Answer:The chemical shift (δ) is 6.88ppm.
Explanation:
We have the following data :
Absorption frequency of the proton in bromoform=2065Hz
Frequency of the NMR spectrometer(instrument)=300MHz
The formula for calculating the chemical shift (δ) in PPM is:
chemical shift (δ)={[Frequency of proton(Hz) -Frequency of reference(Hz]÷Frequency of NMR spectrometer(MHz }
Using the formula for chemical shift we can calculate the value of chemical shift (δ) in ppm
chemical shift (δ) ={[2065Hz-0]÷300×10⁻⁶}
chemical shift (δ) =6.88×10⁶
chemical shift (δ)=6.88ppm for a 300MHz NMR spectrometer
The chemical shift (δ) is a ratio of frequency absorbed by proton with that of NMR spectrometer frequency hence the chemical shift value would remain same what ever NMR spectrometer frequency we use. Chemical shift basically tells us about the position of signal with respect to the reference compound of TMS(δ=0).
chemical shift (δ) is measured in
So the value of (δ) is same for any spectrometer used.
The chemical shift (δ) for a 500MHz NMR spectrometer used would also be 6.88PPM.
Alternatively since the frequency of proton absorbed is directly related to the magnetic field applied that is the frequency of NMR spectrometer hence:
Let the frequency of proton absorbed in 300MHZ=V₁=2065
Let the frequency of proton absorbed in 500MHZ=V₂=?
frequency of proton absorbed∝Applied magnetic field(Frequency of NMR spectrometer)
So V₁/V₂=[300×10⁶]/[500×10⁶]
V₂=[2065×500]÷300
V₂=3441HZ
For a 500 MHz proton the frequency of absorption would be 3441MHz
using this frequency we can calculate chemical shift (δ) using above formula:
(δ)=[3441-0]/[500×10⁻⁶]
(δ)=6.88PPM
Hence we obtain the same value of chemical shift in both the spectrometers.
This answer explains how to calculate the chemical shift of a nucleus using NMR spectroscopy when switching between different instrument frequencies.
Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful tool used by chemists to analyze the structure of molecules. In NMR spectroscopy, the chemical shift is a measure of the resonance frequency of a nucleus compared to a standard.
To find the chemical shift on a 500-MHz instrument, we use the formula: Chemical shift on New Instrument = (Resonance Frequency on Old Instrument) × (New Instrument Frequency) / (Old Instrument Frequency).
Plugging in the values, we get: Chemical shift = 2065 Hz × 500 MHz / 300 MHz = 3441.67 Hz.
Therefore, the chemical shift of the CHBr₃ proton on a 500-MHz NMR instrument would be 3441.67 Hz.
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of 0.5 bar from 100 °C to such a temperature that the specific volume increases by 2.5 times. If the amount of heat that must be added to accomplish this change is 500 kJ, calculate the final temperature of the steam, the expansion work, and the change in internal energy.
Answer:
Final temperature: 659.8ºC
Expansion work: 3*75=225 kJ
Internal energy change: 275 kJ
Explanation:
First, considering both initial and final states, write the energy balance:
[tex]U_{2}-U_{1}=Q-W[/tex]
Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:
[tex]dw=Pdv[/tex]
The pressure is constant, so: [tex]w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}[/tex] (There is a multiplication by 100 due to the conversion of bar to kPa)
So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):
[tex]U_{2}-U_{1}=500-(3*75)=275kJ[/tex]
On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:
[tex]P_{1}V_{1}=nRT_{1}[/tex]
[tex]P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}[/tex]
Subtracting the first from the second:
[tex]1.5P_{1}V_{1}=nR(T_{2}-T_{1})[/tex]
Isolating [tex]T_{2}[/tex]:
[tex]T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}[/tex]
Assuming that it is water steam, n=0.1666 kmol
[tex]V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}[/tex]
[tex]T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76[/tex] ºC
Each of the following substances is a liquid at –50°C. Place these liquids in order of increasing vapor pressure: dimethyl ether (CH3OCH3), propane (C3H8), and ethanol (CH3CH2OH).a) propane < ethanol < dimethyl etherb)ethanol < propane < dimethyl etherc)propane < dimethyl ether < ethanold)ethanol < dimethyl ether < propanee)dimethyl ether < ethanol < propane
Answer:
Increasing vapor pressure order -
propane > dimethyl ether > ethanol
Explanation:
Vapor pressure - the pressure exerted the gaseous molecules , on the walls of the container is called the vapor pressure.
Boiling point - the temperature at which the the vapor pressure of the liquid equals the external atmospheric pressure.
Both boiling point and vapor pressure are linked by the inter molecular forces between the atoms.
The compound with stronger inter molecular forces are tightly held , hence more amount of energy is required to vaporize. Therefore, higher boiling point , and in turn the vapor pressure will be lower .
And the compound with weaker inter molecular forces are loosely held , hence less amount of energy is required to vaporize. Therefore, lower boiling point , and in turn the vapor pressure will be higher .
Therefore,
Ethanol has lowest vapor pressure , because at lower temperature , it has H - bonding , hence, its boiling point is more, therefore, less vapor pressure.
Dimethyl ether has vapor pressure more than ethanol as it has weaker dipole - dipole interactions as compared to strong H - bonding .
Propane has maximum vapor pressure among all , because of lower molecular mass than others , it can easily vaporize into vapors.
Hence,
Increasing vapor pressure order -
propane > dimethyl ether > ethanol
The Increasing vapor pressure order of vapor pressure for the liquids listed in the question is; propane > dimethyl ether > ethanol.
The vapor pressure of a liquid shows how easily the liquid is converted to vapor. Liquids that are easily converted to vapor are said to have a low vapor pressure.
The Increasing vapor pressure order of vapor pressure for the liquids listed in the question is; propane > dimethyl ether > ethanol. This is because, propane has the least intermolecular interaction and ethanol has the highest degree of intermolecular interaction among the molecules in the list.
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16.78 A buffer is prepared by mixing 525 mL of 0.50 M formic acid, HCHO2, and 475 mL of 0.50 M sodium for- mate, NaCHO2. Calculate the pH. What would be the pH of 85 mL of the buffer to which 8.6 mL of 0.15 M hydro- chloric acid had been added?
Answer: Before addition of HCl, the pH is 3.70 and after addition of HCl, the pH is 3.64 .
Explanation: The pH of the buffer solution is calculated by using Handerson equation:
[tex]pH=pKa+log(\frac{base}{acid})[/tex]
Let's calculate the moles of acid and base(salt) present in the original buffer.
mL are converted to L and then multiplied by molarity to get the moles.
[tex]525mL(\frac{1L}{1000mL})(\frac{0.50molHCHO_2}{1L})[/tex]
= [tex]0.2625molHCHO_2[/tex]
[tex]475mL(\frac{1L}{1000mL})(\frac{0.50molNaCHO_2}{1L})[/tex]
= [tex]0.2375molNaCHO_2[/tex]
Total volume of the buffer solution = 0.525 L + 0.475 L = 1.00 L
Since, the total volume is 1.00 L, concentration of base will be 0.2375 M and the concentration of acid will be 0.2625 M.
pKa for formic acid is 3.74. Let's plug in the values in the equation and calculate the pH of the original buffer.
[tex]pH=3.74+log(\frac{0.2375}{0.2625})[/tex]
pH = 3.74 - 0.04
pH = 3.70
Now, we add 8.6 mL of 0.15 M HCl acid to 85 mL of the buffer. Let's calculate the moles of acid and base in 85 mL of the buffer.
[tex]85mL(\frac{1L}{1000mL})(\frac{0.2625molHCHO_2}{1L})[/tex]
= [tex]0.0223molHCHO_2[/tex]
[tex]85mL(\frac{1L}{1000mL})(\frac{0.2375molNaCHO_2}{1L})[/tex]
= [tex]0.0202molNaCHO_2[/tex]
Now, let's calculate the moles of HCl added to the buffer.
[tex]8.6mL(\frac{1L}{1000mL})(\frac{0.15molHCl}{1L})[/tex]
= 0.00129 mol HCl
This added HCl reacts with base(sodium formate) and formic acid is produced.
So, 0.00129 moles of HCl will react with 0.00129 moles of sodium formate to produce 0.00129 moles of formic acid. We can write formate ion in place of sodium formate and hydrogen ion in place of HCl. The equation would be:
[tex]H^++CHO_2^-\rightarrow HCHO_2[/tex]
moles of base after reaction with HCl = 0.0202 mol - 0.00129 mol = 0.01891 mol
moles of acid after addition of HCl = 0.0223 mol + 0.00129 mol = 0.02359 mol
Let's plug in the values again in the Handerson equation and calculate the pH:
[tex]pH=3.74+log(\frac{0.01891}{0.02359})[/tex]
pH = 3.74 - 0.096
pH = 3.64
Which one of the following is an attribute for an extracting solvent?
a. The extracting solvent should not be volatile
b. The extracting solvent should be non-toxic and readily available
c. The extracting solvent should would react with the solute
d. The extracting solvent should be miscible with water
Answer:
The answer is B
[The extracting solvent should be non-toxic and readily available]
Explanation:
The extracting solvent should not pose a risk to life (or the risk have to be minimum) and be available in large quantity.
A. Most of the extracting solvents are volatile, and is a good think because you can remove them by distillation.
C. The extraction solvent should not react with the solute, because you lose the extracted compound.
D. The extraction solvent should not be miscible with water. Usually you extract compounds organic compounds from water using an appropriate extracting solvent.
The correct attribute for an extracting solvent is that it should be non-toxic and readily available, making (b) the right answer. Volatility, reactivity with the solute, and miscibility with water are undesirable features for an extraction solvent.
An extracting solvent is used in liquid-liquid extraction to separate compounds based on their solubility properties. The ideal characteristics of an extracting solvent include being non-toxic, readily available, not miscible with water, and having a different density than water to facilitate easy separation. For example, diethyl ether is an effective extracting solvent because of its ability to dissolve non-ionic organic compounds, immiscibility with water, and the fact that ionic compounds are generally insoluble in it. In contrast, Toluene (C₆H₅-CH₃) is a nonpolar solvent which is ideal for dissolving nonpolar substances like octane , but not polar substances such as water (H₂O) or ionic compounds like sodium sulfate .
In selecting the correct answer to the student's question, the attribute that an extracting solvent should have is that it should be non-toxic and readily available. Solvents that are volatile, reactive with the solute, or miscible with water are not ideal for extraction purposes. Therefore, the correct choice is (b).
A preschool has Campbell's Chunky Beef soup, which contains 2.5 g of fat and 15 mg of cholesterol per serving (cup), and Campbell's Chunky Sirloin Burger soup, which contains 7 g of fat and 15 mg of cholesterol per serving. By combining the soups, it is possible to get 8 servings of soup that will have 29 g of fat and 120 mg of cholesterol. How many cups of each soup should be used?
The problem is solved using system of equations. Considering the given fat and cholesterol content of each soup, we create two equations and solve for x (Chunky Beef soup servings) and y (Chunky Sirloin Burger soup servings), finding that 4 servings of each soup are needed.
Explanation:This problem can be solved using system of equations. Let's define x as the number of servings of Chunky Beef soup and y as the number of servings of the Chunky Sirloin Burger soup. Considering the fat content from both types of soup, we get the equation 2.5x + 7y = 29. The cholesterol content in both soups is the same (15mg per serving), yielding a second equation 15x + 15y = 120. The second equation simplifies to x + y = 8, meaning that we will have a total of 8 servings.
By solving the system of equations, we find that x (the number of cups of Chunky Beef Soup) is 4 and y (the number of cups of the Chunky Sirloin Burger soup) is also 4.
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The solution involves solving a system of equations. The calculation shows that to get 8 servings of soup with 29 g of fat and 120mg of cholesterol, we should mix approximately 3.33 cups of the Chunky Beef soup and 4.67 cups of the Chunky Sirloin Burger soup.
Explanation:This problem can be approached using a system of linear equations. Assume x is the number of cups of Campbell's Chunky Beef soup and y is the number of Campbell's Chunky Sirloin Burger soup.
From the question, we have:
1. x + y = 8 (since there are 8 servings total)
2. 2.5x + 7y = 29 (calculating total fat in grams)
3. 15x + 15y = 120 (calculating total cholesterol in milligrams).
The third equation simplifies to x + y = 8. This is equivalent to the first equation, so it doesn't provide any new information. We can therefore solve the first and second equations simultaneously.
Subtract the first equation from the second to get 4.5y = 21, from which y = 21/4.5 = 4.67 (approximately). Substituting y = 4.67 into the first equation, we get x = 8 - 4.67 = 3.33. Therefore, to get 8 servings of soup with 29 g of fat and 120 mg of cholesterol, we should mix approximately 3.33 cups of Campbell's Chunky Beef soup and 4.67 cups of Campbell's Chunky Sirloin Burger soup.
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You have 3.00 L of a 3.00 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 M solution of AgNO3(aq) called solution B. You mix these solutions together, making solution C. Hint: AgCl is a precipitate. Calculate the concentrations (in M) of the following ions in solution C. NO3-
The concentration of NO3- ions in solution C, which is made from a mixture of NaCl and AgNO3 solutions, is calculated to be 0.80 M.
Explanation:To calculate the concentration of the NO3- in solution C, we consider both AgNO3 and NaCl. In AgNO3( aq), each molecule yields an Ag+ and a NO3-. The total amount of moles of NO3- from this solution can be calculated by multiplying the volume of the solution(2.00 L) by its concentration(2.00 M). This gives 4.00 moles of NO3-.
As for NaCl(aq), it yields Na+ and Cl- ions when dissolved but no NO3- ions. So, NO3- ions will only come from solution B in this case.
The total volume of solution C is the volume of solution A plus the volume of solution B, that is, 3.00L + 2.00L = 5.00L. We can now calculate the concentration of NO3- ions in solution C by dividing the total moles of NO3- (4.00 moles) by the total volume of solution C (5.00 L). Therefore, the concentration of the NO3- ion in solution C is 4.00 moles / 5.00 L = 0.80 M.
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The concentration of [tex]NO3^-[/tex] ions in solution C is 0.800 M.
The concentration of [tex]NO3^-[/tex]ions in solution C can be calculated by considering the moles of [tex]NO3^-[/tex] present in solution B and the final volume of solution C after mixing.
First, let's calculate the moles of [tex]NO3^-[/tex] in solution B:
Moles of [tex]NO3^-[/tex] = Concentration of [tex]NO3^-[/tex] in solution B × Volume of solution B
Moles of [tex]NO3^-[/tex]= 2.00 M × 2.00 L = 4.00 moles
Now, let's find the final volume of solution C after mixing solutions A and B:
Volume of solution C = Volume of solution A + Volume of solution B
Volume of solution C = 3.00 L + 2.00 L = 5.00 L
Next, we calculate the concentration of [tex]NO3^-[/tex] in solution C:
Concentration of [tex]NO3^-[/tex] in solution C = Moles of [tex]NO3^-[/tex]/ Volume of solution C
Concentration of [tex]NO3^-[/tex] in solution C = 4.00 moles / 5.00 L = 0.800 M
Therefore, the concentration of [tex]NO3^-[/tex] ions in solution C is 0.800 M.
The answer is: 0.800.
Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide (NO) to N2 and O2 is second order with a rate constant of 0.0796 M−1⋅s−1 at 737∘C and 0.0815 M−1⋅s−1 at 947∘C. Calculate the activation energy for the reaction in kJ/mol
Answer : The activation energy for the reaction is, 1.151 KJ
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at [tex]737^oC[/tex] = [tex]0.0796M^{-1}s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at [tex]947^oC[/tex] = [tex]0.0815M^{-1}s^{-1}[/tex]
[tex]Ea[/tex] = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]737^oC=273+737=1010K[/tex]
[tex]T_2[/tex] = final temperature = [tex]947^oC=273+947=1220K[/tex]
Now put all the given values in this formula, we get:
[tex]\log (\frac{0.0815M^{-1}s^{-1}}{0.0796M^{-1}s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{1010K}-\frac{1}{1220K}][/tex]
[tex]Ea=1151.072J/mole=1.151KJ[/tex]
Therefore, the activation energy for the reaction is, 1.151 KJ
60.0 mL of 3.00 M Na2CO3 and 40. mL of 2.00 M CaCl2 were mixed together. What is the limiting reagent, and what is the theoretical yield of Ca2CO3 in moles?
Answer: [tex]CaCl_2[/tex] is the limiting reagent and theoretical yield of [tex]CaCO_3[/tex] is 8 grams.
Explanation:
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
Thus [tex]\text{no of moles}of Na_2CO_3={3.0M}\times {0.06 L}=0.18moles[/tex]
Thus [tex]\text{no of moles}of CaCl_2={2.0M}\times {0.04 L}=0.08moles[/tex]
[tex]Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl[/tex]
As 1 mole of [tex]Na_2CO_3[/tex] combines with 1 mole of [tex]CaCl_2[/tex]
Thus 0.08 moles of [tex]CaCl_2[/tex] react with =[tex]\frac{1}{1}\times 0.08=0.08[/tex] moles of [tex]Na_2CO_3[/tex]
Thus [tex]CaCl_2[/tex] is the limiting reagent as it limits the formation of products and [tex]Na_2CO_3[/tex] is an excess reagent.
1 moles of [tex]CaCl_2[/tex] form = 1 mole of [tex]CaCO_3[/tex]
0.08 moles of [tex]CaCl_2[/tex] form =[tex]\frac{1}{1}\times 0.08=0.08[/tex] moles of [tex]CaCO_3[/tex]
[tex]{\text {Mass of}CaCO_3=moles\times {\text {Molar Mass}}=0.08moles\times 100g/mol=8g[/tex]
Thus theoretical yield of [tex]CaCO_3[/tex] is 8 grams.
Given:
60.0 mL of 3.00 M Na₂CO₃ 40.0 mL of 2.00 M CaCl₂Question:
What is the limiting reagent?What is the theoretical yield of CaCO₃ in moles?The Process:
Step-1: prepare moles for each reagent
[tex]\boxed{ \ Molarity = \frac{moles}{volume} \ }[/tex] [tex]\rightarrow \boxed{ \ n = MV \ }[/tex]
[tex]\boxed{ \ Na_2CO_3 \ }[/tex] [tex]\rightarrow \boxed{ \ n = 3 \ \frac{mol}{L} \times 60 \ mL \ }[/tex] [tex]\rightarrow \boxed{ \ n = 180 \ mmol \ }[/tex]
[tex]\boxed{ \ CaCl_2 \ }[/tex] [tex]\rightarrow \boxed{ \ n = 2 \ \frac{mol}{L} \times 40 \ mL \ }[/tex] [tex]\rightarrow \boxed{ \ n = 80 \ mmol \ }[/tex]
Step-2: the ICE table
We use the ICE table to see how the reaction occurs between two salts.
Balanced reaction:
[tex]\boxed{ \ Na_2CO_3 + CaCl_2 \rightarrow CaCO_3 + 2NaCl \ }[/tex]
Initial: 180 80 - -
Change: 80 80 80 80
Equlibrium: 100 - 80 80
Notice that the coefficient ratio between Na₂CO₃ and CaCl₂ is 1 : 1. Remember, the coefficient ratio is also the mole ratio of the reactants.Based on this ratio, it appears that the initial amount of Na₂CO₃ is more than CaCl₂, or the amount of Na₂CO₃ is known to be excessive.CaCl₂ acts as the limiting reagent because the amount does not remain at the end of the reaction.Final step: calculate the theoretical yield of CaCO₃ in moles and grams
From the ICE table above, the theoretical yield of CaCO₃ produced is 80 mmol or 0.08 moles.
Let us try to continue by calculating the mass of CaCO₃ produced.
Relative atomic mass: Ca = 40, C = 12, and O = 16.Relative molecular mass (Mr) of CaCO₃ = 40 + 12 + 3(16) = 100 g/mol.Let us convert mol to grams.
[tex]\boxed{ \ n = \frac{mass}{Mr} \ }[/tex] [tex]\rightarrow \boxed{ \ mass = mol \times Mr \ }[/tex]
[tex]\boxed{ \ CaCO_3 \ }[/tex] [tex]\rightarrow \boxed{ \ mass = 0.08 \ moles \times 100 \ \frac{g}{mol} \ }[/tex]
Thus, the theoretical yield of CaCO₃ in grams equal to 8 grams.
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Answer:
the compound is hexadeca-6,10-diyne.
Explanation:
As given that the compound has two triple bonds and it is giving carboxylic acids on ozonolyis so this is an oxidative ozonolysis.
We are getting only two kinds of products, it means the triple bonds are located at symmetrical position.
The middle carbon chain must have four carbons and the side chains must have six carbons each,
The structure of hydrocarbon is shown in figure.
Draw a structure of glycogen with two alpha (1>4) linkages and two alpha (1>6) 3 linkages between glucose molecules What advantage does a branched structure have over a straight chain polysaccharide? b. Give the primary location of glycogen in the body.
Answer:
Here's what I get
Explanation:
a. Structure
The structure of glycogen is shown below.
The bonds between the glucose units in the horizontal chain are α(1⟶4) linkages.
The upper glucose units are linked to the main chain by α(1⟶6) linkages.
Enzymes attack the end units of a polysaccharide. The branched structure of glycogen provides more end units, so enzymes can break it down into glucose more quickly.
b. Storage site
The primary site for the storage if glycogen is the liver.
The secondary site is muscle tissue. The concentration of glycogen in the liver is five times that in muscle, but there is more glycogen in muscle because the body has a much greater muscle mass.
2.61 kilograms of water in a container have a pressure of 200 kPa and temperature of 200°C . What is the volume of this container? m (Round to three decimal places)
Answer: The volume of the container is [tex]2.8497m^3[/tex]
Explanation:
To calculate the volume of water, we use the equation given by ideal gas, which is:
[tex]PV=nRT[/tex]
or,
[tex]PV=\frac{m}{M}RT[/tex]
where,
P = pressure of container = 200 kPa
V = volume of container = ? L
m = Given mass of water = 2.61 kg = 2610 g (Conversion factor: 1kg = 1000 g)
M = Molar mass of water = 18 g/mol
R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]
T = temperature of container = [tex]200^oC=[200+273]K=473K[/tex]
Putting values in above equation, we get:
[tex]200kPa\times V=\frac{2610g}{18g/mol}\times 8.31\text{L kPa }\times 473K\\\\V=2849.7L[/tex]
Converting this into cubic meter, we use the conversion factor:
[tex]1m^3=1000L[/tex]
So, [tex]\Rightarrow \frac{1m^3}{1000L}\times 2849.7L[/tex]
[tex]\Rightarrow 2.8497m^3[/tex]
Hence, the volume of the container is [tex]2.8497m^3[/tex]
A pump increases the water pressure from 70 kPa at the inlet to 700 kPa at the outlet. Water enters this pump at 15 °C through a 1-cm-diameter opening and exits through a 1.5-cm-diameter opening. Determine the velocity of the water at the inlet and outlet when the mass flow rate through the pump is 0.5 kg/s. How much work is required to operate the pump?
Answer:
[tex]v_{in} =6.373 \frac{m}{s}[/tex]
[tex]v_{out} =2.832 \frac{m}{s}[/tex]
[tex]W=0.63063*0.5=0.3153kW[/tex]
Explanation:
First, we are going to need the water specific volume at 15ºC: v=0.001001 [tex]\frac{m^{3}}{kg}[/tex]. The density '[tex]p[/tex]' of the water is the inverse of the specific volume: [tex]p=999 \frac{kg}{m^{3} }[/tex]
First, consider the mass flow, which is related to the volumetric flow (density and velocity) and the area:
[tex]m=pvA[/tex]
The area of each cross-section is:
[tex]A_{in} =\frac{\pi*0.01^{2} }{4}=7.854*10^{-5}[/tex] (in square meters). Here, the radius was not used but the diameter, which means a division by 4 (2 squared).
[tex]A_{out} =\frac{\pi*0.015^{2} }{4}=1.767*10^{-4}m^{2}[/tex]
From mass flow isolate the velocity and calculate it:
[tex]v=\frac{m}{pA}[/tex]
[tex]v_{in} =\frac{0.5}{999*7.854*10^{-5} }=6.373 \frac{m}{s}[/tex]
[tex]v_{out} =\frac{0.5}{999*1.767*10^{-4} }=2.832 \frac{m}{s}[/tex]
The work of the pump is calculated considering an energy balance on the pump:
[tex]w=h_{in}-h_{out}[/tex]
Considering the isentropic process may give us the relation:
[tex]dh=vdP\\ h_{2} -h_{1}=v*(P_{2} -P_{1})[/tex]
Applying that to the pump,
[tex]w=-0.001001*(700-70)=-0.63063 \frac{kJ}{kg}[/tex]
Multiplying it by the mass flow:
[tex]W=-0.63063*0.5=-0.3153kW[/tex]
The work is negative because it is entering to the system, but the required is positive. (It is just a standard rule)
The question involves applying the continuity equation in physics to calculate the velocities of water at the inlet and outlet of a pump, and computing the work done by the pump in increasing water pressure.
Explanation:To answer the question about the velocity of the water at the inlet and outlet of the pump, we can apply the principle of continuity equation which states that the density times the area times the velocity must remain constant. Since the density of the water does not change, the velocity times the cross-sectional area entering a region must equal the cross-sectional area times the velocity leaving the region.
The work done by the pump can be calculated, knowing that it can be used to increase the kinetic energy and gain potential energy as the pressure increases.
The power required to operate the pump can be computed from these parameters. A substantial amount of power requires a large pump, such as the one in the question which increases the water's pressure from 70 kPa to 700 kPa. This drastic increase in pressure indicates that a significant amount of work is being done on the fluid.
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Complete and balance the precipitation reactions. Include physical states. Refer to the solubility rules as necessary. precipitation reaction: AgNO_{3}(aq) + NaCl(aq) -> AgNO3(aq)+NaCl(aq)⟶ precipitation reaction: K_{3}PO_{4}(aq) + MgCl_{2}(aq) -> K3PO4(aq)+MgCl2(aq)⟶
The balanced chemical reactions are:
[tex]AgNO{_3}(aq) \ + \ NaCl(aq) \rightarrow \ AgCl(s) \ + NaNO_{3}(aq)\\2K_{3}PO_{4}(aq) \ + 3MgCl_{2}(aq) \rightarrow \ 6KCl(aq) \ + Mg_{3}(PO_{4})_{2}(s)[/tex]
Further Explanation:
The following reactions will undergo double displacement where the metal cations in each compound are exchanged and form new products.
For reaction 1, the compounds involved are nitrates and chlorides. To determine the states of the products, the solubility rules for nitrates and chlorides must be followed:
All nitrates are generally soluble.Chlorides are generally soluble except AgCl, PbCl2, and Hg2Cl2.Therefore, the products will have the following characteristics:
silver chloride (AgCl) is insolublesodium nitrate (NaNO3) is solubleFor reaction 2, the compounds involved are phosphates and chlorides. The solubility rules for phosphates and chlorides are as follows:
Phosphates are generally insoluble except for Group 1 phosphates.Chlorides are generally soluble except for AgCl, PbCl2, and Hg2Cl2.Hence, the products of the second reaction will have the following characteristics:
potassium chloride (KCl) is solublemagnesium phosphate is insolubleInsoluble substances are denoted by the symbols (s) in a chemical equation. The soluble substances are denoted as (aq).
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Here are the balanced precipitation reactions with physical states: 1. AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq), 2. K3PO4(aq) + MgCl2(aq) -> 3KCl(aq) + Mg3(PO4)2(s)
To complete and balance the precipitation reactions, we need to determine which combinations of ions will form insoluble salts based on the solubility rules. Insoluble salts will precipitate out of solution. Here are the balanced precipitation reactions: AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)In this reaction, silver chloride (AgCl) is insoluble and forms a white precipitate, while sodium nitrate (NaNO3) remains in the aqueous state. The balanced equation represents the formation of solid AgCl.K3PO4(aq) + MgCl2(aq) -> 3KCl(aq) + Mg3(PO4)2(s)In this reaction, magnesium phosphate (Mg3(PO4)2) is insoluble and forms a solid precipitate, while potassium chloride (KCl) remains in the aqueous state. The balanced equation represents the formation of solid Mg3(PO4)2.These reactions follow the principle of double replacement reactions, where ions in the reactants switch partners to form new compounds. Solubility rules are used to identify which compounds remain soluble (aqueous) and which form insoluble precipitates when combined in solution.For more such questions on physical
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Animals breathe oxygen and release carbon dioxide during cellular respiration according to the equation below. How much oxygen is needed to produce 120 grams of carbon dioxide?
Answer:
14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.
Explanation:
Animals take in oxygen and breathe out carbon dioxide during cellular respiration. The reaction for the metabolism of the food in the animal body is:
[tex]C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + 6H_2O + Energy[/tex]
As can be seen from the reaction stoichiometry that:
6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.
Also,
Given :
Mass of carbon dioxide gas = 120 g
Molar mass of carbon dioxide gas = 44 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus, moles of carbon dioxide are:
[tex]moles_{CO_2} = \frac{120 g}{44 g/mol}[/tex]
[tex]moles_{CO_2} = 2.7273 mol[/tex]
As mentioned:
6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.
1 mole of carbon dioxide gas can be produced from 1/6 mole of oxygen gas in the process of metabolism of glucose.
2.7273 mole of carbon dioxide gas can be produced from [tex] \frac{1}{6} \times 2.7273[/tex] moles of oxygen gas in the process of metabolism of glucose.
Thus, moles of oxygen gas needed = 0.4545 moles
Molar mass of oxygen gas = 32 g/mol
The mass of oxygen gas can be find out by using mole formula as:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Mass\ of\ oxygen\ gas = Moles \times Molar mass}[/tex]
[tex]Mass\ of\ oxygen\ gas = 0.4545 \times 32}[/tex]
[tex]Mass\ of\ oxygen\ gas = 14.544 g[/tex]
14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.
Answer:
87.3 g
Explanation:
The cellular respiration can be represented through the following equation.
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
We can establish the following relations:
The molar mass of CO₂ is 44.01 g/mol.The molar ratio of CO₂ to O₂ is 6:6.The molar mass of O₂ is 32.00 g/mol.The mass of O₂ that produces 120 g of CO₂ is:
[tex]120gCO_{2}.\frac{1molCO_{2}}{44.01gCO_{2}} .\frac{6molO_{2}}{6molCO_{2}} .\frac{32.00gO_{2}}{1molO_{2}} =87.3gO_{2}[/tex]
An electrochemical cell based on the following reaction has a standard cell voltage (E°cell) of 0.48 V: Sn(s) + Cu2+(aq) → Sn2+(aq) + Cu(s) What is the standard reduction potential of tin(II)? (E°(Cu2+/Cu) = 0.34 V)
Answer:
0.82V
Explanation:
0.48V+0.34V=0.82V
Answer:
-.14
Explanation:
5.00 mL is withdrawn from a 1.00L solution of NaCl and diluted to 100.00 mL. If the concentration of the NaCl in the 100.00 mL solution is 0.076, what is the concentration of NaCl in the original 1.00 L solution? answers
Answer: The concentration of NaCl in 1.00 L of solution is 0.0076 M.
Explanation:
To calculate the molarity of the concentrated solution, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution
We are given:
Conversion factor used: 1L = 1000 mL
[tex]M_1=?M\\V_1=1L=1000mL\\M_2=0.076M\\V_2=100mL[/tex]
Putting values in above equation, we get:
[tex]M_1\times 1000=0.076\times 100\\\\M_1=0.0076M[/tex]
Hence, the concentration of NaCl in 1.00 L of solution is 0.0076 M.
Kinetics of chymotrypsin shows a biphasic pattern; the burst phase and the steady-state phase. 一一(T/F)
Answer:
TRUE
Explanation:
The reaction of chymotrypsin with its substrate shows a biphasic pattern.
There an initial burst phase at the beginning of the reaction, followed by a steady-state phase that follows Michaelis-Menten kinetics.
For example, a kinetic plot of the reaction with a coloured substrate looks like the diagram below.
Determine the rate law and the value of k for the following reaction using the data provided.2 NO(g) + O2(g) → 2 NO2(g)[NO]i (M)[O2]i (M)Initial Rate (M-2s-1)0.0300.00558.55 x 10-30.0300.01101.71 x 10-20.0600.00553.42 x 10-2
Answer: Rate law=[tex]Rate=k[NO]^2[O_2]^1[/tex]
Rate law constant is [tex]1727.3L^2mol^{-2}s^{-1}[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]
[tex]Rate=k[NO]^x[O_2]^y[/tex]
k= rate constant
x = order with respect to NO
y = order with respect to [tex]O_2[/tex]
n = x+y = Total order
a) From trial 1: [tex]8.55 x 10^{-3}=k[0.030]^x[0.0055]^y[/tex] (1)
From trial 2: [tex]1.71 x 10^{-2}=k[0.030]^x[0.0110]^y[/tex] (2)
Dividing 2 by 1 :[tex]\frac{1.71\times 10^{-2}}{8.55\times 10^{-3}}=\frac{k[0.030]^x[0.0110]^y}{k[0.030]^x[0.0055]^y}[/tex]
[tex]2=2^y,2^1=2^y[/tex] therefore y=1.
b) From trial 1 :[tex]8.55 x 10^{-3}=k[0.030]^x[0.0055]^y[/tex] (3)
From trial 3:[tex]3.42\times 10^{-2}=k[0.060]^x[0.0055]^y[/tex] (4)
Dividing 4 by 3:[tex]\frac{3.42\times 10^{-2}}{8.55\times 10^{-3}}=\frac{k[0.060]^x[0.0055]^y}{k[0.030]^x[0.0055]^y}[/tex]
[tex]4=2^x,2^2=2^x[/tex], x=2Thus rate law is [tex]Rate=k[NO]^2[O_2]^1[/tex]
Thus order with respect to [tex]NO[/tex] is 2 , order with respect to [tex]O_2[/tex] is 1 and total order is 1+2=3.
Rate law is [tex]Rate=k[NO]^2[O_2]^1[/tex]
b) For calculating k:
Using trial 1: [tex]8.55\times 10^{-3}=k[0.030]^2[0.0055]^1[/tex]
[tex]k=1727.3L^2mol^{-2}s^{-1}[/tex]
The value of rate constant is [tex]1727.3L^2mol^{-2}s^{-1}[/tex]
The combustion of propane (C3H8) produces CO2 and H2O: C3H8 (g) 5O2 (g) → 3CO2 (g) 4H2O (g) The reaction of 2.5 mol of O2 will produce ________ mol of H2O.
To find the amount of water produced when 2.5 mol of O2 is consumed in the combustion of propane, we can use stoichiometry. The combustion of 2.5 mol of O2 will produce 2 mol of H2O.
Explanation:The balanced equation for the combustion of propane is:
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)
To find the amount of water produced when 2.5 mol of O2 is consumed, we need to use stoichiometry. Since the ratio of O2 to H2O is 5:4, we can set up the following proportion:
2.5 mol O2 / 5 mol O2 = x mol H2O / 4 mol H2O
Solving for x, we find that x = 2 mol H2O. Therefore, the combustion of 2.5 mol of O2 will produce 2 mol of H2O.
Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount required to combust all of the ethanol by 10%, and 85% of the ethanol reacts, calculate the mole fraction of oxygen leaving the reactor.
Answer:
[tex]y_{O2} =4.3[/tex]%
Explanation:
The ethanol combustion reaction is:
[tex]C_{2}H_{5} OH+3O_{2}[/tex]→[tex]2CO_{2}+3H_{2}O[/tex]
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:
[tex]x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}[/tex]
Dividing the previous equation by x:
[tex]1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}[/tex]
We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:
[tex]3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )[/tex]
Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:
[tex]3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles[/tex]
Calculate the number of moles of CO2 and water considering the same:
[tex]0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)[/tex]
[tex]0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)[/tex]
The total number of moles at the reactor output would be:
[tex]N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)[/tex]
So, the oxygen mole fraction would be:
[tex]y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3[/tex]%
During heat transfer, thermal energy always moves in the same direction: A. COLDHOT B. HOT COLD C. No movement D. HOTHOT
Answer: Option (B) is the correct answer.
Explanation:
Thermal energy is defined as the energy present within the molecules of a substance.
Also, when two objects that have different temperature and they are in contact with each other then heat will always flow from hot object to cold object.
For example, if a metal spoon is placed in a hot cup of coffee then heat will flow from hot coffee to the metal spoon.
Therefore, we can conclude that during heat transfer, thermal energy always moves in the same direction: HOT COLD.
Final answer:
Thermal energy during heat transfer always moves from a higher-temperature object to a lower-temperature object, as per the second law of thermodynamics.
Explanation:
When considering heat transfer, energy typically moves from a higher-temperature object to a lower-temperature object. This movement is adequately described by the laws of thermodynamics, particularly the second law of thermodynamics, which states that heat transfer flows from a hotter object to a cooler one and that heat energy, in any process, is lost to available work in a cyclical process. For example, in the scenario of heating food in a pot on a stove, the energy in the form of heat is transferred from the hot stove element (higher temperature) to the pot and its contents (lower temperature).
Name a possible product of this reaction in the presence of ether and AlCl3: methylbenzene + 1-chlorodecane.a. 1-methyl-2-decylbenzeneb. 1-methyl-3-decylbenzenec. 1-methyl-4-decylbenzened. None of these
Answer:
None of these
Explanation:
Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.
Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :
Step -1 : Generation of stable carbocation.
Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.
Step-2: Attack of the ring to the carbocation
The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.
The product formed is shown in mechanism does not mention in any of the options.
So, None of these is the answer
Answer:
c. 1-methyl-4-decylbenzene
Explanation:
Hello,
On the attached document you will find the major product for the stated chemical reaction even thought the listed product are possible, nonetheless, by cause of the steric hindrance, the most probable and abundant product is the shown one, 1-methyl-4-decylbenzene, as it has more space for the decyl to become part of the ring. Such reaction is a typical Friedel-Crafts alkylation of an aromatic compound whereas aluminium chloride is used as the catalyst to attach the alkyl chloride to the aromatic ring.
Best regards.
Proteoglycans are part of the extracellular matrix; they provide structure, viscosity and lubrication, and adhesiveness. They are composed of proteins conjugated to carbohydrate components called glycosaminoglycans. The glycosaminoglycan component makes up the majority of the mass of a proteoglycan. Which of the following are possible components of glycosaminoglycans? Select all that apply. Determine which of the following statements about glycosaminoglycans are true. There is more than one true statement; select all the true statements. Dermatan sulfate and keratan sulfate are examples of glycosaminoglycans. Because glycosaminoglycans consist of only two residues, they generally have low molecular weights. Dextran is an example of a glycosaminoglycan. Glycosaminoglycans are heteropolysaccharides composed of repeating disaccharide units. The amino groups of the amino sugar derivatives provide a positive charge that offsets the negative charges from the sulfate or carboxylate groups.
Answer:
Here’s what I find.
Explanation:
A. Dermatan sulfate and keratan sulfate are examples of glycosaminoglycans (GAGs). TRUE.
They are both long unbranched polymers in which the repeating unit is a disaccharide consisting of a uronic sugar and an N-acetylated amino sugar.
D. Glycosaminoglycans are heteropolysaccharides composed of repeating disaccharide units. TRUE.
A heteropolysaccharide contains two or more monosaccharides, and GAGs contain a hexoseuronic acid and an N-acetylhexosamine.
B. FALSE. GAGs consist of only two residues, but they are the disaccharide units in a polymer chain that has a molecular weight greater
than 2.5 × 10^6 u.
C. FALSE. Dextran is a polysaccharide of glucose.
E. FALSE. The amino groups in the amino sugars are acetylated, so they are acetamide derivatives. Amides (Kb = 10^-15) are much less basic than amines (Kb = 10^-5), so they are not protonated.
Dermatan sulfate and keratan sulfate are true examples of glycosaminoglycans. Glycosaminoglycans have high molecular weights and are composed of repeating disaccharide units. Dextran is not a GAG, and GAGs' amino sugar derivatives provide a positive charge to offset negative charges.
Explanation:Glycosaminoglycans (GAGs) are essential components of proteoglycans that are found in the extracellular matrix. These molecules provide structural support and are heavily involved in the regulation of cell behavior. One of the main characteristics of GAGs is that they are composed of repeating disaccharide units. Among the possible options given, dermatan sulfate and keratan sulfate are actual examples of GAGs. They contribute to the viscous and adhesive properties of the extracellular matrix.
It is also important to note that GAGs are not limited to having only two residues; their molecular weight can be quite high due to extensive sulfation and the presence of long chains of repeating units. Consequently, the statement that GAGs have low molecular weights due to having only two residues is false.
Moreover, dextran is not an example of a glycosaminoglycan but rather a separate type of polysaccharide. The last statement about GAGs being heteropolysaccharides composed of repeating disaccharide units is true. Additionally, the amino groups of the amino sugar derivatives indeed contribute to the overall charge by providing a positive charge that can offset negative charges from the sulfate or carboxylate groups within GAGs.
A 10.1 g sample of NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. The temperature increases from 23.0 °C to ________°C. Specific heat of liquid water is 4.18 J/g-K and ΔH for the dissolution of sodium hydroxide in water is 44.4 kJ/mol.
Hey there!:
moles of NaOH = 10.1 / 40 = 0.2525
heat = ΔH x moles
= 44.4 x 0.2525
= 11.21 kJ
total mass = 10.1 + 250 = 260.1 g
Q = m Cp dT
11211 = 260.1 x 4.18 x dT
dT = 10.3
T2 = 10.3 + 23 = 33.3 °C
temperature = 33.3 ºC°
Hope this helps!
Answer:
33.3 °C
Explanation:
You have two heat flows in this experiment.
Heat from solution of NaOH + heat to warm water = 0
q1 + q2 = 0
nΔH + mCΔT = 0
Data:
m(NaOH) = 10.1 g
ΔH = -44.4 kJ/mol
m(H2O) = 250.0 g
C = 4.18 J/(K·mol)
Ti = 23.0 °C
Calculation:
n = 10.1 g NaOH × (1 mol NaOH/40.00g NaOH = 0.2525 mol NaOH
q1 = 0.2525 mol × (-44 400 J/mol) = -11 210 J
m(solution) = m(NaOH) + m(water) = 10.1 + 250.0 = 260.1g
q2 = 260.1 × 4.18 × ΔT = 1087ΔT J
-11 210 + 1087ΔT = 0
1087ΔT = 11 210
ΔT = 11 210/108745 = 10.31 °C
ΔT = T2 - T1 = T2 - 23.0 = 10.73
T2 = 23.0 + 10.31 = 33.3 °C
The temperature increases to 33.3 °C.
A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total internal energy. m3 (Round to four decimal places) kJ (Round to one decimal place)
Answer: The volume of the vessel is [tex]0.1542m^3[/tex] and total internal energy is 162.0 kJ.
Explanation:
To calculate the volume of water, we use the equation given by ideal gas, which is:[tex]PV=nRT[/tex]
or,
[tex]PV=\frac{m}{M}RT[/tex]
where,
P = pressure of container = 700 kPa
V = volume of container = ? L
m = Given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000 g)
M = Molar mass of R-134a = 102.03 g/mol
R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]
T = temperature of container = [tex]60^oC=[60+273]K=333K[/tex]
Putting values in above equation, we get:
[tex]700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L[/tex]
Converting this value into [tex]m^3[/tex], we use the conversion factor:
[tex]1m^3=1000L[/tex]
So, [tex]\Rightarrow (\frac{1m^3}{1000L})\times 154.21L[/tex]
[tex]\Rightarrow 0.1542m^3[/tex]
To calculate the internal energy, we use the equation:[tex]U=\frac{3}{2}nRT[/tex]
or,
[tex]U=\frac{3}{2}\frac{m}{M}RT[/tex]
where,
U = total internal energy
m = given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000g)
M = molar mass of R-134a = 102.03 g/mol
R = Gas constant = [tex]8.314J/K.mol[/tex]
T = temperature = [tex]60^oC=[60+273]K=333K[/tex]
Putting values in above equation, we get:
[tex]U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J[/tex]
Converting this into kilo joules, we use the conversion factor:
1 kJ = 1000 J
So, 161994.6 J = 162.0 kJ
Hence, the volume of the vessel is [tex]0.1542m^3[/tex] and total internal energy is 162.0 kJ.
Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying this transformation is a change in the atomic radius of Fe—from RBCC = 0.12584 nm to RFCC = 0.12894 nm—and, in addition, a change in density (and volume). Compute the percentage volume change associated with this reaction. Indicate a decreasing volume by a negative number.
The percentage volume change associated with the allotropic transformation of iron from BCC to FCC is approximately 6.494%.
We can use the following formula to determine the percentage volume change brought about by the allotropic conversion of iron (Fe) from bcc (α phase) to fcc ( γ phase):
Percentage volume change = [(Vγ - Vα) / Vα] * 100
Where
Vγ is the volume of the FCC phase and
Vα is the volume of the BCC phase.
We can use the following formula to determine the volume:
Volume = [tex](4/3) * \pi * r^3[/tex]
Where r is the radius of the atom.
We can calculate the volume as the atomic radius of Fe changes from RBCC = 0.12584 nm to RFCC = 0.12894 nm:
Vα = (4/3) * π *[tex](RBCC^3)[/tex]
Vγ = (4/3) * π * [tex](RFCC^3)[/tex]
Now that the numbers have been substituted, we can determine the percentage change in volume:
Percentage volume change = [(Vγ - Vα) / Vα] * 100
= [(Vγ / Vα) - 1] * 100
= [(Vγ / Vα) - 1] * 100
Vα = (4/3) * π * [tex](0.12584^3)[/tex] = 0.00167709 [tex]nm^3[/tex]
Vγ = (4/3) * π * [tex](0.12894^3)[/tex] = 0.00178565 [tex]nm^3[/tex]
Percentage volume change = [(0.00178565 / 0.00167709) - 1] * 100
= [(1.06494) - 1] * 100
= 0.06494 * 100
= 6.494%
Hence, the percentage volume change associated with the allotropic transformation of iron from BCC to FCC is approximately 6.494%.
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The percentage volume change associated with given reaction is 7.6 %.
We can calculate the percentage volume change using the formula:
[tex]\% \text { Volume Change }=\frac{V_{\text {final }}-V_{\text {initial }}}{V_{\text {initial }}} \times 100[/tex]
The volume of a unit cell can be calculated using the formula V = [tex]a^3[/tex] , where a is the lattice parameter.
For the BCC (α phase), the initial volume ([tex]V_{\text {initial }}[/tex]) is given by [tex]a_{\mathrm{BCC}}^3[/tex], and for the FCC (γ phase), the final volume ([tex]V_{\text {final }}[/tex]) is given by [tex]a_{\mathrm{FCC}}^3[/tex].
Let's calculate it:
Given that a(BCC) is the lattice parameter for BCC phase (0.12584 nm) and a(FCC) is the lattice parameter for FCC phase (0.12894 nm).
[tex]\% \text { Volume Change }=\frac{(0.12894 \mathrm{~nm})^3-(0.12584 \mathrm{~nm})^3}{(0.12584 \mathrm{~nm})^3} \times 100[/tex]
[tex]\% \text { Volume Change }=\frac{\left(0.0001509298202 \mathrm{~nm}^3\right)}{(0.12584 \mathrm{~nm})^3} \times 100[/tex]
%Volume Change ≈ 0.076×100
%Volume Change ≈ 7.6 %
Boyle's law state that pressure of a gas is directly proportionate to its volume under c (T/F)
Answer:Boyle's law, which stated that the pressure and volume of a gas were inversely proportionate when its temperature was kept constant.
Explanation: In other words, at a constant temperature, when the volume of a gas goes up, the pressure goes down, while the volume drops when the pressure rises.
Arrange the following compounds in order of decreasing basicity: MgO, Cs2O, Cl2O7, SnO2, P4O10
1) MgO > Cs2O > P4O10 > SnO2 > Cl2O7
2) Cs2O > SnO2 > MgO > P4O10 > Cl2O7
3) MgO > Cs2O > SnO2 > P4O10 > Cl2O7
4) Cl2O7 > P4O10 > SnO2 > Cs2O > MgO
5) Cs2O > MgO > SnO2 > P4O10 > Cl2O7
Answer:
5) Cs2O > MgO > SnO2 > P4O10 > Cl2O7
Explanation:
Hello,
In this case, the decreasing basicity order depends on the metal contained into the oxide, in such a way, on the attached picture you will find how the basicity increases leftwards and downwards on the periodic table, as the metallic character does.
In this manner, as the cesium in the Cs₂O is the leftest atom, it is the most basic, next MgO, afterwards SnO₂, then P₄O₁₀ and finally Cl₂O₇, therefore, the order is: 5) Cs₂O> MgO > SnO₂> P₄O₁₀> Cl₂O₇.
Best regards.
The order of decreasing basicity of the compounds listed is; Cs2O > MgO > SnO2 > P4O10 > Cl2O7.
Metals are known to form basic oxides. The oxides of metals react with acids to form salt and water. This is one of the properties of metals.
However, the basicity of metallic oxides decreases across the period. As you go from left to right the metal oxides become less basic. Recall that the oxides of metals are acidic. The react with bases to form salts and water.
The order of decreasing basicity of the compounds listed is; Cs2O > MgO > SnO2 > P4O10 > Cl2O7.
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For which of the following aqueous solutions would one expect to have the largest van’t Hoff factor (i)? a. 0.050 m NaCl b. 0.500 m K2SO4 c. 0.050 m K2SO4 d. 0.500 m NaCl e. 0.500 m C6H12O6 (glucose)
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
The largest van't Hoff factor would be expected for the 0.500 m K2SO4 solution. This is because the van't Hoff factor depends on both the concentration and the degree of ionization, which are highest in this case.
Explanation:The van't Hoff factor (i) is a measure of the number of actual particles (ions or molecules) in solution after a compound has been dissolved compared to the number of formula units originally dissolved. It is used to predict various colligative properties of solutions such as freezing point depression and boiling point elevation.
In general, substances that do not ionize in solution, such as glucose (C6H12O6), have a van’t Hoff factor of 1. Sodium chloride (NaCl), when fully dissociated in water, yields two ions (Na+ and Cl-) and so has a van’t Hoff factor of 2 under ideal conditions. Potassium sulfate (K2SO4), when fully dissociated, yields three ions (2K+ and SO4--) and therefore has a van't Hoff factor of 3 under ideal conditions.
Given the choices provided, the 0.500 m K2SO4 solution would be expected to have the largest van't Hoff factor because both the concentration and degree of ionization are highest in this case.
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Formic acid (HCO2H, Ka = 1.8 × 10-4) is the principal component in the venom of stinging ants. What is the molarity of a formic acid solution if 25.00 mL of the formic acid solution requires 39.80 mL of 0.0567 M NaOH to reach the equivalence point?
Answer : The molarity of a formic acid solution is, 0.0903 M
Explanation :
To calculate the molarity of formic acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of formic acid which is [tex]CH_3COOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of sodium hydroxide base which is NaOH.
As we are given:
[tex]n_1=1\\M_1=?\\V_1=25.00ml\\n_2=1\\M_2=0.0567M\\V_2=39.80ml[/tex]
Now put all the given values in above equation, we get:
[tex]1\times M_1\times 25.0ml=1\times 0.0567M\times 39.80ml\\\\M_1=0.0903M[/tex]
Hence, the molarity of a formic acid solution is, 0.0903 M