Answers
To draw the seven constitutional isomers of the cycloalkane with the formula C6H12, we need to consider different arrangements of carbon atoms in a cyclic structure
To draw the seven constitutional isomers of the cycloalkane with the formula C6H12, we need to consider different arrangements of carbon atoms in a cyclic structure with the given molecular formula:
1. Cyclohexane:
H H H
| | |
H--C--C--C--H
| | |
H H H
2. Methylcyclopentane:
H H H
| | |
H--C--C--C--C--H
| | |
H H H
3. Dimethylcyclobutane:
H H H
| | |
H--C--C--C--C--H
| | | |
H H H H
4. Ethylcyclopentane:
H H H
| | |
H--C--C--C--C--H
| | |
H H H
5. 1,1-Dimethylcyclobutane:
H H H
| | |
H--C--C--C--C--H
| | | |
H H H H
6. 1,2-Dimethylcyclopentane:
H H H
| | |
H--C--C--C--C--C--H
| | |
H H H
7. 1,3-Dimethylcyclobutane:
H H H
| | |
H--C--C--C--C--H
| | |
H H H
For the two stereoisomers of 1,3-dibromocyclobutane, we can have:
1. Cis-1,3-dibromocyclobutane:
Br H
| |
H--C--C--C--Br
| |
H H
2. Trans-1,3-dibromocyclobutane:
Br Br
| |
H--C--C--C--H
| |
H H
These structures represent the seven constitutional isomers of cycloalkane C6H12 and the two stereoisomers of 1,3-dibromocyclobutane.
Related Questions
Determine the minimum work required by an air compressor. At the inlet the conditions are 150 kg/min, 125 kPa and 33 °C. At the exit, the pressure is 550 kPa. Assume air is an ideal gas with MW 29 g/mol, Cp 3.5R (constant).
Answers
Explanation:
The given data is as follows.
MW = 29 g/mol, [tex]C_{p}[/tex] = 3.5 R
Formula to calculate minimum amount of work is as follows.
[tex]W_{s} = C_{p}T_{1}[(\frac{P_{2}}{P_{1}})^{\frac{R}{C_{p}}} - 1][/tex]
= [tex]3.5 \times 8.314 J/k mol \times 306 \times [(\frac{550}{125})^{\frac{1}{3.5}} - 1][/tex]
= 4.692 kJ/mol
Therefore, total work done will be calculated as follows.
Total work done = [tex]m \times W_{s}[/tex]
Since, m = [tex]\frac{150 \times 10^{3}g/min}{29}[/tex]. Therefore, putting these values into the above formula as follows.
Total work done = [tex]m \times W_{s}[/tex]
= [tex]\frac{150 \times 10^{3}g/min}{29} \times 4.692 kJ/min[/tex]
= 24268.96 kJ/min
It is known that 1 kJ/min = 0.0166 kW. Hence, convert 24268.96 kJ/min into kW as follows.
[tex]24268.96 kJ/min \times \frac{0.0166 kW}{1 kJ/min}[/tex]
= 402.86 kW
Thus, we can conclude that the minimum work required by an air compressor is 402.86 kW.
2mL of a serum sample was added to 18mL of phosphate buffered saline (PBS) in Tube 1. 10mL of Tube 1 was added to 40mL of PBS in Tube What is the dilution of serum in Tube 2?
Answers
Answer:
Tube 2 has a total dilution of 1:50
Explanation:
We have a 2 ml serum sample added to a 18 mL phosphate buffered saline sample in tube 1. This means now in tube 1 there is 20 mL.
We have a 1:10 (= 2:20) dilution here.
10 ml of this 1:10 diluted tube 1 is taken and added to a 40 mL of PBS in tube 2.
Now we have 50 mL in tube 2.
This is a 10:50 (= 1:5) dilution.
The total dilution is 10x5 = 50
So the total ditultion has a rate 1:50
Tube 2 has a total dilution of 1:50
Which of the following statements about resonance are correct?
Which of the following statements about resonance are correct? Atoms can move between different resonance structures. Resonance generally involved lone pairs, pi bonds, and formal charges. Each resonance structure must be a valid Lewis structure. Resonance structures are separated by a double-headed arrow: OElectrons can move between different resonance structures. Resonance often involves sp3 hybridized carbon atoms. The actual molecular structure will alternate between all possible resonance structures.
Answers
4) resonance struct are seperated by double arrow
also the last 3 statement are true
A prescription medication requires 5.98 mg per kg of body weight.
A)Convert this quantity to the number of grams required per pound of body weight.
B)Determine the correct dose (in g) for a 191-lb patient. Express your answer with the appropriate units.
Answers
To convert from mg/kg to g/lb, use the conversion factors 1000 mg/g and 2.20462 lbs/kg. For a 191-lb patient, the correct dose is approximately 0.51761 g.
Explanation:To convert the prescription medication requirement from milligrams per kilogram (mg/kg) to grams per pound (g/lb), we first need to know the conversion factors between the given units. There are 453.59237 mg in a gram and 2.20462 pounds in a kilogram. Following the conversion steps:
First, we convert 5.98 mg/kg to g/kg by dividing by 1000 (since there are 1000 mg in 1 g):5.98 mg / kg × (1 g / 1000 mg) = 0.00598 g/kgNext, we convert g/kg to g/lb by dividing by 2.20462 (since there are 2.20462 pounds in 1 kilogram):
0.00598 g/kg × (1 kg / 2.20462 lbs) = 0.00271 g/lb
To find the correct dose for a 191-lb patient, we multiply the medication requirement by the patient's weight:
0.00271 g/lb × 191 lbs = approximately 0.51761 g.
Therefore, the correct dose for a 191-lb patient is 0.51761 g.
A) The number of grams required per pound of body weight is approximately 0.00272 grams per pound. B) The correct dose (in g) for a 191-lb patient is approximately 0.519 grams.
A) To convert the medication dosage from milligrams per kilogram to grams per pound, we need to perform two conversions: one from milligrams to grams and another from kilograms to pounds.
1 milligram is equal to 0.001 grams (since there are 1000 milligrams in a gram).1 kilogram is approximately equal to 2.20462 pounds.Starting with the dosage of 5.98 mg per kg, we convert to grams per kilogram: 5.98 mg/kg = 5.98 x 0.001 g/kg = 0.00598 g/kg
Now, we convert from grams per kilogram to grams per pound:
0.00598 g/kg x 2.20462 kg/lb = 0.013175 g/lb
To simplify the calculation, we can round this to a more convenient number, such as 0.00272 g/lb for practical purposes.
B) To determine the correct dose for a 191-lb patient, we multiply the patient's weight in pounds by the dosage in grams per pound:
Dose (in g) = patient's weight (in lb) * dosage (in g/lb)
Dose (in g) = 191-lb x 0.00272 g/lb
Dose (in g) =0.519 g
Therefore, the correct dose for a 191-lb patient is approximately 0.519 grams.
Carbon burns in the presence of oxygen to give carbon dioxide. Which chemical equation describes this reaction? A. carbon + oxygen + carbon dioxide B. carbon + oxygen → carbon dioxide C. carbon dioxide → carbon + oxygen D. carbon dioxide + carbon → oxygen
Answers
Answer:
Hello my friend! The correct answer to this quastion is "B. carbon + oxygen → carbon dioxide"
Explanation:
Carbon uses oxygen and heat as fuel for the O2 chemical bond breakdown reaction, and the new reaction between carbon and formed oxygen or carbon dioxide.
C + O2 ----> CO2
Carbon burns in the presence of oxygen to give carbon dioxide. The chemical equation describing this reaction is [tex]\text { carbon }+\text { oxygen } \rightarrow \text { carbon dioxide }[/tex]
Answer: Option B
Explanation:
The chemical equations are the representation of a particular reaction.This equation used to avoid the description of the reaction and narrowing it to precise statement. It consists of two parts namely,
The reactants that initially present undergoes mutual reaction.The products, the aftermath of the reaction.Here Carbon and oxygen are the reactants, the arrow symbol shows the direction of the reaction and the product here is carbon dioxide.
Two Carnot engines are operated in series with the exhaust (heat output) of the first engine being the input of the second engine. The upper temperature of this combination is 260F, the lower temperature is 40F. If each engine has the same thermal efficiency, determine the exhaust temperature of the first engine (the inlet temperature of the second engine). Ans: T = 140F 3. A nuclear power plant generates 750 MW of power. The heat engine uses a nuclear reactor operating at 315C as the source of heat. A river is available (at 20C) which has a volumetric flow rate of 165 m/s. If you use the river as a heat sink, estimate the temperature rise in the river at the point where the heat is dumped. Assume the actual efficiency of the plant is 60% of the Carnot efficiency.
Answers
Answer:
(a) 140 F
(b) The temperature rise at the point where the heat is dumped is 2.51 degC
Explanation:
(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:
[tex]\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}[/tex]
The temperatures have to be expressed in Rankine (or Kelvin) degrees
[tex]1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F[/tex]
(b) The Carnot efficiency of the cycle is
[tex]\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502[/tex]
If the efficiency of the plant is 60% of the Carnot efficiency, we have
[tex]\eta=0.6*\eta_{c}=0.6*0.502=0.302[/tex]
The heat used in the plant can be calculated as
[tex]Q_i=W/\eta=750MW/0.302=2483MW[/tex]
And the heat removed to the heat sink is
[tex]Q_o=Qi-W=2483-750=1733MW[/tex]
If the flow of the river is 165 m3/s, the heat per volume in the sink is
[tex]\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3[/tex]
Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is
[tex]\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C[/tex]
Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrical response proportional to the concentration of hydrogen ion. To convert these responses to a pH reading, the electrode must be calibrated against standard solutions of known H+ concentration. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4 · H2O; FW 138) and disodium hydrogen phosphate (Na2HPO4; FW 142) needed to prepare 1 L of a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M
Answers
To prepare a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M, you will need to calculate the amount of sodium dihydrogen phosphate (NaH2PO4 · H2O) and disodium hydrogen phosphate (Na2HPO4) needed.
Explanation:To prepare a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M, you will need to calculate the amount of sodium dihydrogen phosphate (NaH2PO4 · H2O) and disodium hydrogen phosphate (Na2HPO4) needed.
Step 1: Calculate the individual concentrations of NaH2PO4 and Na2HPO4.
Using the molecular weight, you can calculate the number of moles of NaH2PO4 and Na2HPO4 needed to achieve a total phosphate concentration of 0.100 M in 1 L of solution.
NaH2PO4: (0.100 M) * (1 L) = x mol
Na2HPO4: (0.100 M) * (1 L) = y mol
Step 2: Calculate the weight of NaH2PO4 and Na2HPO4.
Using the number of moles calculated in step 1, you can calculate the weight of NaH2PO4 and Na2HPO4 needed.
NaH2PO4: (x mol) * (138 g/mol) = weight in grams
Na2HPO4: (y mol) * (142 g/mol) = weight in grams
By following these steps, you will be able to determine the weight in grams of NaH2PO4 · H2O and Na2HPO4 needed to prepare 1 L of the standard buffer solution.
During the experiment a student precipitated and digested the BaSO4. After allowing the precipitate to settle, they added a few drops of BaCl2 solution, and the previously clear solution became cloudy. Explain what happened.
Answers
Answer:
Incomplete precipitation of barium sulfate
Explanation:
The student has precipitated and digested the barium sulfate on his/her side. But on the addition of [tex]BaCl_2[/tex] in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When [tex]BaCl_2[/tex] is added, there are still sulfate ions present in the solution with combines with [tex]BaCl_2[/tex] and forms [tex]BaSO_4[/tex] and the formation of this precipitate makes the solution cloudy.
The solution became cloudy after adding BaCl2 because the additional Ba2+ ions reacted with any remaining [tex]SO4^2-[/tex]orming more BaSO4 precipitate, indicating incomplete digestion of the initial BaSO4.
Explanation:During the experiment, the student added a few drops of BaCl2 solution to the clear solution that contained digested BaSO4. BaSO4 is known for its low solubility in water; however, it's soluble in solutions containing ions that can form more soluble compounds with the constituent ions. In the experiment, adding more BaCl2 likely introduced additional Ba2+ ions into the solution. If any unreacted sulfate ions (SO42-) were present, these extra Ba2+ ions could have reacted with them, forming additional BaSO4 precipitate, thus causing the solution to become cloudy. This suggests that the digestion process was not complete, leaving some sulfate ions in the solution which reacted with the added barium ions to form more BaSO4 precipitate.
Charles' law relates the way two gas properties change when another property remains the same. What are the two changing properties in Charles' law?
Pressure and temperature
Pressure and volume
Pressure, temperature, and volume
Temperature and volume
ik its not B
Answers
Answer:
The two changing properties in Charles’ law are temperature and volume.
Explanation:
Charles’ law state the presence of direct relationship between temperature and volume of the system of gas molecules at constant pressure condition. In this law, the expansion of gas has been explained with the increase of temperature.
As the temperature is increased or the system of gas molecules are heated, the gas molecules tend to expand their volume to maintain the pressure same.
So the temperature and volume are directly proportional at constant pressure. Thus the two changing properties of Charles’ law is temperature and volume. The mathematical representation of Charles’ law is
V∝T (at constant Pressure)
[tex]V=kT[/tex]
Here k is the non-zero constant and V and T are volume and temperature respectively.
a.) How would you prepare .250 L of a 0.300 M phosphate buffer at pH=3 (H3PO4: pka1 = 2.12, pka2 = 7.21, pka = 12.32) using the appropriate weak acid and conjugate salt. b.) How would you prepare the assigned buffer (given in question #1a) if using combining the weak acid and 0.400 M NaOH? c.) How would you prepare the assigned buffer (given in question #1a) if using combining the weak acid and 0.400 M HCl?
Answers
Answer:
a) 0,857 g of H₃PO₄ with 9,016 g of KH₂PO₄
b) 166 mL of 0,400M NaOH
c) 22 mL of 0,400M HCl
Explanation:
a) The appropriate weak acid and conjugate salt are:
H₃PO₄ ⇄ H₂PO₄⁻ + H⁺ where pka = 2,12
Henderson–Hasselbalch equation finding pH = 3:
3 = 2,12 + log₁₀ [tex]\frac{[H2PO4-]}{[H3PO4]}[/tex]
7,59 = [tex]\frac{[H2PO4-]}{[H3PO4]}[/tex] (1)
If buffer concentration is 0,300M:
0,300 M = [H₃PO₄] + [H₂PO₄⁻] (2)
Replacing (2) in (1):
[H₃PO₄] = 0,035 M
Thus:
[H₂PO₄⁻] = 0,265 M
Thus, to prepare this buffer you need weight:
0,035 M × 0,250 L = 8,75x10⁻³ moles × [tex]\frac{97,994 g}{1mol}[/tex] = 0,857 g of H₃PO₄
And:
0,265 M × 0,250 L = 6,63x10⁻² moles × [tex]\frac{136,086 g}{1mol}[/tex] = 9,016 g of KH₂PO₄
b) Using 0,400 M NaOH the equilibrium is:
H₃PO₄ + NaOH ⇄ H₂PO₄⁻ + H₂O
Knowing the equilibrium concentrations are:
[H₃PO₄] = 0,035 M = 0,300 M - x -Because in the first all 0,300 M must be of H₃PO₄-
[H₂PO₄⁻] = 0,265 M
Thus, x = 0,265 M are NaOH needed to obtain the desire pH. Those are obtained thus:
0,265 mol/ L × 0,250L × [tex]\frac{1L}{0,400mol}[/tex] = 0,166 L ≡ 166 mL of 0,400M NaOH
c) Using 0,400 M HCl the equilibrium is:
H₃PO₄ ⇄ H₂PO₄⁻ + HCl
Knowing the equilibrium concentrations are:
[H₃PO₄] = 0,035 M
[H₂PO₄⁻] = 0,265 M = 0,300 M - x -Because in the first all 0,300 M must be of H₂PO₄-
Thus, x = 0,035 M are HCl needed to obtain the desire pH. Those are obtained thus:
0,035 mol/ L × 0,250L × [tex]\frac{1L}{0,400mol}[/tex] = 0,022 L ≡ 22 mL of 0,400M HCl
I hope it helps!
The surface temperature on Venus may approach 757 K. What is this temperature in degrees Celsius? 757 K = The temperature on Mercury may drop to -261 °F at night. What is this temperature in degrees Celsius? -261 °F =
Answers
Answer:
757 K = 484 °C
-261 °F = -163 °C
Explanation:
The formula to convert Kelvin to degrees Celsius is:
°C = K - 273.15 = 757 - 273.15 = 484 °C
The formula to convert °F to °C is:
°C = 5/9 (°F -32) = 5/9 (-261 - 32) = -163
100.0 kg of liquid methanol and 100.0 kg of liquid water are mixed in a stirred tank.
Assuming volume additivity of methanol and water, determine the moles and volumes of the two substances in the mixture.
M(l) W(l)
MW (kg / kmol) 32.04 18.01
rho (kg / L) .791 1.00
1. kmol of methanol?
2. kmol of water?
3. Liters of methanol?
4. L of water?
Answers
Answer:
1. kmol of methanol= 3.12 Kmol
2. kmol of water= 5.55 Kmol
3. Liters of methanol= 126.4 L
4. L of water= 100 L
Explanation:
1. kmol of methanol?
32.04 kg methanol ______________ 1 kmol of methanol
100 kg of methanol_______________ X= 3.12 kmol ofmethanol
2. kmol of water?
18.01 kg water ______________ 1 kmol of wáter
100 kg of wáter_______________ X= 5.55 kmol of water
3. Liters of methanol?
0.791 kg methanol _______________________1.00 L of methanol
100kg methanol _______________________x= 126.4 L of methanol
4. L of water?
1kg water _______________________1.00 L of water
100kg water _______________________x= 100 L of water
In water, hydroxides of Group 2 metals a. are all strong bases. b. are all weak bases. c. are all acids. d. are nonelectrolytes
Answers
Answer:
The correct answer is: a. are all strong bases
Explanation:
Alkaline earth metals are the chemical elements that belong to the group 2 of the periodic table. The members or elements of this group are all highly reactive metals.
Except beryllium (Be), all the alkaline earth metals react with water to give metal hydroxides. These hydroxides of the alkaline earth metals are highly soluble and very strong bases.
Final answer:
Hydroxides of Group 2 metals in water are all strong bases because they dissociate almost completely into ions, significantly raising the solution's pH by releasing a high concentration of OH- ions.
Explanation:
The question asks about the nature of hydroxides of Group 2 metals when dissolved in water. Hydroxides of the Group 2 metals (the alkaline earth metals) like Ca(OH)2, Sr(OH)2, and Ba(OH)2 are known to be strong bases. They are considered strong bases because they dissociate almost completely into ions when dissolved in water, providing a high concentration of OH- ions that increase the solution's pH markedly. To answer the provided options, a. are all strong bases, matches the description for hydroxides of Group 2 metals in water.
What is the thermodynamic equilibrium constant under standard conditions for the following balanced redox reaction? Zr(s) + O2(g) - ZrO2 (s) Een=2.463 V
Answers
Answer:
Equilibrium constant = [tex]2.23 \times 10^{83}[/tex]
Explanation:
[tex]Zr(s) + O_2(g) \rightarrow ZrO_2(s)[/tex]
[tex]E^0_{cell}[/tex] = 2.463 V
Equilibrium constant is related with [tex]E^0_{cell}[/tex] as
[tex]E_{cell}=E^0_{cell} - \frac{2.303 RT}{nF} ln k_{eq}[/tex]
In standard condition,
T = 25 °C = 25 + 273 = 298 K
F = 96500 C mol^-1
R = 8.314 [tex]J\ K^{-1}mol^{-1}[/tex]
On substituting values, the above expression becomes:
[tex]E_{cell}=E^0_{cell} - \frac{0.059}{n} log k_{eq}[/tex]
n = 2
At equilibrium, [tex]E_{cell}= 0[/tex]
[tex]0=E^0_{cell} - \frac{0.059}{2} log k_{eq}[/tex]
[tex]log k_{eq}=\frac{2 \times 2.463}{0.059}[/tex]
= 83.35
[tex]K_{eq} = antilog 83.35 = 2.23 \times 10^{83}[/tex]
functional group and bond hybridization of vanillin
Answers
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a sp² hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a sp³ hybridization because they are involved only in single bonds.
Define "Anomeric" carbon
Answers
Anomeric carbon is a stereocenter present in the cyclic structures of carbohydrates (mono or polysaccharides). Being a stereocenter, more exactly an epimer, two diastereosoimers derive from it, designated by the letters α and β; these are anomers, and are part of the extensive nomenclature in the world of sugars.
Draw the Lewis Structure for NaCl
Answers
Explanation:
The electronic configuration of sodium with Z = 11 is : 2, 8, 1
The electronic configuration of chlorine with Z = 17 is : 2, 8, 7
The Lewis structure is drawn in such a way that the octet of each atom is complete.
Thus, sodium losses one electron to chlorine and chlorine accepts this electron to form ionic bond.
Thus, the valence electrons are shown by dots in Lewis structure. The structure is shown in image below.
The Lewis structure of NaCl involves sodium losing one electron to become Na+ and chlorine gaining one electron to become Cl-. Draw the ions next to each other to represent the ionic bond. Sodium has no electrons around it while chlorine has a complete octet with the extra electron in brackets.
The Lewis structure of sodium chloride (NaCl), follow these steps:
Identify the valence electrons: Sodium (Na) is in group 1 and has 1 valence electron. Chlorine (Cl) is in group 17 and has 7 valence electrons.Show the transfer of electrons: Sodium will lose its 1 valence electron to achieve a stable electron configuration, becoming a positively charged ion (Na+). Chlorine will gain this electron to complete its octet, becoming a negatively charged ion (Cl-).Represent the ions: Write the Lewis structures of the resulting ions next to each other to indicate the ionic bond:Na: Na+ [ ]
Cl: [ :Cl: ]-
Note that the brackets around the chlorine indicate it has gained an electron and the overall charge of the ion.
The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation 3H2(g)+N2(g)→2NH3(g) The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation. 1.15 g H2 is allowed to react with 9.93 g N2, producing 1.12 g NH3. Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.
Answers
The theoretical yield of NH3 in the Haber-Bosch process under the given conditions is 12.036 grams.
Explanation:Theoretical yield refers to the maximum amount of product that can be obtained in a chemical reaction according to the balanced chemical equation. To calculate the theoretical yield of ammonia in this reaction, you need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be obtained.
In this case, you have 1.15 g of H2 and 9.93 g of N2. To determine the limiting reactant, you can compare the moles of H2 and N2 using their molar masses:
Moles of H2 = (1.15 g H2) / (2 g/mol H2) = 0.575 mol H2
Moles of N2 = (9.93 g N2) / (28 g/mol N2) = 0.354 mol N2
Since the coefficients in the balanced equation are in a 1:1 ratio for H2 and N2, it is clear that the limiting reactant is N2 because there are fewer moles of N2 available.
Now you can use the limiting reactant to calculate the theoretical yield of NH3. According to the balanced equation, the stoichiometric ratio between N2 and NH3 is 1:2. Therefore, moles of NH3 = 2 * moles of N2 = 2 * 0.354 mol = 0.708 mol NH3. Finally, you can convert moles of NH3 to grams using the molar mass of NH3:
Mass of NH3 = (0.708 mol NH3) * (17 g/mol NH3) = 12.036 g NH3
Therefore, the theoretical yield of NH3 under the given conditions is 12.036 grams.
Learn more about Haber-Bosch here:https://brainly.com/question/26667299
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Assuming Ideal Gas Law is applicable, calculate density, specific weight, and specific volume of air at 120 oF and 50 psia.
Answers
Answer:
a) ρ = 3.735 Kg/m³
b) γ = 36.603 N/m³
c) Vw = 2.68 E-4 g/m³
Explanation:
PV = RTn .....ideal gas lawa) ρ = m/V
∴ Mw air = 28.966 g/mol......from literature
⇒ P = RTn / V
∴ n = m / Mw
⇒ P = mRT / Mw.V = m/V * R.T/Mw
ρ = P.Mw / R.T∴ P = 50 psi = 344738 Pa
∴ R = 8.3144 Pa.m³/mol.K
∴ T = 120°F = 48.889 °C = 321.889 K
⇒ ρ = (( 344738 Pa ) * ( 28.996 g/mol )) / (( 8.3144 Pa.m³/mol.K) * ( 321.889K ))
⇒ ρ = 3734.996 g/m³ * ( Kg / 1000g ) = 3.735 Kg/m³
specific weight:
γ = w / V = m.g / V = ρ *g∴ ρ = 3.735 Kg/m³
∴ g = 9.8 m/s²
⇒ γ = 36.603 N/m³
c) specific volume
Vw = R.T / P.Mw⇒ Vw = (( 8.3144 Pa.m³/ mol.K ) * ( 321.889 K )) / ( 344738 Pa )* ( 28.966 g/mol)
⇒ Vw = 2.68 E-4 g/m³
Define ""green chemistry"" and ""ecological footprints"".
Answers
Explanation:
Green chemistry
It is the process of designing a chemical compound via reducing or eliminating the use or generation of the hazardous substances .
It is a eco - friendly method , which does not harm the nature .
Ecological footprints
It is the tool to measure the demand of humans on nature , the quantity of nature humans require to support economy .
It tracks the demand of the humans via ecological accounting system .
pH indicator. A dye that is an acid and that appears as different colors in its protonated and deprotonated forms can be used as a pH indicator. Suppose that you have a 0.001 M solution of a dye with a p Ka of 7.2. From the color, the concentration of the protonated form is found to be 0.0002 M. Assume that the remainder of the dye is in the deprotonated form. What is the pH of the solution? Berg, Jeremy M.. Biochemistry (p. 24). W. H. Freeman. Kindle Edition.
Answers
Answer:
pH = 7.8
Explanation:
The Henderson-Hasselbalch equation may be used to solve the problem:
pH = pKa + log([A⁻] / [HA])
The solution of concentration 0.001 M is a formal concentration, which means that it is the sum of the concentrations of the different forms of the acid. In order to find the concentration of the deprotonated form, the following equation is used:
[HA] + [A⁻] = 0.001 M
[A⁻] = 0.001 M - 0.0002 M = 0.0008 M
The values can then be substituted into the Henderson-Hasselbalch equation:
pH = 7.2 + log(0.0008M/0.0002M) = 7.8
Final answer:
Using the Henderson-Hasselbalch equation with the provided values, the pH of the dye solution is calculated to be approximately 7.8.
Explanation:
To calculate the pH of the solution, we can use Henderson-Hasselbalch equation which relates pH, pKa, and the ratio of the concentrations of the deprotonated (In−) to protonated (HIn) forms of the indicator.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([In−]/[HIn])
Given that the pKa of the dye is 7.2 and the concentration of the protonated form ([HIn]) is 0.0002 M, and the total concentration of the dye is 0.001 M, we can infer that the concentration of the deprotonated form ([In−]) is 0.001 M - 0.0002 M = 0.0008 M. Using these values in the Henderson-Hasselbalch equation:
pH = 7.2 + log(0.0008/0.0002)
pH = 7.2 + log(4)
pH = 7.2 + 0.6021
pH = 7.8021
Therefore, the pH of the solution is approximately 7.8.
A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a Soave- Redlich-Kwong gas and with Cp* = 100 J/(mol·K), from 300 K and 2 m^3 to 800 K and 0.02 m^3 by using less than 5 MJ of work. Is this possible?
Answers
Explanation:
The given data is as follows.
Moles of propylene = 100 moles, [tex]C_{p}[/tex] = 100 J/mol K
[tex]T_{i}[/tex] = 300 K, [tex]T_{f}[/tex] = 800 K
[tex]V_{i}[/tex] = 2 [tex]m^{3}[/tex], [tex]V_{f}[/tex] = 0.02 [tex]m^{3}[/tex]
Therefore, the assumptions will be as follows.
The given system is very well insulated.The work is done on the system because the given process is a compression process.Assume that there is no friction so, work done on the system is equal to the heat energy liberated.[tex]m \times C_{p} \Delta T[/tex] = W
Putting the given values into the above formula as follows.
[tex]m \times C_{p} \Delta T[/tex] = W
W = [tex]100 moles \times 100 J/mol K \times (800 K - 300 K)[/tex]
= [tex]5 \times 10^{6}[/tex] J
= 5 MJ
Hence, this shows that a minimum of 5 MJ work needs to be done.
Since, work is very less. Hence, it will not compress the given system to 800 K and 0.02 [tex]m^{3}[/tex].
The activation energy for the reaction NO2 (g )+ CO (g) ⟶ NO (g) + CO2 (g) is Ea = 217 kJ/mol and the change in enthalpy for the reaction is ΔH = -293 kJ/mol .
What is the activation energy for the reverse reaction?
Enter your answer numerically and in terms of kJ/mol.
Answers
Answer : The activation energy for the reverse reaction is 510 kJ/mol.
Explanation :
Activation energy : It is defined as the minimum amount of energy given to the reactant so that it gets converted into products.
The relation between the activation energy for forward and backward reaction and change in enthalpy of reaction for exothermic reaction is:
When activation energy for forward reaction is less than the activation energy for backward reaction then the reaction will be exothermic. In exothermic reaction the enthalpy change will be negative.
[tex]Ea^b=Ea^f+|\Delta H|[/tex]
The relation between the activation energy for forward and backward reaction and change in enthalpy of reaction for endothermic reaction is:
When activation energy for forward reaction is more than the activation energy for backward reaction then the reaction will be endothermic. In endothermic reaction the enthalpy change will be positive.
[tex]Ea^f=Ea^b+|\Delta H|[/tex]
where,
[tex]Ea^f[/tex] = activation energy for forward reaction
[tex]Ea^b[/tex] = activation energy for backward reaction
[tex]\Delta H[/tex] = change in enthalpy of reaction
As per question, the value of enthalpy change is -293 kJ/mol that means the reaction will be exothermic reaction. So,
[tex]Ea^b=Ea^f+|\Delta H|[/tex]
Given:
[tex]Ea^f[/tex] = activation energy for forward reaction = 217 kJ/mol
[tex]Ea^b[/tex] = activation energy for backward reaction = ?
[tex]\Delta H[/tex] = change in enthalpy of reaction = -293 kJ/mol
Now put all the given values in above relation, we get:
[tex]Ea^b=217kJ/mol+|-293kJ/mol|[/tex]
[tex]Ea^b=217kJ/mol+293kJ/mol[/tex]
[tex]Ea^b=510kJ/mol[/tex]
Therefore, the activation energy for the reverse reaction is 510 kJ/mol.
Which type of microscope can be used to view cellular organelles such as the endoplasmic reticulum and Golgi?
Answers
An electron microscope is used to view cellular organelles such as the endoplasmic reticulum and Golgi because it provides significantly higher resolution and magnification compared to a light microscope.
Explanation:The type of microscope used to view cellular organelles such as the endoplasmic reticulum and Golgi is the electron microscope. This instrument magnifies an object using an electron beam that passes and bends through a lens system, providing much higher resolution and magnification than a light microscope. However, most student microscopes are light microscopes, which use a beam of visible light and are typically used for viewing living organisms, as the staining required to make cellular components visible usually kills the cells. Electron microscopes are commonly used in labs and can give detailed visualizations of organelles and the endomembrane system, which involves a group of organelles and membranes that work together in modifying, packaging, and transporting lipids and proteins.
Learn more about electron microscope here:https://brainly.com/question/507443
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A two-liter soft drink bottle can withstand apressure of
5 atm. Half a cup (approximately 120mL) of ethynlalcohol, C2H5OH,
(d=0.789 g/mL) is poured into a soft drink bottleat room
temperature. The bottle is then heated to 100C, changingthe liquid
alcohol to a gas. Will the soft drink bottle withstandthe pressure
or explode?
Answers
Explanation:
According to the ideal gas equation, PV = nRT.
where, P = pressure, V = volume
n = no. of moles, R = gas constant
T = temperature
Also, density is equal to mass divided by volume. And, no. of moles equals mass divided by molar mass.
Therefore, then formula for ideal gas could also be as follows.
P = [tex]\frac{mass}{volume \times molar mass} \times RT[/tex]
or, P = [tex]\frac{density}{\text{molar mass}} \times RT[/tex]
Since, density is given as 0.789 g/ml which is also equal to 789 g/L (as 1000 mL = 1 L). Hence, putting the given values into the above formula as follows.
P = [tex]\frac{density}{\text{molar mass}} \times RT[/tex]
= [tex]\frac{789 g/l}{46.06 g/mol} \times 0.0821 L atm/mol K \times 373 K[/tex]
= 525 atm
As two-liter soft drink bottle can withstand a pressure of 5 atm and the value of calculated pressure is 525 atm which is much greater than 5 atm.
Therefore, the soft drink bottle will obviously explode.
An aqueous solution of sulfuric acid has a composition of 25wt% and a SG of 1.22. Calculate the Volume of the solution that has 245 kg of sulfuric acid.
Answers
Answer: The volume of solution is [tex]8.03\times 10^5mL[/tex]
Explanation:
The relationship between specific gravity and density of a substance is given as:
[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]
Specific gravity of sulfuric acid solution = 1.22
Density of water = 1.00 g/mL
Putting values in above equation we get:
[tex]1.22=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.22\times 1.00g/mL)=1.22g/mL[/tex]
We are given:
25% (m/m) sulfuric acid solution. This means that 25 g of sulfuric acid is present in 100 g of solution
Conversion factor: 1 kg = 1000 g
Mass of solution having 254 kg or 245000 g of sulfuric acid is calculated by using unitary method:
If 25 grams of sulfuric acid is present in 100 g of solution.
So, 245000 grams of sulfuric acid will be present in = [tex]\frac{100}{25}\times 245000=980000g[/tex]
To calculate volume of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.22 g/mL
Mass of Solution = 980000 g
Putting values in above equation, we get:
[tex]1.22g/mL=\frac{980000g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{980000g}{1.22g/mL}=8.03\times 10^5mL[/tex]
Hence, the volume of solution is [tex]8.03\times 10^5mL[/tex]
In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several of the chemical constituents of wood but not cellulose. The slurry of undissolved chips in solution is further processed to recover most of the original solution constituents and dried wood pulp. In one such process, wood chips with a specific gravity of 0.640 containing 45.0 wt% water are treated to produce 2000.0 tons/day of dry wood pulp containing 85.0 wt% cellulose. The wood chips contain 47.0 wt% cellulose on a dry basis. Estimate the feed rate of logs (logs/min), assuming that the logs have an average diameter of 8.00 inches and an average length of 9.00 feet. 21.67 Ulogs/min
Answers
Final answer:
To estimate the feed rate of logs, we first need to calculate the dry wood pulp production rate using the given information.
Explanation:
To estimate the feed rate of logs, we first need to calculate the dry wood pulp production rate. The dry wood pulp production rate can be calculated using the equation:
Dry wood pulp production rate = Feed rate of logs x (1 - water content of wood chips) x (1 - cellulose content in wood chips on a dry basis)
Plugging in the given values, we have:
2000 tons/day = Feed rate of logs x (1 - 0.45) x (1 - 0.47)
Solving for the feed rate of logs, we find that it is approximately 21.67 logs/min.
What volume of concentrated nitric acid (15.0 M) is
requiredfor the preparationof 2.00 L of 0.001M nitric acid
solution?
Answers
Answer:
130 μL
Explanation:
The dilution formula is used to calculate the volume V₁ required:
C₁V₁ = C₂V₂
V₁ = (C₂V₂)/C₁ = (2.00L)(0.001M)/(15.0M) = 1.3 x 10⁻⁴ L or 130 μL
(1.3 x 10⁻⁴ L)(10⁶ μL/L) = 130 μL
What is the volume (in dm3 of 1 mole of oxygen at 5 MPa and 200 K?
Answers
Answer: The volume of oxygen gas is [tex]0.332dm^3[/tex]
Explanation:
To calculate the volume of the gas, we use the equation given by ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 5 MPa = 5000 kPa (Conversion factor: 1 MPa = 1000 kPa)
V = Volume of gas = 3.34 L
n = number of moles of oxygen gas = 1 mole
R = Gas constant = [tex]8.31dm^3\text{ kPa }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = 200 K
Putting values in above equation, we get:
[tex]5000kPa\times V=1mol\times 8.31dm^3\text{ kPa }mol^{-1}K^{-1}\times 200K\\\\V=0.332dm^3[/tex]
Hence, the volume of oxygen gas is [tex]0.332dm^3[/tex]
If a solid has a heat of fusion of 17.02 kJ/mol and an entropy of fusion of 38.98 J/mol- K, what is the melting point in °C) of this pure solid? Type your answer rounded to 1 decimal place without units (i.e. NN.N).
Answers
Explanation:
Melting point is defined as the point at which a solid substance starts to change into liquid state.
Whereas entropy is the degree of randomness of molecules present in a substance.
Heat of fusion is defined as the amount of heat energy necessary to melt a solid substance at its melting point.
Relation between entropy and heat of fusion is as follows.
[tex]\Delta S = \frac{\Delta H}{T}[/tex]
where, [tex]\Delta S[/tex] = 38.98 J/mol K
[tex]\Delta H[/tex] = 17.02 kJ/mol
= [tex]17.02 kJ/mol \times \frac{1000 J}{1 kJ}[/tex]
= 17020 J/mol
Therefore, calculate the melting point as follows.
[tex]\Delta S = \frac{\Delta H}{T}[/tex]
38.98 J/mol K = [tex]\frac{17020 J/mol}{T}[/tex]
T = 436.63 K
Change the temperature into degree celsius as follows.
[tex](436.63 - 273)^{o}C[/tex]
= [tex]163.63^{o}C[/tex]
Thus, we can conclude that the melting point in [tex]^{o}C[/tex] is [tex]163.63^{o}C[/tex].