Answer:
i) The pCa before initiating the titration is 2.6
ii) The pCa is 6.67
Explanation:
please look at the solution in the attached Word file
Draw the correct structure for (3S,4S)-3,4-dimethylhexane. Show stereochemistry clearly. To ensure proper grading, explicitly draw all four groups, including wedge/dash bonds, around a chirality center. Indicate whether the compound could exist in an optically active form.
Find the figure in the attachment
Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that you have 125 mL of a buffer that is 0.360 M in both acetic acid ( CH 3 COOH ) and its conjugate base ( CH 3 COO − ) . Calculate the maximum volume of 0.300 M HCl that can be added to the buffer before its buffering capacity is lost.
Answer:
The maximum volume is 122.73 mL
Explanation:
125 mL of butter that is 0.360 M in CH₃COOH and CH₃COO⁻.
pKa of acetic acid = 4.76
Using Henderson Hasselbalch equation.
pH =pKa + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]
= 4.76 + [tex]log \frac{0.3600}{0.3600}[/tex]
= 4.76
Assuming the pH of the buffer changes by a unit of 1 , then it will lose its' buffering capacity.
When HCl is added , it reacts with CH₃COO⁻ to give CH₃COOH.
However, [CH₃COOH] increases and the log term results in a negative value.
Let assume, the new pH is less than 4.76
Let say 3.76; calculating the concentration when the pH is 3.76; we have
3.76 = 4.76 + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]
log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 4.76 - 3.76
log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 1.00
[tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 0.1
Let number of moles of acid be x (i.e change in moles be x); &
moles of acetic acid and conjugate base present be = Molarity × Volume
= 0.360 M × 125 mL
= 45 mmol
replacing initial concentrations and change in the above expression; we have;
[tex]\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.1[/tex]
[tex]\frac{45-x}{45+x} =0.1[/tex]
0.1(45 + x) = 45 - x
4.5 + 0.1 x = 45 -x
0.1 x + x = 45 -4.5
1.1 x = 40.5
x = 40.5/1.1
x = 36.82
So, moles of acid added = 36.82 mmol
Molarity = 0.300 M
So, volume of acid = [tex]\frac{moles}{molarity }[/tex] = [tex]\frac{36.82 \ mmol}{0.300}[/tex]
= 122.73 mL
An acetate buffer solution is prepared by combining 50. mL of a 0.20 M acetic acid, HC2H3O2 (aq), and 50. mL of 0.20 M sodium acetate, NaC2H3O2 (aq). A 5.0 mL sampleof 0.10 M NaOH (aq) is added to the buffer solution. Which of the following is a correct pairing of the acetate species present in greater concentration and of the pH of the solution after the NaOH (aq) is added? (the pKa of acetic acid is 4.7)
A. HC2H3O2 pH < 4.7
B. HC2H3O2 pH >4.7
C. C2H3O2- pH < 4.7
D. C2H3O2- pH > 4.7
Answer:
B. CH3COOH pH > 4.7 (4.8)
Explanation:
CH3COOH + NaOH ↔ CH3COONa + H2OCH3COONa + NaOH ↔ CH3COONa∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol
⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol
⇒ C CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)
⇒ C CH3COOH = 0.0905 M
∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol
⇒ C CH3COONa = (0.01 mol + 5 E-4 mol) / (0.105 L )
⇒ C CH3COONa = 0.1 M
∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5
⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])
⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]
⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0
⇒ [H3O+] = 1.5835 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = - Log (1.5835 E-5)
⇒ pH = 4.8004 > 4.7
Final answer:
After adding NaOH to the acetate buffer, the acetate ion concentration increases and the pH of the solution will be greater than the pKa of acetic acid (4.7). Option D, with C2H3O2- and pH > 4.7, is correct.
Explanation:
The question involves calculating the pH of an acetate buffer solution after the addition of sodium hydroxide, NaOH. The buffer consists of acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2). The pKa of acetic acid is 4.7, and equal concentrations of the buffer components are typically ideal. When NaOH is added, it neutralizes some of the acetic acid, converting it to acetate and increasing the pH slightly.
Before the addition of NaOH, the buffer has equal concentrations of acetic acid and acetate, so its pH is close to the pKa of acetic acid, which is 4.7. After adding NaOH, more acetic acid is converted to acetate, but the pH will still be maintained close to the pKa since the buffer action resists changes in pH. Therefore, the correct pairing is that the acetate ion (C2H3O2-) will be present in greater concentration and the pH of the solution will be greater than 4.7 after NaOH is added, making option D correct.
The following information is given for ether, C2H5OC2H5, at 1atm: boiling point = 34.6 °C Hvap(34.6 °C) = 26.5 kJ/mol specific heat liquid = 2.32 J/g°C /At a pressure of 1 atm, what is H in kJ for the process of condensing a 22.5 g sample of gaseous ether at its normal boiling point of 34.6 °C.
Answer:
The heat is [tex]H= -8.044KJ[/tex]
Explanation:
From the question we are told that
The pressure is [tex]P = 1 \ atm[/tex]
The boiling point is [tex]B_P = 34^oC[/tex]
The heat of vaporization at 34°C is = [tex]26.5 kJ/mol[/tex]
The specific heat of the liquid is [tex]c_p = 2.32 J/g^oC[/tex]
The mass is [tex]m = 22.5g[/tex]
The no of moles of the sample of [tex]C_2 H_5OC_2H_5[/tex] is given as
[tex]No \ mole (n) = \frac{Mass \ of \ sample }{Molar \ mass }[/tex]
The Molar mass for [tex]C_2 H_5OC_2H_5[/tex] is a value = [tex]= 74.12 g/mol[/tex]
Substituting the value into the above equation
[tex]n = \frac{22.5}{74.12}[/tex]
[tex]= 0.30356 \ mol[/tex]
The heat H is mathematically as
[tex]H =- nH_{vap}[/tex]
The negative sign show that the heat is for condensing
[tex]H = 0.30356 * 26.5 *10^{3} J/mol[/tex]
[tex]H= -8.044KJ[/tex]
The enthalpy change for condensing a 22.5 g sample of gaseous ether at its normal boiling point of 34.6 °C is approximately -8.045 kJ.
The question asks about the enthalpy change (H) during the condensation of a sample of ether. To calculate the enthalpy change for condensing 22.5 g of diethyl ether at its boiling point, we'll use the given enthalpy of vaporization ([tex]H_{vap}[/tex]) and the mass of the ether. Considering that the enthalpy of vaporization ([tex]H_{vap}[/tex]) is the amount of energy required to vaporize 1 mole of a substance at its boiling point, the enthalpy of condensation will be the negative of this value since condensation is the reverse process.
First, we convert the mass of ether to moles:
22.5 g * (1 mol / 74.12 g/mol) ≈ 0.3036 mol
Next, multiply the moles by the enthalpy of vaporization (Hvap) but with a negative sign for condensation:
-0.3036 mol * 26.5 kJ/mol ≈ -8.045 kJ
This is the enthalpy change for the condensation process of the ether sample.
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) a. What is the limiting reagent when 9.05 g of NH3 reacted with 45.2 g of CuO?(5 points) b. How many grams of N2 can be made?(10 points) c. If 4.61 g of N2 are made, what is the percent yield? (5 points) d. What is the mass of the excess reactant that remains after the reaction. (10 points)
Answer:
a. Copper (II) oxide is the limiting reactant.
b. [tex]m_{N_2}=5.30g[/tex]
c. [tex]Y=87\%[/tex]
d. [tex]m_{NH_3}=2.60gNH_3[/tex]
Explanation:
Hello,
In this case, for the given reaction:
[tex]2 NH_3(g) + 3 CuO(s) \rightarrow N_2(g) + 3 Cu(s) + 3 H_2O(l)[/tex]
a. The limiting reactant is identified by computing the available moles of ammonia and the moles of ammonia that react with 45.2 g of copper (I) oxide as shown below:
[tex]n_{NH_3}^{Available}=9.05gNH_3*\frac{1molNH_3}{17gNH_3}=0.532molNH_3\\n_{NH_3}^{Reacted}=45.2gCuO*\frac{1molCuO}{79.545gCuO}*\frac{2molCuO}{3molCuO} =0.379molNH_3[/tex]
In such a way, as there more ammonia available than that is reacted, we say it is in excess and the copper (II) oxide the limiting reactant.
b. Here, with the reacting moles of ammonia, we compute the yielded grams of nitrogen:
[tex]m_{N_2}=0.379molNH_3*\frac{1molN_2}{2molNH_3}*\frac{28gN_2}{1molN_2}\\m_{N_2}=5.30g[/tex]
c. Now, since the 5.30 g of nitrogen are the expected grams of it, the percent yield of nitrogen is compute by dividing the real obtained mass over the theoretical previously computed mass:
[tex]Y=\frac{4.61g}{5.30g} *100\%\\Y=87\%[/tex]
d. Finally, as 0.532 moles of ammonia are available, but just 0.379 moles react, the unreacted moles are:
[tex]n_{NH_3}=0.532molNH_3-0.379molNH_3=0.153molNH_3[/tex]
That in grams are:
[tex]m_{NH_3}=0.153molNH_3*\frac{17gNH_3}{1molNH_3}=2.60gNH_3[/tex]
Best regards.
The limiting reactant is CuO, the percent yield of nitrogen gas is 29%.
The equation of the reaction is;
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
Number of moles of NH3 = 9.05 g/17 g/mol = 0.53 moles
Number of moles of CuO = 45.2 g/80 g/mol = 0.565 moles
Since 2 moles of NH3 reacts with 3 moles of CuO
0.53 moles of NH3 reacts with 0.53 moles × 3 moles/ 2 moles
= 0.795 moles
We can see that there is not enough CuO in the system hence it is the limiting reactant.
Number of moles of N2 produced = 0.565 moles × 28 g/mol =15.82 g of N2
Percent yield = Actual yield/Theoretical yield × 100/1
Percent yield = 4.61 g/15.82 g × 100/1
Percent yield = 29%
If 2 moles of NH3 reacts with 3 moles of CuO
x moles of NH3 reacts with 0.565 moles of CuO
x = 2 moles × 0.565 moles/3 moles
x = 0.38 moles
Number of moles of excess reactant left over = 0.53 moles - 0.38 moles
= 0.15 moles
Mass of excess reactant left over = 0.15 moles × 17 g/mol = 2.55 g
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Show that the ideal c/a ratio (height of unit cell divided by edge length) for the HCP unit cell is 1.633. (You may wish to refer to Exercise E.2 in the text page GL 1-6) Comment on the fact that real HCP metals display c/a ratios varying from 1.58 (for Be) to 1.89 (Cd). The atomic radius of HCP Mg is 0.1605 nm. Find the lattice constants, c and a, the c/a ratio and theoretical density for Mg.
Answer:
The structure in the first image file attached below shows the arrangement of atoms in hexagonal close packing
we need to show that the ratio between the height of the unit cell divided by its edge length is 1.633
In the structure, the two atoms are shown apart. But in fact the two atoms are touching. Therefore, the edge length is the sum of the radius of two atoms. If we assume r as radius then the expression for edge length will be as follows.
a = 2r ..............(1)
other attached files show additional solutions in steps
The ideal c/a ratio in an HCP unit cell is 1.633, depicting the most compact arrangement of spheres. However, real HCP metals, influenced by bonding characteristics and atomic sizes, display varying c/a ratios. For Mg with known atomic radius in HCP arrangement, the lattice constants, c/a ratio, and theoretical density can be calculated.
Explanation:The HCP unit cell (hexagonal-close-packed) is one of the simplest lattice types in crystal structures. It includes two types of atoms, one at the corners (and center) of a hexagonal prism and the other at the top and bottom of the prism. The ideal ratio of c/a (height of the unit cell c divided by edge length a) corresponds to the most compact possible arrangement of spheres, and is theoretically calculated to be √(8/3), or approximately 1.633.
In real-world applications, various HCP metals display c/a ratios varying slightly, such as 1.58 for Be and 1.89 for Cd. This depends on bonding characteristics and varying atomic sizes in the metal crystal structure. For example, for Mg (magnesium) which has a known atomic radius of 0.1605 nm in HCP arrangement, we can calculate the lattice constants, c and a, and thus the c/a ratio. Theoretical density can be determined using the mass of atoms in a unit cell, Avogadro's number, and the volume of unit cell.
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6. Air at 300°C and 130kPa ows through a horizontal 7-cm ID pipe at a velocity of 42.0m/s. (a) Calculate _ Ek W, assuming ideal-gas behavior. (b) If the air is heated to 400°C at constant pressure, what is Δ _ Ek _ Ek 400°C _ Ek 300°C? (c) Why would it be incorrect to say that the rate of transfer of heat to the gas in Part (b) must equal the rate of change of kinetic energy?
Answer:
Check the explanation
Explanation:
kindly check the attached image below to see the answer to question A and B
(c) The energy balance equation if the pressure is constant (ΔE=0) is,
ΔE= ΔU + Δ[tex]E_{k}[/tex]
From the above relation, it is clear that some heat energy used to raises the temperature of the air.
Hence, the internal energy is not equal to zero. Therefore, the rate of transfer of heat to air is not equal to the rate of thane in kinetic energy of the air.
That is ΔE ≠ Δ[tex]E_{k}[/tex]
Answer:
a) [tex]\delta E_k[/tex] = 112.164 W
b) [tex]\delta E_k[/tex] = 42.567 W
c) From what we've explained in part b; increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.
Explanation:
The given data include:
Inlet diameter [tex](d_1)[/tex] = 7 cm = 0.7 m
Inlet velocity [tex](v_1)[/tex] = 42 m/s
Inlet pressure [tex](P_1)[/tex] = 130 KPa
Inlet temoerature [tex](T_1)[/tex] = 300°C = (300 + 273.15) = 573.15 K
a) Assuming Ideal gas behaviour
Inlet Volumetric flowrate [tex](V_1)[/tex] = Inlet velocity [tex](v_1)[/tex] × area of the tube
[tex]= v_1*(\frac{\pi}{4})0.07^2\\\\= 0.161635 \ m^3/s[/tex]
Using Ideal gas law at Inlet 1
[tex]P_1V_1 =nRT_1[/tex]
where ; n = molar flow rate of steam
making n the subject of the formula; we have:
[tex]n = \frac{P_1V_1}{RT_1}[/tex]
[tex]n = \frac{130*42}{8.314*573.15}[/tex]
[tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]
Moleular weight of air = 28.84 g/mol
The mass flow rate = molar flowrate × molecular weight of air
[tex]= 4.4096*10^{-3} \ * 28.84\\= 0.12717 \ kg/s[/tex]
Finally: the kinetic energy at Inlet [tex](\delta E_k) = \frac{1}{2}*mass \ flowrate *v_1^2[/tex]
= [tex]0.5*0.12717*42^2[/tex]
= 112.164 W
b) If the air is heated to 400°C;
Then temperature at 400°C = (400 + 273.15)K = 673.15 K
Thee pressure is also said to be constant ;
i.e [tex]P_1= P_2[/tex] = 130 KPa
Therefore; the mass flow rate is also the same ; so as the molar flow rate:
Thus; [tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]
Using Ideal gas law at Inlet 2
[tex]P_2V_2 = nRT_2[/tex]
making [tex]V_2[/tex] the subject of the formula; we have:
[tex]V_2 = \frac{nRT_2}{P_2}[/tex]
[tex]V_2 = \frac{4.4906*10^{-3}*8.314*673.15}{130}[/tex]
[tex]V_2 = 0.189836 \ m^3/s[/tex]
Assuming that the diameter is constant
[tex]d_1 = d_2 = 0.07 \ cm[/tex]
Now; the velocity at outlet = [tex]\frac{V_2}{\frac{\pi}{4}d_2^2}[/tex]
= [tex]\frac{0.0189836}{\frac{\pi}{4}(0.07)^2}[/tex]
= 49.33 m/s
Change in kinetic energy [tex]\delta E_k[/tex] = [tex]\frac{1}{2}*mass \ flowrate * \delta V[/tex]
= [tex]0.5*0.12717 *(49.33^2 -42^2)[/tex]
= 42.567 W
c).
From what we've explained in part b; increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.
What is the molar mass of Cro2
Answer: The Molar mass of Cro2 (or Chromium (IV) oxide) is 83.9949 g/mol.
Explanation: The atomic mass of Chromium is 51.9961 and the atomic mass of Oxygen is 15.9994. You then add both numbers and multiply by 2. Put it in the calculator like this: 51.9961 + 15.9994 x 2 and you will get 83.9949 g/mol.
How many moles are in 12.0 grams of O2
Answer:
Moles = 0.375
Explanation:
Moles= m/M
= 12/32 = 0.375mol
There are approximately 0.375 moles of O₂ in 12.0 grams of O₂.
To find the number of moles in a given mass of a substance, one can use the formula:
[tex]\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \][/tex]
The molar mass of oxygen (O₂) is approximately 32.0 g/mol, since one oxygen atom has an atomic mass of approximately 16.0 g/mol and O₂ has two oxygen atoms.
Given the mass of O₂ is 12.0 g, we can calculate the moles as follows:
[tex]\[ \text{moles of \ } {O_2} = \frac{12.0 \text{ g}}{32.0 \text{ g/mol}} \][/tex]
[tex]\[ \text{moles of} \ {O_2} = 0.375 \text{ mol} \][/tex]
Therefore, there are 0.375 moles of O₂ in 12.0 grams of O₂.
2. A mixture of 60 wt% benzene, 35% toluene, and 5% naphthalene is being distilled. The distillate product should be 99.5 wt% benzene. Also, 99% of benzene fed should be recovered in the distillate, Determine/calculate: a. Distillate and bottom product flow rates (D and B) per 1 kg of feed b. compositions of distillate and bottom product c. fraction of toluene fed that is recovered in the bottom product
Answer:
(a) Distillate (D) = 0.597 kg
Bottom product (B) = 0.403 kg
(b) Compositions of distillate product:
Benzene = 0.995
Toluene = 0.05
Compositions of bottom product:
Benzene = 0.0149
Naphthalene = 0.1241
Toluene = 0.8610
(c) Fraction of toluene fed that is recovered in the bottom product = 99.14%
Explanation:
See the attached file for the calculation
Could the structures below undergo a Fischer esterification reaction? Reaction scheme of benzoic acid and thionyl chloride to form an unknown product. No, the structures above cannot undergo a Fischer esterification reaction to form an ester. Yes, if the carboxylic acid is heated in the SOCl2 reagent and a small amount of H2SO4 is used as a catalyst.
Answer:
The correct answer is the first option. No, the structures above cannot undergo a Fischer esterification reaction to form an ester.
Explanation:
The reaction that will take place can be found in the attached file. The reaction does not require any catalyst and it cannot undergo a Fischer esterification reaction to form an ester.
The accurate answer is the first option which is No, the structures above cannot undergo a Fischer esterification response to form an ester.
What is Fischer esterification reaction?
The reaction that will take the spot can be found in the connected file. The reaction does not require any catalyst and it cannot experience a Fischer esterification reaction to create an ester.
Therefore the correct option is No, the structures above cannot experience a Fischer esterification reaction to form an ester.
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Construct a simulated 1H NMR spectrum for 1-chloropropane by dragging and dropping the appropriate splitting patterns into the boxes on the chemical shift baseline, and by dragging integration values into the small box above each signal. Items may be used more than once. Peak heights do not represent integration.
To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Follow these steps: identify the different types of hydrogen atoms, determine the chemical shifts, assign integration values, and drag and drop splitting patterns.
Explanation:To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Here's a step-by-step guide:
Identify the different types of hydrogen atoms in 1-chloropropane. In this molecule, there are three types: H on the methyl group, H on the second carbon, and H on the third carbon.Determine the chemical shifts of the hydrogen atoms. The chemical shift values for H on the methyl group, H on the second carbon, and H on the third carbon are typically around 0.9-1.2 ppm, 1.2-1.4 ppm, and 2.5-3.0 ppm, respectively.Assign integration values to the signals. The integration values represent the relative number of hydrogen atoms in each type. For example, the H on the methyl group typically has an integration value of 3, indicating that there are three hydrogen atoms in the methyl group.Drag and drop the appropriate splitting patterns into the boxes on the chemical shift baseline. The splitting pattern depends on the neighboring hydrogen atoms and follows the n+1 rule. For example, if a hydrogen atom has two neighboring hydrogen atoms, it will show a triplet splitting pattern.Learn more about simulated 1H NMR spectrum for 1-chloropropane here:https://brainly.com/question/9812005
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An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm absolute. State a quick way to show thatthe system has two phases. Estimate (i) the molar compositions of each phase, (ii) thepercentage of the total moles in the vessel that are in the vapor phase, and (iii) thepercentage of the vessel volume occupied by the vapor phase.
The question is incomplete, the table of the question is given below
Answer:
I) xA= 0.34, yA= 0.55
ii) 76.2 mole % vapor
iii) Percentage of vapor volume = 98%
Explanation:
i) xA= 0.34, yA= 0.55
xA= 0.34, yA= 0.55
ii) 0.50 = 0.55 nv + 0.34 nL
Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid
This shows 76.2 mole % vapor
iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3
Therefore, ρ = 0.790 g/cm3
Now, we have:
MA = 58.08 g/mol and ME= 46.07 g/mol
So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol
1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor
Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3
Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3
Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %
Trehalose, sophorose and turanose are disaccharides. Trehalose is found in the cocoons of some insects, sophorose turns up in a few bean varieties, and turanose is an ingredient in low-grade honey made by bees with indigestion from a diet of pine tree sap. Identify trehalose, sophorose, and turanose (may be none of the above) among the following structures on the basis of the following information.
a. Turanose and sophorose are reducing sugars. Trehalose is non-reducing.
b. Upon hydrolysis, sophorose and trehalose give two molecules each of aldoses. Turanose gives one molecule of an aldose and one molecule of a ketose.
c. The two aldoses that constitute sophorose are anomers to each other.
Answer:
First structure is sophorose, second structure is turanose, third structure is trehalose, fourth structure is none of the above.
Trehalose is a non-reducing disaccharide, similar to sucrose, with two glucose units linked at their anomeric carbons. Sophorose is a reducing sugar yielding two glucose anomers upon hydrolysis. Turanose is also a reducing sugar but yields one molecule each of glucose (aldose) and fructose (ketose) upon hydrolysis.
Explanation:Identifying trehalose, sophorose, and turanose among the disaccharides requires understanding their structural characteristics and behavior upon hydrolysis. Trehalose is a non-reducing sugar, which means it doesn't have a free anomeric carbon that can form an open-chain aldehyde or ketone upon hydrolysis. From the given information, we know that trehalose upon hydrolysis gives two molecules of aldoses, which means it must have two glucose units. As trehalose is a non-reducing sugar, it is similar to sucrose in having a glycosidic bond between the two anomeric carbons.
Sophorose is a reducing sugar that upon hydrolysis yields two molecules of an aldose that are anomers of each other, implying the presence of two glucose molecules. Given that it is reducing, at least one of the anomeric carbons must not be involved in the glycosidic linkage, allowing it to open and reduce other compounds.
Turanose, like sophorose, is a reducing sugar, but it gives one molecule of an aldose and one molecule of a ketose upon hydrolysis. This indicates that turanose is composed of one glucose and one fructose molecule. The glucose component would be linked through its anomeric carbon, leaving the fructose part with a potential ketose structure.
Calculate the amount of energy in kilojoules needed to change 369 gg of water ice at −−10 ∘C∘C to steam at 125 ∘C∘C. The following constants may be useful: Cm (ice)=36.57 J/(mol⋅∘C)Cm (ice)=36.57 J/(mol⋅∘C) Cm (water)=75.40 J/(mol⋅∘C)Cm (water)=75.40 J/(mol⋅∘C) Cm (steam)=36.04 J/(mol⋅∘C)Cm (steam)=36.04 J/(mol⋅∘C) ΔHfus=+6.01 kJ/molΔHfus=+6.01 kJ/mol ΔHvap=+40.67 kJ/molΔHvap=+40.67 kJ/mol Express your answer with the appropriate units.
Answer:
We need 1136.4 kJ energy
Explanation:
Step 1: Data given
The mass of water = 369 grams
the initial temperature = -10°C
The finaltemperature = 125 °C
Cm (ice)=36.57 J/(mol⋅∘C)
Cm (water)=75.40 J/(mol⋅∘C)
Cm (steam)=36.04 J/(mol⋅∘C)
ΔHfus=+6010 J/mol
ΔHvap=+40670 J/mol
Step 2: :Calculate the energy needed to heat ice from -10 °C to 0°C
Q = n*C*ΔT
⇒Q = the energy needed to heat ice to 0°C
⇒with n = the moles of ice = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of ice = 36.57 / J/mol°C
⇒ ΔT = the change of temperature = 10°C
Q = 20.48 moles * 36.57 J/mol°C * 10°C
Q = 7489.5 J = 7.490 kJ
Step 3: calculate the energy needed to melt ice to water at 0°C
Q = n* ΔHfus
Q = 20.48 moles * 6010 J/mol
Q = 123084.8 J = 123.08 kJ
Step 4: Calculate energy needed to heat water from 0°C to 100 °C
Q = n*C*ΔT
⇒Q = the energy needed to heat water from 0°C to 100 °C
⇒with n = the moles of water = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of water =75.40 J/(mol⋅∘C)
⇒ ΔT = the change of temperature = 100°C
Q = 20.48 moles * 75.40 J/mol°C* 100°C
Q = 154419.2 J = 154.419 kJ
Step 5: Calculate energy needed to vapourize water to steam at 100°C
Q = 20.48 moles * 40670 J/mol
Q = 832921.6 J = 832.922 kJ
Step 6: Calculate the energy to heat steam from 100 °C to 125 °C
⇒Q = the energy needed to heat steam from 100 °C to 125 °C
⇒with n = the moles of steam = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of steam = 36.04 J/(mol⋅∘C)
⇒ ΔT = the change of temperature = 25 °C
Q = 20.48 moles * 36.04 J/mol*°C * 25°C
Q = 18452.5 J = 18.453 kJ
Step 7: Calculate total energy needed
Q = 7489.5 J + 123084.8 J + 154419.2 J + 832921.6 J + 18452.5 J
Q = 1136367.6 J
Q = 1136.4 kJ
We need 1136.4 kJ energy
The amount of energy needed is [tex]\( \\1120 \text{ kJ}} \).[/tex]
To calculate the amount of energy required to change 369 g of water from ice at[tex]\(-10^\circ \text{C}\)[/tex] to steam at [tex]\(125^\circ \text{C}\),[/tex] we need to consider the different phases of water and the energy required for each phase change and temperature change.
Let's break down the calculation step by step.
1. Energy to heat ice from [tex]\(-10^\circ \text{C}\) to \(0^\circ \text{C}\).[/tex]
First, calculate the energy required to heat the ice initially at[tex]\(-10^\circ \text{C}\)[/tex] to its melting point [tex]( \(0^\circ \text{C}\) ).[/tex]
Specific heat capacity of ice [tex](\(C_m\)): \(36.57 \text{ J/(mol} \cdot \text{°C)}\).[/tex]
Molar mass of water (H₂O). [tex]\(18.015 \text{ g/mol}\).[/tex]
Amount of ice in moles:
[tex]\[ \text{moles of ice} = \frac{369 \text{ g}}{18.015 \text{ g/mol}} \approx 20.49 \text{ mol} \][/tex]
Energy required to heat ice.
[tex]\[ Q_1 = n \times C_m \times \Delta T = 20.49 \text{ mol} \times 36.57 \text{ J/(mol} \cdot \text{°C)} \times (0^\circ \text{C} - (-10^\circ \text{C})) = 7485.39 \text{ J} \][/tex]
[tex]\( Q_1 = 7.48539 \text{ kJ} \)[/tex]
2. Energy to melt the ice at [tex]\(0^\circ \text{C}\).[/tex]
Latent heat of fusion [tex](\(\Delta H_{\text{fus}}\)) for ice: \(6.01 \text{ kJ/mol}\)[/tex].
Energy required to melt ice:
[tex]\[ Q_2 = \text{moles of ice} \times \Delta H_{\text{fus}} = 20.49 \text{ mol} \times 6.01 \text{ kJ/mol} = 123.05 \text{ kJ} \][/tex]
[tex]\( Q_2 = 123.05 \text{ kJ} \)[/tex]
3.Energy to heat water from[tex]\(0^\circ \text{C}\) to \(100^\circ \text{C}\)[/tex].
Specific heat capacity of water[tex](\(C_m\))[/tex]: [tex]\(75.40 \text{ J/(mol} \cdot \text{°C)}\)[/tex].
Energy required to heat water:
[tex]\[ Q_3 = \text{moles of water} \times C_m \times \Delta T = 20.49 \text{ mol} \times 75.40 \text{ J/(mol} \cdot \text{°C)} \times (100^\circ \text{C} - 0^\circ \text{C}) = 155,297.80 \text{ J} \][/tex] [tex]\( Q_3 = 155.2978 \text{ kJ} \)[/tex]
4. Energy to vaporize water at [tex]\(100^\circ \text{C}\)[/tex]:
Latent heat of vaporization [tex](\(\Delta H_{\text{vap}}\)) for water: \(40.67 \text{ kJ/mol}\).[/tex]
Energy required to vaporize water:
[tex]\[ Q_4 = \text{moles of water} \times \Delta H_{\text{vap}} = 20.49 \text{ mol} \times 40.67 \text{ kJ/mol} = 833.34 \text{ kJ} \][/tex]
[tex]\( Q_4 = 833.34 \text{ kJ} \)[/tex]
5. Energy to heat steam from [tex]\(100^\circ \text{C}\) to \(125^\circ \text{C}\):[/tex]
Specific heat capacity of steam [tex](\(C_m\)): \(36.04 \text{ J/(mol} \cdot \text{°C)}\).[/tex]
Energy required to heat steam:
[tex]\[ Q_5 = \text{moles of steam} \times C_m \times \Delta T = 20.49 \text{ mol} \times 36.04 \text{ J/(mol} \cdot \text{°C)} \times (125^\circ \text{C} - 100^\circ \text{C}) = 1831.79 \text{ J} \][/tex]
[tex]\( Q_5 = 1.83179 \text{ kJ} \)[/tex]
6. Total energy required.
Summing up all the energy contributions.[tex]\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 7.48539 \text{ kJ} + 123.05 \text{ kJ} + 155.2978 \text{ kJ} + 833.34 \text{ kJ} + 1.83179 \text{ kJ} = 1120.00598 \text{ kJ} \][/tex]
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant for AgCl at 298 K is 1.77 × 10 − 10 and the standard reduction potential of the half‑reaction Ag + ( aq ) + e − − ⇀ ↽ − Ag ( s ) is + 0.799 V .
Answer: The value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.
Explanation:
Equation for solubility equilibrium is as follows.
[tex]AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)[/tex]
Its solubility product will be as follows.
[tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]
Cell reaction for this equation is as follows.
[tex]Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)[/tex]
Reduction half-reaction: [tex]Ag^{+} + 1e^{-} \rightarrow Ag(s)[/tex], [tex]E^{o}_{Ag^{+}/Ag} = 0.799 V[/tex]
Oxidation half-reaction: [tex]Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}[/tex], [tex]E^{o}_{AgCl/Ag}[/tex] = ?
Cell reaction: [tex]Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)[/tex]
So, for this cell reaction the number of moles of electrons transferred are n = 1.
Solubility product, [tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]
= [tex]1.77 \times 10^{-10}[/tex]
Therefore, according to the Nernst equation
[tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]
At equilibrium, [tex]E_{cell}[/tex] = 0.00 V
Putting the given values into the above formula as follows.
[tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]
[tex]0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}[/tex]
[tex]E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}[/tex]
= [tex]0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}[/tex]
= 0.577 V
Hence, we will calculate the standard cell potential as follows.
[tex]E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}[/tex]
[tex]0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}[/tex]
[tex]0.577 V = 0.799 V - E^{o}_{AgCl/Ag}[/tex]
[tex]E^{o}_{AgCl/Ag}[/tex] = 0.222 V
Thus, we can conclude that value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.
The E° for the half-reaction can be calculated by using the Nernst equation, solubility product constant (Ksp) and the standard reduction potential for Ag+(aq) + e− ⇀ ↽ − Ag(s) which is +0.799 V.
Explanation:The E° for the half-reaction, AgCl (s) + e− ⇀ ↽ − Ag (s) + Cl− (aq) can be calculated using the Nernst equation, considering the relationship between standard reduction potential and solubility product constant.
First, we can rewrite the half-reaction for the dissolution of AgCl: AgCl(s) ⇀ ↽ − Ag+(aq) + Cl−(aq). The solubility product constant (Ksp) for this reaction is 1.77 × 10^−10.
We also know that the standard reduction potential of Ag+(aq) + e− ⇀ ↽ − Ag(s) is +0.799 V.
We can relate this data through the Nernst equation: E=E°−(RT/nF)lnQ. Here, we can set Q as Ksp and adjust the equation to solve for our wanted E.
Using these values in Nernst equation would yield the E for the half-reaction.
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What parts of a circuit is a conductor?
Answer:
copper
Explanation:
These are usually copper wires with no insulation. They make the path through which the electricity flows. One piece of the wire connects the current from the power source (cell) to the load
Write a balanced equation for the decomposition reaction described, using the smallest possible integer coefficients. When calcium carbonate decomposes, calcium oxide and carbon dioxide are formed.
Answer:
1. The balanced equation is given below:
CaCO3 —> CaO + CO2
2. The coefficients are: 1, 1, 1
Explanation:
Step 1:
The word equation is given below:
calcium carbonate decomposes to produce calcium oxide and carbon dioxide.
Step 2:
The elemental equation. This is illustrated below:
Calcium carbonate => CaCO3
calcium oxide => CaO
carbon dioxide => CO2
CaCO3 —> CaO + CO2
A careful observation of the above equation shows that the equation is balanced.
The coefficients are: 1, 1, 1
Final answer:
The balanced equation for the decomposition reaction of calcium carbonate is CaCO3(s) → CaO(s) + CO2(g), representing the formation of calcium oxide and carbon dioxide gas from calcium carbonate.
Explanation:
The decomposition reaction of calcium carbonate forming calcium oxide and carbon dioxide can be represented by the following balanced chemical equation:
CaCO3(s) → CaO(s) + CO2(g)
This equation indicates that one mole of calcium carbonate (CaCO3) decomposes to form one mole of calcium oxide (CaO) and one mole of carbon dioxide (CO2) gas upon heating. The reactants and products are represented using their smallest possible integer coefficients, reflecting the principle of conservation of mass and the stoichiometry of the decomposition reaction.
In this chemical Formula for Ammonia, the Subscripts indicate what?
A. When added together, we know that Nitrogen and Hydrogen combined create four molecules of Ammonia
B. There are two atoms of Nitrogen and two atoms of Hydrogen combined to make Ammonia.
C. There are two molecules of Nitrogen and two molecules of Hydrogen combined to make Ammonia
D. The compound N2 and the compound H2 make Ammonia
Answer:
B
Explanation:
B. There are two atoms of Nitrogen and two atoms of Hydrogen combined to make Ammonia.
a sample of a radioactive isotope initially contains 20 x 10^10 atoms. After 16 days, 5 x 10^10 atoms remain. what is the half-life of the isotope?
A)12 days
B)20 days
C)10 days
D)16 days
E)8 days
The half-life of the radioactive isotope is 8 days because after 16 days (which is two half-lives), the sample is reduced from 20 x 10^10 atoms to 5 x 10^10 atoms.
Explanation:The question deals with the concept of the half-life of a radioactive isotope, which is the time required for half of the radioactive atoms in a sample to decay. According to the problem statement, an initial sample containing 20 x 1010 atoms decays to 5 x 1010 atoms in 16 days. To find the half-life (t1/2), we can use the fact that after one half-life, the number of atoms would be halved. Starting with 20 x 1010 atoms, after one half-life, there would be 10 x 1010 atoms, and after two half-lives, there would be 5 x 1010. This indicates that two half-lives have passed in 16 days, making the half-life 8 days.
The correct answer is 8 days, option E.
I’m confused. Need any help.
Answer:
Problem 1 => 8778 joules
Problem 2 => 14,630 joules
Explanation:
Reference the Heating Curge of Water Problem posted earlier. These are temperature change fragments of that type problem. As you read the problem note the 'temperature change' phrase in the problem. This should signal you to use the q = m·c·ΔT expression as opposed to the q = m·ΔH expression where no temperature change is noted; i.e., melting or evaporation/boiling.
For the listed problems...
Problem 1: Amount of heat needed to heat 150 grams water from 21.0 to 35.0 Celcius.
Note the temperature change in the problem context => use q = m·c·ΔT ...
q = (150g)(4.18j/g·°C)(35.00°C - 21.0°C) = 8778 joules (4 sig. figs. based on the 150.0g value of water)
Problem 2: Amount of heat needed to heat 250.0 grams water from 31.0 to 45.0 Celcius.
Same type problem. Note temperature change in text of problem => use q = m·c·ΔT ...
q = (250.0g)(4.18j/g·°C)(45.0°C - 31.0°C) = 14,630 joules
Hope this helps. Doc
In a Grignard reagent, the carbon bonded to the magnesium has a partial ________ charge, because carbon is ________ electronegative than magnesium. This makes this carbon of the Grignard ________. In a Grignard reagent, the carbon bonded to the magnesium has a partial ________ charge, because carbon is ________ electronegative than magnesium. This makes this carbon of the Grignard ________. negative, less, electrophilic positive, less, electrophilic positive, more, electrophilic negative, more, nucleophilic positive, less, nucleophilic
Answer:
In a Grignard reagent, the carbon bonded to the magnesium has a partial negative charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard nucleophilic.
Explanation:
Organometallic compounds have covalent bonds between carbon atoms and metallic atoms. Organometallic reagents are useful because they have nucleophilic carbon atoms, in contrast to the electrophilic carbon atoms of the alkyl halides. The majority of metals are more electropositive than carbon, and the bond is biased with a positive partial charge on the metal and a negative partial charge on the carbon.
Answer:
In a Grignard reagent, the carbon bonded to the magnesium has a partial negative charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard nucleophilic In a Grignard reagent, the carbon bonded to the magnesium has a partial positive charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard less electrophilic.
Explanation:
The Grignard reagent is a highly reactive organomagnesium compound formed by the reaction of a haloalkane with magnesium in an ether solvent. The carbon atom of a Grignard reagent has a partial negative charge. Hence a Grignard reagent will have a general formula RMgX. Where R must contain a carbanion.
Grignard reagents are versatile reagents used in many applications in organic chemistry.
A chemist must prepare 800mL of potassium hydroxide solution with a pH of 13.90 at 25C.?
He will do this in three steps.
1. fill a 800mL volumetric flask about halfway with distilled water.
2. weigh out a small amount of solid potassium hydroxide and add it to the flask.
3. fill the flask to the mark with distilled water.
Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step.
Answer:
35.6 g
Explanation:
Step 1: Calculate the [H⁺]
We use the following expression
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -13.90 = 1.259 × 10⁻¹⁴ M
Step 2: Calculate the [OH⁻]
We use the ionic product of water (Kw).
Kw = 10⁻¹⁴ = [H⁺] × [OH⁻]
[OH⁻] = 0.7943 M
Step 3: Calculate the moles of OH⁻
[tex]\frac{0.7943mol}{L} \times 0.800 L = 0.635 mol[/tex]
Step 4: Calculate the required moles of KOH
KOH is a strong base that dissociates according to the following equation.
KOH → K⁺ + OH⁻
The molar ratio of KOH to OH⁻ is 1:1. Then, the required moles of KOH are 0.635 moles.
Step 5: Calculate the mass of KOH
The molar mass of KOH is 56.11 g/mol.
[tex]0.635 mol \times \frac{56.11g}{mol} =35.6 g[/tex]
Consider the reaction: 2N2(g) + O2(g)2N2O(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.19 moles of N2(g) react at standard conditions. S°system = J/K
Answer:
ΔS = -114.296 J/K
Explanation:
Step 1: Data given
Temperature = 298 K
Number of moles N2 = 2.19 moles
S°(N2) = 191.6 J/K*mol
S°(O2)= 161.1 J/k*mol
S°(N2O) = 219.96 J/K*mol
Step 2: The balanced equation
2N2(g) + O2(g) ⇆ 2N2O(g)
Step 3: Calculate ΔSrxn
ΔSrxn = ∑S°products - ∑S°reactants
ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol
ΔSrxn = -104.38 J/K
Step 4: Calculate ΔS for 2.19 moles
The reaction is for 2 moles N2
ΔS = -114.296 J/K
Final answer:
The entropy change for a chemical reaction can be determined using standard absolute entropies. In this case, the given reaction of N2(g) and O2(g) to form N2O(g) can be analyzed using these principles.
Explanation:
The entropy change for the given reaction can be calculated using the standard absolute entropies at 298K. The formula to calculate the entropy change is ΔS = ΣnS(products) - ΣnS(reactants). In this case, for the reaction 2N2(g) + O2(g) → 2N2O(g), the entropy change can be calculated using the standard absolute entropies of each component.
Step 1: Data given
Temperature = 298 K
Number of moles N2 = 2.19 moles
S°(N2) = 191.6 J/K*mol
S°(O2)= 161.1 J/k*mol
S°(N2O) = 219.96 J/K*mol
Step 2: The balanced equation
2N2(g) + O2(g) ⇆ 2N2O(g)
Step 3: Calculate ΔSrxn
ΔSrxn = ∑S°products - ∑S°reactants
ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol
ΔSrxn = -104.38 J/K
Step 4: Calculate ΔS for 2.19 moles
The reaction is for 2 moles N2
ΔS = -114.296 J/K
What is the daily cost of incandescent lamp
Draw the curved arrow(s) to depict the formation of the keto form of an enolate ion via a strong base, B. Complete the resonance structures of the enolate anion\'s keto and enolate forms with bonds, charges, and nonbonding electron pairs. Use curved arrows to show how the keto form resonates to the enolate form.
The enolate ion is formed from a carbonyl function via a strong base, leading to resonance structures between keto and enolate forms. The resonance involves shifting of electron pairs, with the enolate form being more prevalent under basic conditions.
Explanation:An enolate ion is formed when a strong base, B, abstracts a proton α to a carbonyl function. The negatively charged oxygen donates a pair of electrons, pushing the electrons from the double bond into the alpha carbon to form the enolate form. The curved arrow represents the movement of a pair of electrons.
The resonance of the keto form to the enolate form involves shifting of electron pairs, with the electron pair on the carbon moving to form a C=O bond and the π bond between carbon and oxygen moving to the oxygen, forming a p-orbital with a lone pair of electrons and making the oxygen negatively charged.
By this mechanism, the keto and enolate forms are interconvertible with the enolate form being more prevalent under basic conditions. The full resonance structures for both the keto and enolate forms would consist of all their bonds, charges, and nonbonding electron pairs.
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g A mass spectrum. The peak at mass 100 has a 8% relative abundance. THe peak with mass 85 has a 40% abundance. The peak at 71 has a 3% abundance. The peak at 57 has a 30% abundance. The peak at mass 43 has a 100% abundance. The peak at mass 29 has a 18% abundance. In this mass spectrum, which peak most likely represents the molecular ion
Answer:
The peak at mass 100 with a 8% relative abundance is the molecular ion peak
Explanation:
Molecular ion peak has the highest charge to mass ratio,
Mass of 100 is same as mass to charge ratio =100
2. Calculate the mass of 3.47x1023 gold atoms.
3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams.
Explanation:
Data given:
number of atoms of gold = 3.47 x [tex]10^{23}[/tex]
mass of the gold in given number of atoms = ?
atomic mass of gold =196.96 grams/mole
Avagadro's number = 6.022 X [tex]10^{23}[/tex]
from the relation,
1 mole of element contains 6.022 x [tex]10^{23}[/tex] atoms.
so no of moles of gold given = [tex]\frac{3.47 X 10^{23} }{6.022 X 10^{23} }[/tex]
0.57 moles of gold.
from the relation:
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
rearranging the equation,
mass = number of moles x atomic mass
mass = 0.57 x 196.96
mass = 113.44 grams
thus, 3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams
The mass of 3.47 x 10^23 gold atoms calculated using Avogadro's number and the molar mass of gold equals approximately 113.52 grams. To calculate this, the given number of atoms was first converted into moles and the number of moles were then multiplied by the molar mass.
Explanation:In chemistry, to calculate the mass of a certain number of atoms, we use the concept of a mole and Avogadro's number (6.022 x 1023). For gold atoms, the molar mass is 197 g/mol. Given that the number of gold atoms is 3.47 x 1023, we can use the relationship that 1 mole of gold atoms = 6.022 x 1023 gold atoms. Therefore, the mass of 3.47 x 1023 gold atoms can be calculated as follows:
First, we convert the number of atoms to moles using Avogadro's number: 3.47 x 1023 atoms / 6.022 x 1023 atoms/mol = 0.576 moles of gold.
Then, we multiply this moles by the molar mass of gold to get the mass in grams: 0.576 moles * 197 g/mol = 113.52 g.
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A 900.0 mLmL sample of 0.18 MHClO4MHClO4 is titrated with 0.27 MLiOHMLiOH. Determine the pHpH of the solution after the addition of 600.0 mLmL of LiOHLiOH (this is the equivalence point). A 900.0 sample of 0.18 is titrated with 0.27 . Determine the of the solution after the addition of 600.0 of (this is the equivalence point). 11.24 13.03 2.76 0.97 7.00
Answer: pH of the solution after the addition of 600.0 ml of LiOH is 7.
Explanation:
The given data is as follows.
Volume of [tex]HClO_{4}[/tex] = 900.0 ml = 0.9 L,
Molarity of [tex]HClO_{4}[/tex] = 0.18 M,
Hence, we will calculate the number of moles of [tex]HClO_{4}[/tex] as follows.
No. of moles = Molarity × Volume
= 0.18 M × 0.9 L
= 0.162 moles
Volume of NaOH = 600.0 ml = 0.6 L
Molarity of NaOH = 0.27 M
No. of moles of NaOH = Molarity × Volume
= 0.27 M × 0.6 L
= 0.162 moles
This shows that the number of moles of [tex]HClO_{4}[/tex] is equal to the number of moles of NaOH.
Also we know that,
[tex]HClO_{4} + NaOH \rightarrow NaClO_{4} + H_{2}O[/tex]
As 1 mole of [tex]HClO_{4}[/tex] reacts with 1 mole of NaOH then all the hydrogen ions and hydroxide ions will be consumed.
This means that pH = 7.
Thus, we can conclude that pH of the solution after the addition of 600.0 ml of LiOH is 7.
The reaction:L-glutarmate + pyruvate → α-ketoglutarate + L-alanine is catalyzed by the enzyme L-glutamate-pyruvate aminotransferase. At 300 K, the equilibrium constant for the reaction is 1.1. Predict whether the forward reaction (left to right) will occur spontaneous if the concentrations of the reactants and products are [L-glutarmate] = 3.0 x 10-5 M, [pyruvate] = 3.3 x 10-4 M,[α-ketoglutarate] = 1.6 x 10-2 M, and [L-alanine] = 6.25 x 10-3 M.
Answer:
Reaction Quotient, Kq = {[a-ketoglutarate]x[L-alanine]}/{[L-glutamate]x[pyruvate]}
or, Kq = {(1.6x10-2)x(6.25x10∧-3)}/{(3x10∧-5)x(3.3x10-4)} = 1.01 x 10∧4
Since Kq > Keqb ; therefore the reaction will proceed in the backward direction, in order words the reaction will not occur in forward direction. i.e formation of reactants will be favored.
Explanation: