Answer: The current needed, in mA, to plate the statue in 3.5 hr is 20 mA
Explanation:
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}=9.6352\times 10^4C[/tex]
[tex]Au^++e^-\rightarrow Au[/tex]
197 g of gold is deposited by = 96500 C of electricity
Thus 0.60 g of gold is deposited by =[tex]\frac{96500}{197}\times 0.60=294 C[/tex] of electricity
To calculate the current required, we use the equation:
[tex]I=\frac{q}{t}[/tex]
where,
I = current passed = ?
q = total charge = [tex]294C[/tex]
t = time required = 3.5 hrs =[tex]3.5\times 3600=12600s[/tex]
Putting values in above equation, we get:
[tex]I=\frac{294C}{12600}\\\\I=0.02A=20mA[/tex]
Hence, the current needed, in mA, to plate the statue in 3.5 hr is 20 mA
Propose a mechanism to account for the formation of a cyclic acetal from 4-hydroxypentanal and one equivalent of methanol. If the carbonyl oxygen of 4-hydroxypentanal is enriched with oxygen-18, do you predict that the oxygen label appears in the cyclic acetal or in the water
Answer:
If carbonyl oxygen of 4-hydroxypentanal is enriched with [tex]O^{18}[/tex], then the oxygen label appears in the water .
Explanation:
In the first step, -OH group at C-4 gives intramolecular nucleophilic addition reaction at carbonyl center to produce a cyclic hemiacetal.Then, one equivalent of methanol gives nucleophilic substitiution reaction by substituting -OH group in cyclic hemiacetal to produce cyclic acetal.If carbonyl oxygen of 4-hydroxypentanal is enriched with [tex]O^{18}[/tex], then the oxygen label appears in the water produced at the end of reaction.Full reaction mechanism has been shown below.A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. After 25.0 mL of base is added, the pH of the solution is 3.42.Estimate the pKa of the weak acid.
Final answer:
The estimated pKa of the weak monoprotic acid is 3.42, as this value is equal to the pH at the halfway point to the equivalence point during a titration with a strong base where the amounts of the weak acid and its conjugate base are equal.
Explanation:
The question addresses the titration of a weak monoprotic acid with a strong base and involves finding the pKa of the acid using pH measurements.
To estimate the pKa of the weak acid, we use the information that at the halfway point of the titration (when half the equivalent amount of base has been added), the pH of the solution equals the pKa of the acid. This is because, at the halfway point, the concentrations of the acid (HA) and its conjugate base (A-) are equal.
Given that 25.0 mL of 0.100 M NaOH is the halfway point since it takes 50.0 mL of NaOH to reach the equivalence point, and the pH at this stage is 3.42, we can directly say that the pKa of the weak acid is approximately 3.42.
This is because at the halfway point of the titration, the amount of acid that has been neutralized by the base is equal to the amount of acid that remains un-neutralized. In such a scenario, according to the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]) and since [A-] = [HA] at the halfway point, we get:
Thus, the estimated pKa value of the weak acid is 3.42.
Predict the sign of ΔS° for 2NO2(g) LaTeX: \longrightarrow⟶ N2O4(g) CaCO3(s) + 2HCl(aq) LaTeX: \longrightarrow⟶ CaCl2(aq) + H2O(l) +CO2(g) Ag+(aq) + Cl-(aq) LaTeX: \longrightarrow⟶ AgCl(s)
Answer: a. [tex]2NO_2(g)\rightarrow N_2O_4(g)[/tex]: [tex]\Delta S[/tex] is negative
b. [tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)[/tex] : [tex]\Delta S[/tex] is negative
c. [tex]Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)[/tex]: [tex]\Delta S[/tex] is negative
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa
a) [tex]2NO_2(g)\rightarrow N_2O_4(g)[/tex]
In this reaction 2 moles of gaseous reactants are converting to 1 mole of gaseous products. The randomness will decrease and hence entropy will also decrease. Thus [tex]\Delta S[/tex] is negative.
b) [tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)[/tex]
In this reaction solid reactants are converting to aqueous products. The randomness will increase and hence entropy will also increase. Thus [tex]\Delta S[/tex] is positive.
c) [tex]Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)[/tex]
In this reaction aqueous reactants are converting to solid products. The randomness will decrease and hence entropy will also decrease. Thus [tex]\Delta S[/tex] is negative
The sign of ΔS° for the given reactions can be determined based on the change in the number of moles of gas. In general, an increase in the number of moles of gas results in a positive ΔS°, while a decrease in the number of moles of gas results in a negative ΔS°.
Explanation:ΔS° represents the change in entropy. To predict the sign of ΔS° for a reaction, we can consider the number of moles of gas formed or consumed. In general, an increase in the number of moles of gas will result in a positive ΔS°, indicating an increase in entropy. On the other hand, a decrease in the number of moles of gas will result in a negative ΔS°, indicating a decrease in entropy.
In the reaction 2NO2(g) → N2O4(g), the number of moles of gas decreases from 2 to 1. Therefore, ΔS° for this reaction is expected to be negative.
In the reaction CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g), the number of moles of gas increases from 0 to 1. Therefore, ΔS° for this reaction is expected to be positive.
In the reaction Ag+(aq) + Cl-(aq) → AgCl(s), the number of moles of gas remains the same. Therefore, ΔS° for this reaction is expected to be close to zero.
A(g) + 2B(g) → C(g) + D(g)
If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of Kp for the reaction at the temperature the reaction was run?
Answer:
0.169
Explanation:
Let's consider the following reaction.
A(g) + 2B(g) ⇄ C(g) + D(g)
We can find the pressures at equilibrium using an ICE chart.
A(g) + 2 B(g) ⇄ C(g) + D(g)
I 1.00 1.00 0 0
C -x -2x +x +x
E 1.00-x 1.00-2x x x
The pressure at equilibrium of C is 0.211 atm, so x = 0.211.
The pressures at equilibrium are:
pA = 1.00-x = 1.00-0.211 = 0.789 atm
pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm
pC = x = 0.211 atm
pD = x = 0.211 atm
The pressure equilibrium constant (Kp) is:
Kp = pC × pD / pA × pB²
Kp = 0.211 × 0.211 / 0.789 × 0.578²
Kp = 0.169
Consider the mechanism. Step 1: A + B − ⇀ ↽ − C equilibrium Step 2: C + A ⟶ D slow Overall: 2 A + B ⟶ D Determine the rate law for the overall reaction, where the overall rate constant is represented as
The rate law for the overall reaction derived from a two-step reaction mechanism is determined by the slowest (rate-determining) step. In this case, the rate law would be rate = k * K_eq * [A]^2[B], where k is the rate constant for the slow step, K_eq is the equilibrium constant for the fast step, and [A] and [B] represent the concentrations of A and B, respectively.
Explanation:In the context of chemistry, one step in a multistep reaction mechanism is often significantly slower than the others. This slow step is known as the rate-determining step or the rate-limiting step. The reaction cannot proceed faster than this slowest step.
In your specific example, the given mechanism consists of two steps. The rate law for each step is generally expressed in terms of the concentration of the reactants involved in that step.
For Step 1 (A + B ⇄ C), assuming that the reaction reaches an equilibrium, the concentrations of A, B, and C would remain constant over time, and won't affect the overall rate. Therefore, we would ignore this step when deriving the rate law for the overall reaction.
On the other hand, Step 2 (C + A ⟶ D) is the slow step, and thus determines the rate of the overall reaction. The rate law for this step would be rate = k * [C][A]. But since C is also a product of Step 1, we need to express C in terms of A and B. From equilibrium of Step 1, we know [C] = K_eq*[A][B] (where K_eq is equilibrium constant).
Substituting this in rate law of step 2, we get rate = k * K_eq * [A]^2[B] for the overall reaction.
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The rate law for the overall reaction 2A + B ⟶ D, considering Step 2 is the rate-determining step, is calculated by using the equilibrium from Step 1 to express the concentration of intermediate C in terms of A and B. After substitution and simplification, the rate law for the overall reaction is rate = k [A]²[B], indicating second-order dependence on A and first-order dependence on B.
Explanation:To determine the rate law for the overall reaction, we must look at the mechanism provided. Since Step 2, which involves the conversion of C and A to D, is the rate-determining step, the rate law for the overall reaction will reflect this slowest step. However, since C is an intermediate that we cannot measure directly, we must use the equilibrium established in Step 1 to express the concentration of C in terms of the concentrations of A and B.
Assuming Step 1 is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and thus:
Rate of forward reaction (k₁ [A][B]) = Rate of reverse reaction (k₁₁ [C])
We can rearrange this to solve for [C]:
[C] = k₁/k₁₁ [A][B]
Now, since the rate-determining step is Step 2, we write the rate law based on this step:
rate = k₂ [C][A]
Substituting in the expression for [C] gives us:
rate = k₂ (k₁/k₁₁ [A][B])[A]
rate = (k₂ * k₁/k₁₁) [A]²[B]
Thus, after simplifying and combining the rate constants into a single overall rate constant (k), we have:
rate = k [A]²[B]
This shows that the reaction is second order with respect to A and first order with respect to B.
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Which compartment has the higher osmotic pressure? Which compartment has the higher osmotic pressure? 1%% (m/vm/v) starch solution 10%% (m/vm/v) starch solution
Answer : The compartment that has the higher osmotic pressure is, 10 % (m/v) starch solution.
Explanation :
Formula used for osmotic pressure :
[tex]\pi=\frac{nRT}{V}\\\\\pi=\frac{wRT}{MV}[/tex]
where,
= osmotic pressure
V = volume of solution
R = solution constant = 0.0821 L.atm/mole.K
T= temperature of solution = [tex]25^oC=273+25=298K [/tex]
M = molar mass of solute
w = mass of solute
Now we have to determine the osmotic pressure for the following solution.
For 1 % (m/v) starch solution :
1 % (m/v) starch solution means that 1 grams of starch present in 100 mL or 0.1 L of solution.
Molar mass of starch = 692.7 g/mol
[tex]\pi=\frac{(1g)\times (0.0821Latm/moleK)\times (298K)}{(692.7g/mol)\times (0.1L)}[/tex]
[tex]\pi=0.353atm[/tex]
For 10 % (m/v) starch solution :
10 % (m/v) starch solution means that 10 grams of starch present in 100 mL or 0.1 L of solution.
Molar mass of starch = 692.7 g/mol
[tex]\pi=\frac{(10g)\times (0.0821Latm/moleK)\times (298K)}{(692.7g/mol)\times (0.1L)}[/tex]
[tex]\pi=3.53atm[/tex]
From this we conclude that, 10 % (m/v) starch solution has the higher osmotic pressure as compared to 1 % (m/v) starch solution.
Hence, the compartment that has the higher osmotic pressure is, 10 % (m/v) starch solution.
The 10% (m/v) starch solution will have a higher osmotic pressure compared to the 1% (m/v) starch solution because osmotic pressure increases with solute concentration.
Explanation:The osmotic pressure is a property that depends on the solute concentration in a solution. In comparing a 1% (m/v) starch solution to a 10% (m/v) starch solution, the solution with the higher solute concentration is the one with higher osmotic pressure. Therefore, the 10% (m/v) starch solution will have a higher osmotic pressure because it has a greater concentration of starch molecules.
The osmotic pressure of a solution can be calculated using the formula Π = MRT, where Π is the osmotic pressure, M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin. As solute concentration increases, so does the osmotic pressure, provided that temperature and the gas constant remain the same. Consequently, hypertonic solutions will have higher osmotic pressure than hypotonic solutions.
5. Phosphoric acid (H3PO4) is a triprotic acid with three ionizable protons. Write a balance equation for the neutralization of phosphoric acid with NaOH. How many milliliters of 0.120 M NaOH would be required to completely neutralize 35.0 ml of 0.0440 M H3PO4
Answer:
1. H3PO4 + 3NaOH —> Na3PO4 + 3H2O
2. 38.5mL
Explanation:
1. We'll begin by writing a balanced equation for the reaction. This is illustrated below:
H3PO4 + 3NaOH —> Na3PO4 + 3H2O
2. H3PO4 + 3NaOH —> Na3PO4 + 3H2O
From the equation above, the following data were obtained:
nA (mole of the acid) = 1
nB (mole of the base) = 3
Data obtained from the question include:
Vb (volume of base) =?
Mb (Molarity of base) = 0.120 M
Va (volume of acid) = 35.0 mL
Ma (Molarity of acid) = 0.0440 M
Using the formula MaVa/MbVb = nA/nB, the volume of the base (i.e NaOH) can be obtained as follow:
MaVa/MbVb = nA/nB
0.0440 x 35/ 0.120 x Vb = 1/3
Cross multiply to express in linear form as shown below:
0.120 x Vb = 0.0440 x 35 x 3
Divide both side by 0.120
Vb = (0.0440 x 35 x 3) /0.120
Vb = 38.5mL
Therefore, 38.5mL of 0.120 M NaOH is needed for the complete neutralization.
Answer:
We need 38.5 mL of NaOH to neutralize the H3PO4 solution
Explanation:
Step 1: Data given
Molarity of NaOH = 0.120 M
Volume of H3PO4 = 35.0 mL = 0.035 L
Molarity of H3PO4 = 0.0440 M
Step 2: The balanced equation
H3PO4 + 3NaOH —> Na3PO4 + 3H2O
Step 3: Calculate the volume of NaOH
b*Ca*Va = a *Cb*Vb
⇒with b = the coefficient of NaOH = 3
⇒with Ca = the concentration of H3PO4 = 0.0440 M
⇒with Va = the volume of H3PO4 = 35.0 mL = 0.0350 L
⇒with a = the coefficient of H3PO4 = 1
⇒with Cb = the concentration of NaOH = 0.120 M
⇒with Vb = the volume of NaOH = TO BE DETERMINED
3*0.0440 * 0.0350 = 0.120 * Vb
Vb = 0.0385 L = 38.5 mL
We need 38.5 mL of NaOH to neutralize the H3PO4 solution
The properties of elements are different than the compound that the elements form. Is this statement True or False
Answer:
true
Explanation:
The pKa of an acid (one dissociable hydrogen) is -5.7. To the nearest ones, what is the Ka of this acid? Please note that we should use scientific notation and fewer significant figures, but Canvas is not configured to use scientific notation.
Answer:
The value of dissociation constant of an acid [tex]5.0\times 10^{5}[/tex].
Explanation:
The [tex]pK_a[/tex] of an acid = -5.7
The dissociation constant of the reaction = [tex]K_a[/tex]
The relation between [tex]pK_a[/tex] and [tex] K_a[/tex] is given by ;
[tex]pK_a=-\log[K_a][/tex]
[tex]-5.7=-\log[K_a][/tex]
[tex]K_a=10^{-(-5.7)}=5.012\times 10^{5}\approx 5.0\times 10^{5}[/tex]
The value of dissociation constant of an acid [tex]5.0\times 10^{5}[/tex].
Compositional analysis of a certain lipid shows that it has exactly one mole of fatty acid per mole of inorganic phosphate. Which of the lipids listed below is the lipid with this composition?
a. galactolipid
b. glycerophospholipid
c. sphingolipid
d. triacylglycerol
The sphingolipid is the lipid with this composition.
Option: C
Explanation:
A biomolecule which is soluble in non-polar solvents like hydrocarbons but remain insoluble in water is understood as a lipid. Its function is: energy storage, signaling and working as structural components of cell membranes. A class of lipids, which is structured from sphingoid base backbone, a group of aliphatic amino alcohols including sphingosine, is known as Sphingolipids. It was discovered in brain extracts during 1870s and named after the mythological sphinx for their mysterious existence.Complicated sphingolipids in animal cells' plasma membrane, particularly nerve cells, have a structural role and are thought to shield the cell surface from unfavorable external factors.The complex lipid with one mole of fatty acid per mole of inorganic phosphate has been Sphingolipid. Thus, option C is correct.
The lipids are biomolecules with the ability to bind with the nonpolar surfaces. The lipids have been bound with other biomolecules to form complex lipids.
The galactolipid has been the complex of fatty acid with galactose.Glycerophospholipid has been the composed of glycerophosphate along with the long chain constituted fatty acid and small alcohols.Sphingolipid has been the sphingosine backbone lipid with fatty acid. The sphingosine has been constituted of long amino acid chain. It has been present in cell membrane with one mole of fatty acid per mole phosphate.Triacylglycerol has been acidified fatty acid with three fatty acid to one glycerol.Thus, the complex lipid with one mole of fatty acid per mole of inorganic phosphate has been Sphingolipid. Thus, option C is correct.
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The decay constant for 14C is .00012 In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.
The question is incomplete, here is the complete question:
The decay constant for 14-C is [tex]0.00012yr^{-1}[/tex] In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14-C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.
Answer: The formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
Explanation:
Carbon-14 isotope is a radioisotope and its decay process follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}[/tex]
t = time taken for decay process = ? yr
[tex][A_o][/tex] = initial amount of the sample = 100 grams
[A] = amount left after decay process = (100 - 20) = 80 grams
Putting values in above equation, we get:
[tex]1.2\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{20}\\\\t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
Hence, the formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
Estimate the following: a) The volume occupied by 18 kg of ethylene at 55°C and 35 bar. b) The mass of ethylene contained in a 0.25-m3 cylinder at 50°C and 115 bar.
The estimated volume occupied by 18 kg of ethylene at 55℃ and 35 bar is approximately 476.6 liters. The mass of ethylene contained in a [tex]0.25m^3[/tex] cylinder at 50℃ and 115 bar is approximately 3.03 kg.
To estimate the volume occupied by 18 kg of ethylene at 55℃ and 35 bar, we can use the ideal gas law expressed as PV = nRT. First, we need to calculate the number of moles (n) using the molar mass of ethylene ([tex]C_2H_4[/tex]) which is approximately 28 g/mol. Therefore, n = 18000 g / 28 g/mol = 642.86 moles. The gas constant (R) is 0.08314 L*bar/mol*K. Converting the temperature to Kelvin, T = 55 + 273 = 328 K. Now, we can solve for V:
V = (nRT) / P = (642.86 mol * 0.08314 L*bar/mol*K * 328 K) / 35 bar = 476.6 L
To determine the mass of ethylene in a 0.25-[tex]m^3[/tex] cylinder at 50℃ and 115 bar, we apply the ideal gas law again. We first convert the volume to liters (V = 0.25 [tex]m^3[/tex] * 1000 L/[tex]m^3[/tex] = 250 L), and temperature to Kelvin (T = 50 + 273 = 323 K). The pressure is already in bar. Now, we calculate the number of moles, solving for n:
n = PV / RT = (115 bar * 250 L) / (0.08314 L*bar/mol*K * 323 K) = 108.14 moles
The mass of ethylene is then m = n * molar mass = 108.14 moles * 28 g/mol = 3027.92 g or approximately 3.03 kg of ethylene.
How many joules of heat are required to heat 110 g of aluminum from 52.0 oC to 91.5 oC?
Answer:
We need 3910.5 joules of energy
Explanation:
Step 1: Data given
Mass of aluminium = 110 grams
Initial temperature = 52.0 °C
Final temperature = 91.5 °C
Specific heat of aluminium = 0.900 J/g°C
Step 2: Calculate energy required
Q = m*c*ΔT
⇒with Q = the energy required = TO BE DETERMINED
⇒with m = the mass of aluminium = 110 grams
⇒with c = the specific heat of aluminium = 0.900 J/g°C
⇒with ΔT = the change in temperature = T2 - T1 = 91.5 °C - 52.0 °C = 39.5 °C
Q = 110 grams * 0.900 J/g°C * 39.5
Q = 3910.5 J
We need 3910.5 joules of energy
1) β-galactosidase is a unique enzyme, in that it can have multiple substrates. What are some other substrates for β-galactosidase? What are some other inhibitors for β-galactosidase?
The natural substrate of beta glycosidase enzyme is Ganglioside GMI, Lactosylceramide, lactose and glycoprotein.
Inhibitors of beta galactosidase enzyme is 1,4-dithiothreitol, beta marcaptoethanol, 4-chloromercurobenzoic acid and Acid-beta galactosidase.
Explanation:
Beta galactosidase enzyme performs the hydrolysis of beta galactosides into monosaccharides. It acts on aryl, amino, alkyl beta glycosidic linkages also.
The enzyme attacks on the bond formed between organic entity and galactose sugar.
It can act on multiple substrates. Some substrates are :
Ganglioside GMI
Lactosylceramide
Lactose : Enzyme beta galactosidase enzyme is a boon for lactose intolerant people because it breaks lactose in yoghurt, sour cream and cheese and makes it easy for consumption to such people.
glycoprotein : These have glycosidic bonds on which enzyme works to break the bonds.
The inhibitors for the β-galactosidase are :
4-dithiothreitol
beta marcaptoethanol
4-chloromercurobenzoic acid
Acid-beta galactosidase.
When inhibitors are bind to enzyme it breaks down the inhibitor and reaction does not takes place.
β-galactosidase can act on different substrates like lactose, ONPG and IPTG, demonstrating its flexibility. Its activity can be regulated by inhibitors such as glucose and PETG, affecting enzyme function either competitively or non-competitively.
Explanation:The enzyme β-galactosidase is unique as it can act on various substrates and is influenced by several inhibitors. This enzyme mainly acts on lactose but can also interact with several structurally related substrates such as o-nitrophenyl-β-D-galactoside (ONPG) and isopropyl β-D-1-thiogalactopyranoside (IPTG), exemplifying the flexibility of β-galactosidase.
The activity of β-galactosidase is subject to regulation by various inhibitors. These inhibitors could exert their effect by binding to the enzyme's active site (competitive inhibition), or noncompetitively by interacting with the enzyme's allosteric site--an alternate part where non-substrate molecules can attach. Examples of inhibitors include glucose and phenylethyl β-D-thiogalactoside (PETG).
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When you decrease the diameter of the stationary phase particles and make them more uniform, in HPLC, which term or terms of the van Deemter Equation is or are minimized?
Answer:
C-term
Explanation:
A correlation between plate height and mobile phase velocity is known as the van Deemter equation.
[tex]H = A + \frac{B}{u}+ C*u[/tex]
Where;
H is the plate height
A is the eddy diffusion term
B is the longitudinal diffusion term
C is the resistance to mass transfer coefficient
u is the linear velocity
Here C-term describes the mass transfer of sample components between the stationary phase and the mobile phase during separation. One of the major factor affecting C-term is the particle size, changing the diameter equally changes the particle size.
The answer is C-term
E. Both A and Cux are terms of the van Deemter Equation.
The van Deemter equation describes the relationship between the height equivalent to a theoretical plate (HETP) and the linear velocity (u) of the mobile phase in chromatography. The equation is:
[tex]\[H = A + \frac{B}{u} + C \cdot u\][/tex]
When the diameter of the stationary phase particles is decreased and made more uniform in High-Performance Liquid Chromatography (HPLC), the effects on the terms of the van Deemter equation are as follows:
1. Eddy Diffusion (A term):
- Eddy diffusion occurs due to the multiple pathways available for the analyte molecules through the packed column. Smaller and more uniform particles reduce the multiple pathways and hence minimize the A term.
2. Longitudinal Diffusion (B/u term):
- Longitudinal diffusion is the spreading of the analyte band due to diffusion along the length of the column. It is generally more significant at low flow rates and is not directly affected by the particle size.
3. Mass Transfer (C * u term):
- The mass transfer term relates to the time it takes for the analyte molecules to equilibrate between the stationary and mobile phases. Smaller particles decrease the distance the analytes must travel in and out of the stationary phase, thereby reducing the C.u term.
Therefore, decreasing the diameter of the stationary phase particles and making them more uniform primarily minimizes the A and [tex]\(C \cdot u\)[/tex] terms.
The correct answer is: E. Both A and Cux
Complete Question:
When you decrease the diameter of the stationary phase particles and make them more uniform, in HPLC, which term or terms of the van Deemter Equation is or are minimized?
A. A
B. B/ux
C. Cux
D. Both A and B/ux
E. Both A and Cux
F. Both B/ux and Cux
G. None of the terms
H. All of the terms
The charges and sizes of the ions in an ionic compound affect the strength of the electrostatic interaction between the ions and thus the strength of the lattice energy of the ionic compound. Arrange the compounds according to the magnitudes of their lattice energies based on the relative ion charges and sizes.a. SrO b. CsIc. RbFd. SrF2
Answer:
the correct answer:
C, B, D, A.
Explanation:
The chemical compounds named are compounds that have strong chemical bonds, forming cubic structures and crystals with very high boiling and melting points to less solid structures.
some are oxides, others salts, others are even used to emit radiation.
Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g of NaCl, and you isolate 52.1 g of PbCl2, what is your percent yield?
Answer:
[tex]\large \boxed{84.7 \, \%}[/tex]
Explanation:
Mᵣ: 58.44 278.11
Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃
m/g: 26.3
1. Moles of NaCl
[tex]\text{Moles of NaCl} = \text{26.3 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}[/tex]
(b) Moles of PbCl₂
[tex]\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} \times \dfrac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}[/tex]
(c) Theoretical yield of PbCl₂
[tex]\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} \times \dfrac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}[/tex]
(d) Percent yield
[tex]\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \,\% = \dfrac{\text{52.1 g}}{\text{61.52 g}} \times 100 \, \% = \mathbf{84.7 \,\%}\\\\\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}[/tex]
Final answer:
In a reaction where lead (II) nitrate reacts with sodium chloride to form lead (II) chloride, given 26.3 g of NaCl and 52.1 g of PbCl2 produced, the percent yield is calculated to be 83.27%.
Explanation:
The student is performing a reaction where lead (II) nitrate reacts with sodium chloride to produce lead (II) chloride and sodium nitrate. To calculate the percent yield, we use the actual mass of PbCl2 obtained from the experiment (52.1 g), and compare it with the theoretical mass that should have been produced if the reaction were 100% efficient.
First, you need to calculate the moles of NaCl:
Molecular weight of NaCl = 22.99 (Na) + 35.45 (Cl) = 58.44 g/molMoles of NaCl = mass (g) / molar mass (g/mol) = 26.3 g / 58.44 g/mol = 0.45 molesBased on the stoichiometry of the balanced equation, 2 moles of NaCl will produce 1 mole of PbCl2 (2:1 ratio). Therefore, 0.45 moles of NaCl should theoretically produce 0.225 moles of PbCl2.
Now, calculate the theoretical yield:
Molecular weight of PbCl2 = 207.2 (Pb) + 2*35.45 (Cl) = 278.1 g/molTheoretical yield (g) = moles * molar mass = 0.225 moles * 278.1 g/mol = 62.57 g of PbCl2To find the percent yield:
Percent yield = (actual yield / theoretical yield) * 100% = (52.1 g / 62.57 g) * 100% ≈ 83.27%The percent yield of the reaction is therefore 83.27%.
A car starts at mile marker 145 on a highway and drives at 55 mi/hrmi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours
Answer:
135 mile marker will the car reach after 2 hours.
Explanation:
Speed of the car = 55 mile/hour
Distance covered in 2 hours = d
[tex]Speed=\frac{Distance}{Time}[/tex]
[tex]55 mile/hour=\frac{d}{2 hour}[/tex]
[tex]d=55 mile/hour\times 2 hour=110 mile[/tex]
The direction of the car is in decreasing marker numbers which mienas that car had started from end where 145 mile marker was present.
So, the marker appearing after travelling 2 hours will be:
145 - 110 = 135
135 mile marker will the car reach after 2 hours.
Explain why a fossil record is not 100% accurate.
Consider the following metals: Ag, Au, Mg, Ni, and Zn.
Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank?
Steel is mostly iron, so use −0.447 V as the standard reduction potential for steel.
Answer:
Mg and Zn
Explanation:
In cathodic protection, the sacrificial anode corrodes instead of the cathode which it protects. The anode is usually higher than the cathode in the electrochemical series. This also means that the reduction potential of the sacrificial anode is more negative than that of the cathode. Consider the reduction potentials of the metals listed in the question:
Mg=-1.185V
Zn= -0.7618V
Ag= +0.7996V
Au= +1.629V
Ni= -0.251V
The reduction potential of the cathode stated in the question is -0.447V hence only magnesium and zinc can function as sacrificial anode.
The metals that can be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are silver (Ag), magnesium (Mg), nickel (Ni), and zinc (Zn).
Explanation:The metals that can be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are the ones with a lower standard reduction potential (E°) than steel. In this case, we need to compare the reduction potentials of steel to the standard reduction potentials of the given metals.
Silver (Ag) has a standard reduction potential of -0.799 V, which is lower than the E° for steel (-0.447 V), so it can be used as a sacrificial anode.Gold (Au) has a standard reduction potential of +1.40 V, which is higher than the E° for steel, so it cannot be used as a sacrificial anode.Magnesium (Mg) has a standard reduction potential of -2.37 V, which is significantly lower than the E° for steel, so it can be used as a sacrificial anode.Nickel (Ni) has a standard reduction potential of -0.257 V, which is lower than the E° for steel, so it can be used as a sacrificial anode.Zinc (Zn) has a standard reduction potential of -0.763 V, which is lower than the E° for steel, so it can be used as a sacrificial anode.Therefore, the metals that could be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are silver (Ag), magnesium (Mg), nickel (Ni), and zinc (Zn).
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A student fills her burette with NaOH to the 2.5 mL mark. She titrated her sample of the NaOH until she reaches the endpoint (i.e. all the acid has been neutralized by the NaOH). The volume marking on her burette - at the endpoint - is 52.5 mL. How many mL of NaOH did she use to reach the endpoint
Answer:
She used 50 mL of NaOH to reach the endpoint.
Explanation:
Assuming the burette is filled to the point marked 3.30 ml, You would record the initial point as 3.30 ml:
If at the end of the titration the level of the NaOH is at 20.30 mL; Subtract the initial reading from the final burette reading to get how many mL was used to reach an end point.
That is 20.3 - 3.3 = 17.00 mL
Therefore, the titration would have required 17.00 mL.
Remember that you should read the number that is at the bottom of the meniscus and at an eye level in order to avoid error.
Initial mark = 2.5 mL
Final mark = 52.5 mL.
volume used = 52.5 - 2.5
= 50 mL
what is the density of a substance that has a mass of 2.0 g, and when placed in a graduated cylinder the volume rose from 70 mL to 75 mL
Answer:
0.4g/mL
Explanation:
The following data were obtained from the question:
Mass = 2g
Volume = 75 — 70 = 5mL
Density =?
Density = Mass /volume
Density = 2/5
Density = 0.4g/mL
The density of the substance is 0.4g/mL
The density of a substance that has a mass of 2.0 g, and when placed in a graduated cylinder the volume rose from 70 mL to 75 mL is 0.4 g/mL.
To find the density of the substance, we need to calculate the volume it displaces and then divide its mass by that volume.
The volume displaced by the substance can be found by subtracting the initial volume in the graduated cylinder from the final volume after the substance is added. The initial volume is 70 mL and the final volume is 75 mL. Therefore, the volume displaced by the substance is:
Volume displaced = Final volume - Initial volume
Volume displaced = 75 mL - 70 mL
Volume displaced = 5 mL
Now, we have the mass of the substance, which is 2.0 g, and the volume it displaces, which is 5 mL. The density [tex]\rho[/tex] is calculated by dividing the mass (m) by the volume (V):
[tex]\rho = \frac{m}{V}[/tex]
[tex]\rho = \frac{2.0 \text{ g}}{5 \text{ mL}}[/tex]
[tex]\rho = 0.4 g/mL[/tex]
This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH_3CO_2H An analytical chemist has determined by measurements that there are 0.054 moles of oxygen In a sample of acetic acid. How many moles of hydrogen are in the sample? Be sure your answer has the correct number of significant digits.
Answer:
0.108mol of Hydrogen
Explanation:
The formula for the compound is: CH3COOH
From the formula of the compound,
There are 2moles of oxygen and 4moles of Hydrogen.
If for every 2moles of oxygen, 4moles of Hydrogen is present.
Then, for 0.054 moles of oxygen = (0.054 x 4)/2 = 0.108mol of Hydrogen is present
Answer:
For 0.027 moles CH3COOH we have 0.108 moles H ≈ 1.1 *10^-1 moles H
Explanation:
Step 1: Data given
Acetic acid = CH3COOH
Number of moles oxygen in the sample = 0.054 moles
Step 2: calculate moles CH3COOH
In 1 mol CH3COOH we have 2 moles O
For 0.054 moles Oxygen we have 0.054/2 = 0.027 moles CH3COOH
Step 3: Calculate moles H
In 1 mol CH3COOH we have 4 moles H
For 0.027 moles CH3COOH we have 4*0.027 = 0.108 moles H ≈ 1.1 *10^-1 moles
A chemist determined by measurements that 0.050 moles of aluminum participated in a chemical reaction. Calculate the mass of aluminum that participated in the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
13.5 g
Explanation:
This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.
Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.
number of moles = n = mass of Al / Atomic Weight Al
⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹
= 13.5 g
We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.
Mass of aluminum that participated in the chemical reaction is 1.35 grams.
Number of Moles:
Number of moles is defined as the ratio of given mass to the molar mass.
Given:
Moles of Aluminum = 0.050 moles
To find:
Mass of Aluminum=?
As we know, Molar mass of Aluminum = 27g/mol
On substituting the values:
[tex]\text{Number of Moles}=\frac{\text{Given mass}}{\text{Molar mass}} \\\\ \text{Given mass}= \text{Number of Moles}*\text{Molar mass}\\\\ \text{Given mass}= 0.050*27\\\\ \text{Given mass}=1.35\text{ grams}[/tex]
Thus, the mass of aluminum that participated in the chemical reaction is 1.35 grams.
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Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution, You can assume the volume of the solution doesn't change shen th s sove m Be sure your answer has the correct number of significant digits
Calculating the final molarity of iodide anion in the solution, convert the grams of iron (II) iodide to moles, convert the volume of the silver nitrate solution to liters, use the stoichiometry of the reaction, calculate the moles of iodide anion, and divide the moles by the final volume of the solution.
Explanation:To calculate the final molarity of iodide anion in the solution, we need to use the stoichiometry of the reaction between iron (II) iodide and silver nitrate. Given that there is 0.981 g of iron (II) iodide and 150 mL of a 35.0 mM aqueous solution of silver nitrate, we can determine the moles of iron (II) iodide and the moles of iodide anion produced. Finally, we can calculate the final molarity of iodide anion by dividing the moles by the final volume of the solution.
Convert the grams of iron (II) iodide to moles by dividing by its molar mass: 0.981 g / (55.85 g/mol + 2 (126.9 g/mol)) = 0.007 mol. Convert the volume of silver nitrate solution to liters: 150 mL = 0.150 L. Use the stoichiometry of the reaction 2 AgNO3 + FeI2 → 2 AgI + Fe(NO3)2. Since the volume of the solution does not change, the moles of iodide anion produced will be the same as the moles of iron (II) iodide used. Calculate the moles of iodide anion: 0.007 mol I-. Calculate the final molarity of iodide anion by dividing the moles by the final volume: 0.007 mol / 0.150 L = 0.0467 M (rounded to the correct number of significant digits, which is 3). Learn more about Calculating the final molarity of iodide anion in a solution here:
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onsider the following reaction at equilibrium: CO(g) + Cl2(g)=======COCl2(g) Predict whether the reaction will shift left,shift right, or remain the unchanged upon each of the following disturbances.a) COCl2 is added to thereaction mixtureb) Cl2 is added to thereaction mixturec) COCl2 is removed fromthe reaction mixture
Explanation:
CO(g) + Cl2(g) ⇄ COCl2(g)
This question is based on Le Chatelier's principle.
Le Chatelier's principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.
a) COCl2 is added to the reaction mixture
COCL2 is a product in the reaction. If we add additional product to a system, the equilibrium will shift to the left, in order to produce more reactants. The reaction would shift to the left.
b) Cl2 is added to the reaction mixture
if we add reactants to the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.
c) COCl2 is removed from the reaction mixture
if we remove products from the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.
Chemical reactions at equilibrium, such as CO(g) + Cl2(g) ⇌ COCl2(g), shift in response to changes to re-establish equilibrium. If a product or reactant is added, the reaction will respectively shift towards reactants or products. Similarly, if a product or reactant is removed, the reaction will respectively shift towards products or reactants.
Explanation:In chemistry, chemical reactions at equilibrium respond to disturbances according to Le Châtelier's principle: the system shifts in a way that counters the disturbance and re-establishes equilibrium. Now, let's consider the reaction: CO(g) + Cl2(g) ⇌ COCl2(g).
(a) When COCl2 is added, it will increase the product's concentration, thus, the reaction will shift to the left (towards the reactants) to re-establish equilibrium. (b) When Cl2 is added, it will increase the reactant's concentration. The reaction will shift to the right (towards the products) to counter this and re-establish equilibrium. (c) When COCl2 is removed, it decreases the product's concentration. To counteract this, the reaction shifts to the right (towards the products) to re-establish equilibrium.Learn more about Chemical equilibrium here:
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When making a solution of sodium hydroxide and water, a student weighed out an certain amount of sodium hydroxide pellets and dissolved them in an certain amount of water. However, the sodium hydroxide concentration of the resulting solution was lower than the concentration that the student thought they made. What was the problem?
Hygroscopic nature of NaOH is the main reason for the lower concentration of the sodium hydroxide in the solution.
Explanation:
It is a well known fact that Sodium hydroxide pellets are hygroscopic in nature, which clearly means that the sodium hydroxide pellets absorbs moisture from the air, so that it becomes deliquescent. When the NaOH crystals are weighed, the crystals absorb moisture from the surroundings, and so the weight of the crystals may change, so the concentration of the solution was lower than the required one.So it is not possible to prepare NaOH solution under normal room temperature. So, while preparing the solution of NaOH we have to be more careful.The main reason for the lower concentration of the sodium hydroxide in the solution is Hygroscopic nature of NaOH.
Nature of Sodium hydroxide:It's obviously true that Sodium hydroxide pellets are hygroscopic in nature, which plainly implies that the sodium hydroxide pellets retains dampness from the air, so it becomes deliquescent. Whenever the NaOH precious stones are gauged, the gems assimilate dampness from the environmental factors, thus the heaviness of the gems might change, so the centralization of the arrangement was lower than the expected one. So it is absurd to expect to get ready NaOH arrangement under ordinary room temperature. In this way, while setting up the arrangement of NaOH we must be more cautious.
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The Br⊘⊘nsted-Lowry model focuses on the transfer of _______ in an acid-base reaction. The Brnsted-Lowry model focuses on the transfer of _______ in an acid-base reaction. electrons neutrons OH−OH− orbitals H+H+
Answer:
The Brönsted-Lowry model focuses on the transfer of H⁺ in an acid-base reaction.
Explanation:
According to the theory of Brönsted-Lowry , an acid is a chemical substance capable of releasing hydrogen ions, while a base is that chemical substance capable of accepting hydrogen ions. That is, acids are substances capable of yielding protons (hydrogen H + ions) and substance bases capable of accepting them.
On the other hand, the conjugate base of a Brønsted-Lowry acid is the species that forms after an acid donated a proton. The conjugate acid of a Brønsted-Lowry base is the species that forms when a base accepts a proton.
Thus, the acid-base reaction is one in which the acid transfers a proton to a base (proton transfer H⁺).
This theory, unlike another theory like Arrhenius, does not require the presence of water as a means of reaction for the transfer of H⁺.
The Brönsted-Lowry model focuses on the transfer of H⁺ in an acid-base reaction.
Which of the following is TRUE? Group of answer choices None of the above is true. The equivalence point is where the amount of acid equals the amount of base during any acid-base titration. An indicator is not pH sensitive. A titration curve is a plot of pH vs. the [base]/[acid] ratio. At the equivalence point, the pH is always 7.
Answer:
TRUE: The equivalence point is where the amount of acid equals the amount of base during any acid-base titration.
Explanation:
The point on the titration curve where the number of base equivalents added equals the number of acid equivalents is the equivalence point or neutralization point.
Chemical indicators are substances that change color thanks to a chemical change, depending on the pH of the medium, and thus indicate the end point or point of equivalence of an acid-base volumetry.
A titration curve occurs by representing the measured pH as a function of the added volume of titrant, where the rapid change in pH for a given volume is observed. The inflection point of this curve is called the equivalence point and its volume indicates the volume of titrant consumed to fully react with the analyte.
In some cases, there are multiple equivalence points that are multiples of the first equivalence point, as in the valuation of a diprotic acid, which indicates that its pH value will not always be 7.
Calculate the pH for each case in the titration of 50.0 mL of 0.220 M HClO ( aq ) with 0.220 M KOH ( aq ) . Use the ionization constant for HClO . What is the pH before addition of any KOH ?
Answer:
Before adding any KOH, the pH is 4.03
Explanation:
Step 1: Data given
Volume of a 0.220 M HClO = 50.0 mL = 0.050 L
Molarity of KOH = 0.220 M
The ionization constant for HClO is 4.0*10^–8
Step 2: The balanced equation
HClO + KOH → KClO + H2O
Step 3: pH before any addition of KOH
When no KOH is added, we only have HClO, a weak acid.
To calculate the pH of a weak acid, we need the Ka
Ka = [H+] / [acid]
4.0*10^-8 = [H+]² / 0.220
[H+]² = (4.0*10^-8 ) * 0.220
[H+]² = 8.8*10^-9
[H+] = √( 8.8*10^-9)
[H+] = 9.38*10^-5 M
pH = -log [H+]
pH = -log(9.38*10^-5)
pH = 4.03
Before adding any KOH, the pH is 4.03
Final answer:
To calculate the pH before the addition of KOH in the titration of HClO with KOH, we can use the ionization constant for HClO (Ka). The initial concentration of HClO can be used to calculate the concentration of H+, which in turn can be used to calculate the pH using the pH formula.
Explanation:
The pH before the addition of any KOH can be calculated using the ionization constant for HClO. HClO is a weak acid, so we can use the expression for the acid dissociation constant, Ka, to calculate the pH. The expression for Ka for HClO is:
Ka = [H+][ClO-] / [HClO]
Since we know the initial concentration of HClO, we can assume that the concentration of H+ is equal to the initial concentration of HClO. Therefore, we can rewrite the expression for Ka as:
Ka = [H+]² / [HClO]
Now we can calculate the concentration of H+ using the given initial concentration of HClO:
[H+] = sqrt(Ka * [HClO])
Finally, we can use the concentration of H+ to calculate the pH using the pH formula:
pH = -log[H+]