86, we can now use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:
pH = pKa + log([A-]/[HA])
pH = 3.86 + log(0.80/1.0)
pH = 3.86 + log(0.80)
pH ≈ 3.86 + (-0.09691)
pH ≈ 3.76
Therefore, the pH of the buffer solution is approximately 3.76.
What mass (in g) of solute is contained in 764.9 mL of a 0.137 M solution of glucose, C6H12O6? Enter your answer with 3 significant figures and no units.
Answer:
18.9
Explanation:
Given data
Volume (V): 764.9 mL = 0.7649 LMolar concentration (C): 0.137 M = 0.137 mol/LWe can find the moles (n) of glucose using the following expression.
n = C × V
n = 0.137 mol/L × 0.7649 L
n = 0.105 mol
The molar mass of glucose is 180.16 g/mol. The mass corresponding to 0.105 moles is:
m = 0.105 mol × 180.16 g/mol = 18.9 g
When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answer may be correct.
1)You should not use an open flame to heat the solvent.
2)You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
3)You should ensure that no one else is using an open flame near your experiment.
Explanation:
A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.
Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.
Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.
And, you should ensure that no one else is using an open flame near your experiment.
Thus, we can conclude that following statements are correct:
You should heat the solvent in a stoppered flask to keep vapor away from any open flames. You should ensure that no one else is using an open flame near your experiment.When heating a flammable or volatile solvent for recrystallization, one should not do the heating using an open flame or heat near any open flame in the laboratory.
What are flammable or volatile liquids?They are liquids that can catch flame easily either directly or through their vapors.
This means that such liquids should not be heated using an open flame or near any open flame to prevent fire accidents.
Heating near an open flame may cause the vapor to catch fire. Heating in a stoppered flask is not ideal because the main aim of heating is to allow the vapor to escape.
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g An example of an element is ____.
a) chicken noodle soup
b) powerade
c) air inside a balloon
d) lead pipe
e) baking soda (NaHCO)
You recrystallized acetanilide from water and found out that the final mixture contains 90 % of acetanilide by weight while the rest is water.
Calculate the molar percentage of acetanilide in the mixture if the molecular weight of acetanilide is 135 g/mol and the molecular weight of water is 18 g/mol. Round your answer to the nearest tenth, e.g. 1.1%.
Answer: 54.5%
Explanation:Please see attachment for explanation
Assuming the boiling point increased by 2 °C, what is the approximate molality of NaCl when the video ends? Kbp(water) = 0.512 °C/(mol/kg)
Answer : The molality of NaCl is, 1.95 mol/kg
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=i\times k_b\times m[/tex]
where,
[tex]\Delta T_b[/tex] = change in boiling point = [tex]2^oC[/tex]
[tex]k_b[/tex] = boiling point constant for water = [tex]0.512^oC/m[/tex]
m = molality
i = Van't Hoff factor = 2 (for electrolyte)
The dissociation [tex]NaCl[/tex] will be,
[tex]NaCl\rightarrow Na^++Cl^{-}[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]Na^++Cl^{-}[/tex] = 1 + 1 = 2
Now put all the given values in the above formula, we get:
[tex]2^oC=2\times (0.512^oC/m)\times m[/tex]
[tex]m=1.95mol/kg[/tex]
Therefore, the molality of NaCl is, 1.95 mol/kg
The molality of NaCl is, 1.95 mol/kg
The calculation is as follows;[tex]\Delta T _ b = i \times k_b \times m\\\\2^{\circ}C = 2\times (0.512^{\circ} C/m) \times m[/tex]
m = 1.95 mol/kg
Here m represent the molarity
[tex]\Delta T_b[/tex] represents a change in boiling point i.e. [tex]2^{\circ}C[/tex]
[tex]k_b[/tex] represents the constant boiling point for water
i = Van't Hoff factor = 2 (for electrolyte)
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A student is carrying out a biological preparation that requires 1 M NaCl to maintain an ionic strength of 1.0. The student chooses to use 1.0 M ammonium sulfate instead. Why is this a serious error?
Explanation:
It is known that the ionic strength of a solution, I, is represented as follows.
[tex]I = \frac{1}{2} \sum m_{i}z^{2}_{i}[/tex]
where, [tex]m_{i}[/tex] = the concentration of the ion,
[tex]z_{i}[/tex] = the charge of the ion
Now, for 1 M NaCl solution, the ionic strength will be calculated as follows.
I = [tex]\frac{1}{2} [(1 \times 1^{2}) + (1 \times 1^{2})][/tex]
= 1.00
And, for 1 M [tex](NH_{4})_{2}SO_{4}[/tex], concentration of [tex][NH_{4}][/tex] = 2 M and [tex][SO_{4}][/tex] = 1 M.
I = [tex]\frac{1}{2} [(2 \times 1^{2}) + ((1 \times 2^{2})][/tex]
= 3.00
As the ionic strength of 1 M [tex](NH_{4})_{2}SO_{4}[/tex], is different therefore it is a serious error.
Using 1.0 M ammonium sulfate instead of 1 M NaCl is a serious error in the biological preparation due to the differences in dissociation and specific properties of the compounds.
Explanation:The student's choice to use 1.0 M ammonium sulfate instead of 1 M NaCl is a serious error because they are not equivalent compounds. Ammonium sulfate dissociates into two ions (NH4+ and SO42-), while NaCl only dissociates into two ions (Na+ and Cl-). Therefore, the ionic strength of the solution will not be the same.
In addition, the presence of different ions in ammonium sulfate will affect the specific properties and behavior of the solution compared to NaCl. For example, the presence of ammonium ions can affect the pH and the precipitation of certain compounds.
It is important to use the correct reagents in a biological preparation to maintain the desired conditions and ensure accurate results.
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If you are trying to pipet an unknown liquid with a pipetman and the liquid keeps running out of the tip before you can transfer it, what are two possible reasons for this? What can you do to remedy the situation?
Answer:
This is due to the physical properties of the sample, since it affects the volume dispensed.
Explanation:
For example, in the case of very dense samples, selected samples to adhere to the surface of the tip, dispensing more slowly. In contrast, ethanol samples are less viscous and more volatile and are dispensed more rapidly. Some of the ways to minimize these inconveniences are the use of ultra low retention pipette tips, since they have a hydrophobic plastic additive that prevents the liquid from adhering to the inside of the tip.
Another way is to use the reverse pipetting.
Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfred Nobel (1833-1896) founded the Nobel Prizes with a fortune he made by inventing dynamite, a mixture of nitroglycerin and inert ingredients that was safe to handle.
1. Write a balanced chemical equation, including physical state symbols, for the decomposition of liquid nitroglycerin ( C 3 H 5 NO 3 3 ) into gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon dioxide.
2. Suppose 69.0 L of carbon dioxide gas are produced by this reaction, at a temperature of − 5.0 °C and pressure of exactly 1 atm . Calculate the mass of nitroglycerin that must have reacted. Be sure your answer has the correct number of significant digits.
Answer:
1. 4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)
2. 238 g of C₃H₅(NO₃)₃ has been reacted.
Explanation:
This is the chemical reaction:
4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)
For the second part, let's apply the Ideal Gases Law to find out the moles of CO₂ that were produced.
P . V = n . R . T
1 atm . 69L = n . 0.082 L.atm/mol.K . 268K
(1 atm . 69L) / (0.082 L.atm/mol.K . 268K) = n → 3.14 moles
In the equation, ratio between nitroglycerin and CO₂ is 12:4.
12 moles of CO₂ were produced by 4 moles of C₃H₅(NO₃)₃
Then, 3.14 moles of CO₂ would have been produced by (3.14 .4) / 12 = 1.04 moles of C₃H₅(NO₃)₃
Let's convert the moles to mass, to find out the mass of nitroglycerin that must have reacted (mol . molar mass)
1.04 mol . 227.08 g/mol = 238 g
"An orange compound was added to the top of a chromatography column. Solvent was added immediately, with the result that the entire volume of solvent in the solvent reservoir turned orange. No separation could be obtained from the chromatography experiment. What went wrong?
Answer:The result indicates that the polar component of the mixture fail to separate. In a case like this a third liquid like acid, base or complexing agent is often added to retain more water in the organic solvent.
Explanation:
Answer:
Seems to me that the sample had not been properly adsorbed, did not establish a true equilibrium, and therefore washed through the column rather that separated. The sample should be allowed to stand a short time before elution rather than putting the solvent in immediately.
Explanation:
Problem: In each of the following situations, what type of filtration device would you use?
A. Remove powdered decolorizing charcoal from 20mL of solution?
B. Collect crystals obtained from crystallizing a substance from about 1 mL of solution.
C. Remove a very small amount of dirt from 1 mL of liquid?
D. Isolate 2.0 g of crystals from about 50 mL of solution after performing a crystallization?
E. Remove dissolved colored impurities from about 3 mL of solution?
F. Remove solid impurities from 5 mL of liquid at room temperature?
Final answer:
Filtration devices used in different situations include membrane filtration, gravity filtration, microfiltration, vacuum filtration, activated carbon filtration, and simple filtration.
Explanation:
For A, to remove powdered decolorizing charcoal from 20mL of solution, you can use membrane filtration with a syringe filter. It allows you to push the solution through by depressing the syringe's plunger.
For B, to collect crystals obtained from crystallizing a substance from about 1 mL of solution, you can use gravity filtration. It involves pouring the solution through a funnel with filter paper to separate the crystals.
For C, to remove a very small amount of dirt from 1 mL of liquid, you can use microfiltration. It uses a membrane filter with a pore size small enough to capture the dirt particles.
For D, to isolate 2.0 g of crystals from about 50 mL of solution after performing a crystallization, you can use vacuum filtration. It involves connecting a filtration unit to a vacuum to draw the solution through the filter.
For E, to remove dissolved colored impurities from about 3 mL of solution, you can use activated carbon filtration. It utilizes activated carbon to adsorb the impurities.
For F, to remove solid impurities from 5 mL of liquid at room temperature, you can use simple filtration. It involves pouring the liquid through a filter to separate the solid impurities.
A graduated cylinder contains 20.8 mL of water. What is the new water level, in milliliters, after 35.2 g of silver metal is submerged in the water?
Answer: The new water level of the cylinder is 24.16 mL
Explanation:
To calculate the volume of water displaced by silver, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of silver = 10.49 g/mL
Mass of silver = 35.2 g
Putting values in above equation, we get:
[tex]10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL[/tex]
We are given:
Volume of graduated cylinder = 20.8 mL
New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver
New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL
Hence, the new water level of the cylinder is 24.16 mL
A gram of gasoline produces 45.0 kJ of energy when burned. Gasoline has a density of 0.77 g/ml. How would you ca the amount of energy produced by burning 35. L of gasoline? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
Explanation:
The given data is as follows.
Density of gasoline = 0.77 g/ml
Volume of gasoline = 35 L = 35000 ml (as 1 L = 1000 ml)
As we know that density of a substance is equal to its mass divided by its volume.
Mathematically, Density = [tex]\frac{mass}{volume}[/tex]
Hence, calculate the mass of given gasoline as follows.
Density = [tex]\frac{mass}{volume}[/tex]
0.77 g/ml = [tex]\frac{mass}{35000 ml}[/tex]
mass = 26950 g
Also, it is given that 1 g of gasoline on combustion produces 45.0 kJ of energy.
Therefore, energy produced by 26950 g of combustion of gasoline will be as follows.
[tex]45.0 \times 26950[/tex]
= 1212750 kJ
Thus, we can conclude that the amount of energy produced by burning 35 L of gasoline is 1212750 kJ.
A flash distillation chamber operating at 101.3 kpa is separating an ethanol water mixture the feed mixture contains z weight ethanol and a molar flow rate of a feed is
Answer:
The answer is [tex][\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454[/tex] mol/hr
[tex][\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454[/tex]mol/hr
Explanation:
For flash distillation
F = V+L
[tex]\frac{V}{F} + \frac{L}{F} = 1[/tex]
[tex]\frac{F}{V} -\frac{L}{V} = 1[/tex]
Fz = Vy+Lx
Y = [tex]\frac{F}{V}\times Z - \frac{L}{V}\times X[/tex] let, [tex]\frac{V}{F} = F[/tex]
[tex]y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X[/tex]
Highlighted reading
F = 299; [tex]\frac{V}{F}[/tex] = 0.85 ; z = 0.36
y = [tex]\frac{0.36}{0.85} - (-0.15)\times X[/tex]
= 0.423 + 0.15x ------------(i)
[tex]y^{*}[/tex] = -43.99713[tex]x^{6}[/tex] + 148.27274[tex]x^{5}[/tex] - 195.46[tex]x^{4}[/tex]+127.99[tex]x^{3}[/tex]-43.3[tex]x^{2}[/tex]+ 7.469[tex]x^{}[/tex]+ 0.02011
At equilibrium, [tex]y^{*}[/tex] = y
0.423+0.15[tex]x^{}[/tex] = [tex]y^{*}[/tex]
-43.99713[tex]x^{6}[/tex]+ 148.27274[tex]x^{5}[/tex] - 195.46[tex]x^{4}[/tex]+127.99[tex]x^{3}[/tex]-43.3[tex]x^{2}[/tex]+ 7.319[tex]x^{}[/tex]-0.403
F(x) for Newton's Law
Let [tex]x_{0} = 0[/tex]
[tex]x_{1}[/tex] = [tex]\frac{0-[{-0.403}]}{7.319}[/tex]
= 0.055
[tex]x_{2}[/tex] = [tex]\frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}[/tex]
= [tex]\frac{{(0.055)}-(-0.11)}{3.59}[/tex]
= 0.085
[tex]x^{3}[/tex] = [tex]\frac{{0.085}-(0.024)}{2.289}[/tex]
= 0.095
[tex]x^{4}[/tex] = [tex]\frac{{0.095}-(-0.0353)}{-1.410}[/tex]
= 0.07
From This x and y are found from equation (i) and L and V are obtained from [tex]\frac{V}{F}[/tex] and F values
[tex][\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454[/tex] mol/hr
[tex][\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454[/tex]mol/hr
The question relates to flash distillation with a focus on an ethanol-water azeotropic mixture. It addresses the limitation of distillation due to the azeotrope formation where the maximum ethanol concentration is about 95.6%. Understanding vapor-liquid equilibrium and distillation curves is essential for this process.
Explanation:The question is concerned with the principles of flash distillation and how it relates to azeotropic mixtures, specifically an ethanol-water mixture. When dealing with flash distillation, we are interested in how the composition of the mixture changes as it goes through vaporization and condensation cycles. Flash distillation is often used for the separation of components in a liquid mixture that have significantly different volatilities. The crux of this process is that the vapor formed will be richer in the more volatile components, in this case, ethanol.
Regarding the ethanol-water azeotrope, the mixture cannot be distilled beyond a certain concentration of ethanol due to the constant boiling nature at that specific composition. An azeotrope is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. In the case of ethanol and water, the maximum concentration of ethanol that can be obtained by simple distillation is approximately 95.6%. This is because, at this concentration, the mixture behaves as if it were a single component with a unique boiling point, and further separation by boiling alone is not possible without additional chemical agents or advanced separation techniques like extractive distillation.
In chemical engineering, it's also important to consider the vapor-liquid equilibrium and how it is represented by phase diagrams, such as the distillation curve given in this problem. Understanding these diagrams is critical for designing and operating distillation processes, whether in the context of a laboratory setup or an industrial scale operation.
Coal, which is primarily carbon, can be converted to natural gas, primarily CH4, by the following exothermic reaction: C(s)+2H2(g)⇌CH4(g) A. What is the effect of adding more C to the reaction mixture? B. What is the effect of adding more H2 to the reaction mixture? C. What is the effect of raising the temperature of the reaction mixture? D. What is the effect of lowering the volume of the reaction mixture? E. What is the effect of adding a catalyst to the reaction mixture? F. Which of the following will favor CH4 at equilibrium?
catalyst to the reaction mixture", "raising the temperature of the reaction mixture", "both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture", "adding more C to the reaction mixture"]
Final answer:
The effects on the equilibrium of converting carbon to methane through the reaction C(s) + 2H2(g) ⇌ CH4(g) depend on changes in the system. Adding more carbon or hydrogen, or lowering volume, favors the formation of methane, while raising the temperature favors reactants. Adding a catalyst speeds up equilibrium but doesn't affect its position.
Explanation:
Effects of Changes on a Chemical Equilibrium
The conversion of carbon to methane in the presence of hydrogen gas is described by the exothermic reaction C(s) + 2H2(g) ⇌ CH4(g). This process is subjected to changes that can affect its equilibrium according to Le Châtelier's principle:
Adding more C to the reaction would shift the equilibrium to the right, favoring the formation of more CH4, as the system tries to counteract the change.Adding more H2 would also shift the equilibrium to the right for the same reason, resulting in more CH4 production.Raising the temperature of an exothermic reaction will shift the equilibrium to the left, favoring the reactants and thus producing less CH4.Lowering the volume of the reaction mixture increases pressure, shifting the equilibrium toward the side with fewer gas molecules, which in this case is the right side, favoring CH4.Adding a catalyst to the reaction mixture does not shift the equilibrium; it only speeds up the rate at which equilibrium is reached.Regarding which options favor CH4 at equilibrium:
A catalyst does not affect the position of the equilibrium.Raising the temperature favors the reactants (not CH4) in an exothermic reaction.Adding more H2 or lowering the volume favors CH4 formation.Adding more C would also favor CH4 production.Thus, both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture would favor the formation of CH4 at equilibrium.
Adding more H₂ or lowering the volume of the reaction mixture will shift the equilibrium to favor CH₄ formation. Raising the temperature will decrease CH₄ production and adding a catalyst or more C will not change the equilibrium position of the reaction.
The reaction given, C(s) + 2H₂(g) ⇌ CH₄(g), is governed by the principles of chemical equilibrium. Let’s analyze each part:
Effect of adding more C: Adding more solid carbon does not affect the equilibrium position because solids are not included in the expression for the equilibrium constant. Therefore, the amount of CH₄ will remain unchanged.Effect of adding more H₂: Adding more hydrogen gas will shift the equilibrium to the right, increasing the production of CH₄ due to Le Chatelier’s principle.Effect of raising the temperature: Since the reaction is exothermic, raising the temperature will shift the equilibrium to the left, decreasing the amount of CH₄ formed.Effect of lowering the volume: Lowering the volume will increase the pressure. Since there are fewer moles of gas on the right (1 mole) than on the left (2 moles), the equilibrium will shift to the right, increasing CH₄ production.Effect of adding a catalyst: Adding a catalyst will increase the rate at which equilibrium is reached but will not affect the equilibrium position itself, so the amount of CH₄ remains the same.Which will favor CH₄ at equilibrium: The options “both adding more H₂ to the reaction mixture and lowering the volume of the reaction mixture” will favor CH₄ formation