The freezing point of the solution made by dissolving the sodium chloride in the liquid is 2.48 °C, calculated using the freezing point depression equation and given constants.
Explanation:This problem involves the Physical Chemistry concept of freezing point depression.
First, we need to calculate the molality of the solution. Molality (m) is defined as moles of solute (mol) divided by mass of solvent (kg). Given 7.57g of NaCl, and we know the molar mass of NaCl is approximately 58.44 g/mol, we therefore have 7.57g / 58.44 g/mol = 0.1297 mol. So, the molality (m) is 0.1297 mol / 0.350 kg = 0.370 mol/kg.
Next, we use the freezing point depression equation: ΔTF = i * K * m, where ΔTF is the change in freezing point, i is the number of ions produced per formula unit (for NaCl, i=2 because NaCl dissociates into 2 ions), K is the cryoscopic constant (given as 6.9 ºC-kg/mol) and m is the molality. Substituting the values we get ΔTF = 2 * 6.9 °C-kg/mol * 0.370 mol/kg = 5.118 °C.
Finally, we subtract this value from the original freezing point of the liquid X (7.60 °C) to get the new freezing point: FP_new =FP_old - ΔTF = 7.60°C - 5.118°C = 2.482°C. However, rounding to three significant figures, it should be 2.48 °C.
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The freezing point of the solution made of 7.57 g of NaCl dissolved in 350 g of liquid X is approximately [tex]\( 5.04 \, \text{degree\ C} \).[/tex]
The freezing point of the solution, we'll use the formula for freezing point depression.
[tex]\[ \Delta T_f = K_f \cdot m \][/tex]
where.
[tex]- \( \Delta T_f \)[/tex] is the freezing point depression,
[tex]- \( K_f \)[/tex] is the freezing point depression constant of the solvent liquid X.
[tex]- \( m \)[/tex] is the molality of the solution.
Let's find the molality [tex](\( m \))[/tex] of the solution.
1. Calculate moles of NaCl.
Molar mass of NaCl = [tex]\( 22.99 \, \text{g/mol} + 35.45 \, \text{g/mol} = 58.44 \, \text{g/mol} \)[/tex].
Moles of NaCl [tex]\( = \frac{7.57 \, \text{g}}{58.44 \, \text{g/mol}} = 0.1295 \, \text{mol} \)[/tex].
2. Calculate molality [tex](\( m \)).[/tex]
[tex]\( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)[/tex]
Mass of solvent [tex](liquid X) \( = 350 \, \text{g} = 0.350 \, \text{kg} \)[/tex].
3. Calculate the freezing point depression[tex](\( \Delta T_f \)).[/tex]
Given [tex]\( K_f = 6.90 \, \text{degree\ C-kg/mol} \),[/tex]
[tex]\( \Delta T_f = K_f \cdot m = 6.90 \, \text{°C-kg/mol} \cdot 0.370 \, \text{mol/kg} \)[/tex]
[tex]\( \Delta T_f = 2.557 \, \text{degree\ C} \)[/tex]
4. Calculate the freezing point of the solution.
Freezing point of solution = Normal freezing point of solvent -[tex]\( \Delta T_f \)[/tex]
Normal freezing point of X = 7.60 °C
Freezing point of solution [tex]\( = 7.60 \, \text{degree\ C} - 2.557 \, \text{degree\ C} \)[/tex]
Freezing point of solution [tex]\( = 5.043 \, \text{degree\ C} \)[/tex]
Round your answer to 3 significant digits [tex]\[ \\5.04 \, \text{degree\ C}} \][/tex]
Define ecology, environmental science and environmentalism.
Explanation:
Ecology -
It is the branch of biology , which studies the interactions between the organisms and their biophysical environment , including both abiotic and biotic components .
Environmental science -
It is an interdisciplinary field , of information science , physical , biological science , which helps to study the environment and the environmental problems .
Environmentalism -
It is the rights regarding the concern for the protection of the environment and the to improve the health of the environment .
The main focus on environment and nature-related aspects of green ideology and politics .
.Answer:
Explanation:
Environmentalism -
It is the rights regarding the concern for the protection of the environment and the to improve the health of the environment .
The main focus on environment and nature-related aspects of green ideology and politics.
Ecology -
It is the branch of biology , which studies the interactions between the organisms and their biophysical environment , including both abiotic and biotic components .
Environmental science -
It is an interdisciplinary field , of information science , physical , biological science , which helps to study the environment and the environmental problems .
Question:Dimethylhydrazine, the
fuel used in the Apollo lunar descentmodule, has a molar mass of
60.10 g/mol. It is made up of carbon,hydrogen, and nitrogen atoms.
The combustion of 2.859g of the fuelin excess oxygen yields 4.190g
of carbon dioxideand 3.428g ofwater. What are the simplest and
molecular formulas fordimethylhydrazine?
Answer:
Explanation:
Given parameters:
Molar mass of compound = 60.1g/mol
Mass of fuel used in the combustion process = 2.859g
Mass of carbon dioxide produced = 4.190g
Mass of oxygen produced= 3.428g
Unknown parameters;
Empirical and molecular formula of the compound
Solution
The empirical formula of a compound is its simplest formula. The molecular formula is the actual formula of the compound showing the proportions of the atoms.
We can derive the empirical formula of a compound from its molecular formula and vice versa.
Here, we have to work from empirical formula to molecular formula.
Solving
We know that the compound contains C, H and N atoms. We need to first find the masses of these atoms in the compoundFor Carbon, we can determine the mass from the amount of carbon dioxide produced:
mass of carbon in compound = [tex]\frac{12}{44}[/tex] x 4.19g = 1.14g
For Hydrogen, we can determine the mass from the amount of water produced:
mass of hydrogen in compound = [tex]\frac{2}{18}[/tex] x 3.428g = 0.38g
To determine the mass of N in the compound:
mass of compound = mass of C + mass of H + mass of N
mass of N = mass of compound - (mass of C + mass of H)
mass of N = 2.859g - (1.14g + 0.38g) = 1.34g
2. we now proceed to find the empirical formula using the process below:
Elements C H N
mass of the
elements 1.14 0.38 1.34
Atomic mass
of elements 12 1 14
Number of
moles 1.14/12 0.38/1 1.34/14
0.095 0.38 0.095
Dividing by
the smallest 0.095/0.095 0.38/0.095 0.095/0.095
1 4 1
The empirical formula of the compound is CH₄N
To obtain the molecular formula, we need to find the number of times the empirical formula must have repeated itself in the original form.
Molar mass of CH₄N = 12 + 4 + 14 = 30g/mol
Ratio = [tex]\frac{molar mass of molecular formula}{molar mass of empirical formula}[/tex] = [tex]\frac{60.1}{30}[/tex] = 2
Molecular formula = 2(CH₄N) = C₂H₈N₂
Answer the following using the following information: ∆Hfus=6.02 kJ/mol; ∆Hvap= 40.7 kJ/mol; specific heat of water is 4.184 J/g∙˚C; specific heat of ice is 2.06 J/g∙˚C; specific heat of water vapor is 2.03 J/g∙˚C.
A. How much heat is required to vaporize 25 g of water at 100˚C?
B. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your answer to three significant figures)?
C. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)?
Answer:
A. 56 kJ
B. 75.8 kJ
C. 11 ˚C
Explanation:
A. The heat of vaporization, ∆Hvap = 40.7 kJ/mol, gives the amount of energy per mole of water required to vaporize water to steam. The molar mass of water is 18.02 g/mol.
Q = M·∆Hvap = (25 g)(mol/18.02g)(40.7 kJ/mol) = 56 kJ
B. Five steps are necessary in this process. First, the ice will be warmed to 0 °C, then melted to water. The water will be heated to 100 °C, then vaporized. Finally, the vapor will be heated from 100 °C to 105 °C.
We calculate the heat required to warm the ice from -4.0 °C to 0 °C:
Q₁ = mcΔt = (25 g)(2.06 J∙g⁻¹˚C⁻¹)(0 °C - (-4.0 °C)) = 206 J
Then we calculate the heat required to melt the ice to water:
Q₂ = M∙∆Hfus = (25 g)(mol/18.02 g)(6.02 kJ/mol) = 8.35 kJ
Then, we calculate the heat required to warm the water from 0 °C to 100 °C.
Q₃ = mcΔt = (25 g)(4.184 J∙g⁻¹˚C⁻)(100 °C - 0 °C) = 10460 J
Then we calculate the heat required to vaporize the water:
Q₄ = M∙∆Hvap = (25 g)(mol/18.02 g)(40.7 kJ/mol) = 56.5 kJ
Finally, the vapor is heated from 100 °C to 105 °C.
Q₅ = mcΔt = (25 g)(2.03 J∙g⁻¹˚C⁻)(105 °C - 100 °C) = 254 J
The total heat required is the sum of Q₁ through Q₅
Qtotal = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qtotal = (206 J)(1 kJ/1000J) + 8.35 kJ + (10460 J)(1 kJ/1000J) + 56.5 kJ + (254 J)(1 kJ/1000J)
Qtotal = 75.8 kJ
C. The heat required to melt the ice is provided by the water as it decreases in temperature.
First, we calculate the energy required to melt ice to water
Q = M∙∆Hfus = (8.32 g)(mol/18.02 g)(6.02 kJ/mol) = 2.779 kJ
There are at least two ways to solve this problem. Here, we will calculate the heat lost when all the water is brought to a temperature of 0 °C:
Q = mc∆t = (55 g)(4.184 J∙g⁻¹˚C⁻¹)(25 °C - 0°C) = 5753 J
We see that the water has enough energy to melt all of the ice. The residual heat energy of the water after melting all the ice is:
5753 J - (2.779 kJ)(1000J/kJ) = 2974 J
Now the problem becomes that we have (8.32 g + 55 g) = 63.32 g of water at 0 °C that will be raised to some final temperature by the residual heat of 2974 J:
Q = mcΔt ⇒ Δt = Q/(mc)
Δt = (2974 J) / (63.32 g)(4.184 J∙g⁻¹˚C⁻¹) = 11 ˚C
T(final) - T(inital) = 11 ˚C
T(final) = 11 ˚C + T(inital) = 11 ˚C + 0 ˚C = 11 ˚C
Thus, the final temperature will be 11 ˚C.
The heat required to vaporize 25g of water at 100˚C is 56.529 kJ. To convert 25g of ice at -4.0˚C to water vapor at 105˚C, approximately 83.8 kJ of heat is required. The final temperature of the water sample after the melting of an 8.32g ice cube is about 24.856˚C.
Explanation:A. To find the heat required to vaporize 25 g of water at 100˚C, we need to use the given heat of vaporization (∆Hvap) which is 40.7 kJ/mol. However, we need the molar mass of water which is approximately 18.015 g/mol. The calculation is as follows: (25 g) / (18.015 g/mol) * (40.7 kJ/mol) = 56.529 kJ. Therefore, the heat needed is 56.529 kJ.
B. To convert 25 g of ice at -4.0˚C to water vapor at 105˚C we first calculate the heat needed to melt the ice to water at 0˚C, then to heat the water from 0˚C to 100 ˚C, and finally to vaporize the water. When all of these are added up, the total heat can be calculated as approximately 83.8 kJ.
C. To find the final temperature of the entire water sample after the ice is melted, we set the heat gained by the ice equal to the heat lost by the water. Solving gives a final temperature of about 24.856˚C.
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Buffer preparation. You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acid plus acetate concentration of 250 mM and a pH of 5.0. What concentrations of acetic acid and sodium acetate should you use? Assuming you wish to make 2 liters of this buffer, how many moles of acetic acid and sodium acetate will you need? How many grams of each will you need (molecular weights: acetic acid 60. 05 g mol 1, sodium acetate, 82. 03 g mol 1)?
Answer:
0.182 moles of acetic acid are needed, this means 10.93 g.
0.318 moles of sodium acetate are needed, this means 26.08 g.
Explanation:
The Henderson–Hasselbalch (H-H) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:
pH = pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.
From the problem we know that
[A⁻] + [HA] = 250 mM = 0.250 M eq. 1
We use the H-H equation, using the data we know, to describe [A⁻] in terms of [HA]:
5.0 = 4.76 + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
[tex]0.24=log\frac{[A^{-} ]}{[HA]}\\\\10^{0.24}=\frac{[A^{-} ]}{[HA]}\\ 1.74 [HA] = [A^{-}][/tex] eq.2
Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:
1.74 [HA] + [HA] = 0.250 M
[HA] = 0.091 M
Then we calculate [A⁻]:
[A⁻] + 0.091 M = 0.250 M
[A⁻] = 0.159 M
Using the volume, we can calculate the moles of each substance:
moles of acetic acid = 0.091 M * 2 L = 0.182 molesmoles of sodium acetate = 0.159 M * 2 L = 0.318 molesUsing the molecular weight, we can calculate the grams of each substance:
grams of acetic acid = 0.182 mol * 60.05 g/mol = 10.93 ggrams of sodium acetate = 0.318 mol * 82.03 g/mol = 26.08 gAn unstable isotope that naturally decays CAN emit which of the following? Choose all correct responses, there are at least two. {Hint: probably are more than two}
Question options:
A) Alpha particle
B) Beta particle
c) Radiation
D) Positron
E) Gamma rays
F) Visible light
Answer: Isotopes are the species which have same number of protons but differs in the number of neutrons. Or in other words isotopes have same atomic number but different mass number. Isotopes are generally unstable and decays to an stable isotope. When the isotope is unstable it releases energy in the form of radiations. It emits alpha particle, gamma rays, and beta particles.
Therefore, the correct option is (A),(B) and (C)
Find the work done using data from the PV plot for an isothermal system.
P1= 100 kPa, P2= 300 kPa
V1= 25 m^3 , V2= 10 m^3
B.) Is the work done by the system or on the system?
Answer:
W = - 500 KJ
∴ the work is done on the system
Explanation:
isothermal system:
∴ ΔU = 0; ⇒ Q = W
∴ W = P1V1 -P2V2
⇒ W = ((100KPa)*(25m³)) - ((300KPa)*(10m³))
⇒ W = 2500KPa.m³ - 3000KPa,m³
⇒ W = - 500 KPa.m³ = - 500 KJ
∴ W (-) the work is done on the system
In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.501 L flask at 1043 K. At equilibrium, the flask contains 0.169 mol of CO gas, 0.257 mol of H2 gas, and 0.255 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1043 K)? Enter to 4 decimal places. HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the K expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into K expression, solve for the unknown. (MTS 5/16/2018)
Answer:
[ H2O ]eq = 0.298 mol/L
Explanation:
CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)∴ V = 0.501 L
∴ T = 1043 K
at equilibrium:
∴ n CH4 = 0.255 mol
⇒ [ CH4 ]eq = 0.255 mol / 0.501 L = 0.509 mol/L
∴ n CO = 0.169 mol
⇒ [ CO ]eq = 0.169 / 0.501 = 0.337 mol/L
∴ n H2 = 0.257
⇒ [ H2 ] eq = 0.257 / 0.501 = 0.513 mol/L
∴ Kc = [ H2 ]³ * [ CO ] / [ CH4 ] * [ H2O ] = 0.30
⇒ [ H2O ] = [ H2 ]³ * [ CO ] / [ CH4 ] * 0.30
replacing the value of the concentration in Kc:
⇒ [ H2O ] = ( 0.513 )³ * ( 0.337 ) / ( 0.509 ) * 0.30
⇒ [ H2O ] = 0.298 mol/L
The equilibrium concentration of water vapor in the conversion of methane to other fuels at 1043 K is determined to be 0.6196 M by using the equilibrium constant expression and the known concentrations of the other substances.
Explanation:The student's question involves determining the water concentration at equilibrium for the conversion of methane to other fuels involving a reaction of methane (CH4) and water vapor (H2O). First, we need to write a balanced chemical equation for the reaction:
CH4(g) + 2H2O(g) ⇌ CO2(g) + 4H2(g)
To find the water concentration, we use the equilibrium constant expression (Kc):
Kc = [tex][CO]^1[H2]^4 / [CH4]^1[H2O]^2[/tex]
Given that Kc = 0.30 at 1043 K, and the equilibrium amounts of the substances are CH4 = 0.255 mol, CO = 0.169 mol, and H2 = 0.257 mol in a 0.501 L flask, we can calculate their concentrations and solve for [H2O]:
[CH4] = 0.255 mol / 0.501 L = 0.509 M
[CO] = 0.169 mol / 0.501 L = 0.337 M
[H2] = 0.257 mol / 0.501 L = 0.513 M
Plugging these into the Kc expression and solving for [H2O], we get:
0.30 = (0.337)[([tex]0.513)^4] / [(0.509)([H2O]^2)][/tex]
After calculations, the water concentration at equilibrium, [H2O], is found to be approximately 0.6196 M.
Radon-219 decays to radon-218 by releasing... a. a positron b.a muon O c. a neutron O d. an electron O e. a proton
Final answer:
Radon-219 decays into radon-218 by emitting an alpha particle, a process that reduces its atomic mass and number, leading to a new isotope.
Explanation:
Radon-219 decays to radon-218 by releasing an alpha particle (Helium nucleus). When radium-226 undergoes alpha decay, it forms radon-222 and an alpha particle. The process involves the nucleus of an atom releasing two protons and two neutrons, which together form an alpha particle. In the case of radon decay, the alpha particle is emitted, decreasing the atomic mass by 4 units and atomic number by 2, resulting in a new isotope. Therefore, the correct answer is b. a muon. The emission of an alpha particle is a common mode of decay for heavy, unstable nuclei such as radon-219.
Perform the following unit conversions:
a)
60 feet of water to Pa
b)
220 psi to lbf/ft^2
c)
120 torr to Pa
d)
1.0 atm to inches of glycerin
e)
1050 mm Hg to lbf/ft^2
Answer :
(a) The 60 feet of water is equal to 179344.2 Pa.
(b) The 220 psi is equal to [tex]31680lbf/ft^2[/tex]
(c) The 120 torr is equal to 15998.6 Pa.
(d) The 1.0 atm to inches of glycerin is equal to 323.07 inches.
(e) The 1050 mm Hg is equal to [tex]376.95lbf/ft^2[/tex]
Explanation :
(a) The conversion used from feet to pascal is:
[tex]1\text{ feet}=2989.07Pa[/tex]
As we are given that 60 feet of water. Now we have to convert into Pa.
As, [tex]1\text{ feet}=2989.07Pa[/tex]
So, [tex]60\text{ feet}=\frac{60\text{ feet}}{1\text{ feet}}\times 2989.07Pa=179344.2Pa[/tex]
The 60 feet of water is equal to 179344.2 Pa.
(b) The conversion used from [tex]psi[/tex] to [tex]lbf/ft^2[/tex] is:
[tex]1\text{ psi}=144lbf/ft^2[/tex]
As we are given that 220 psi. Now we have to convert into [tex]lbf/ft^2[/tex].
As, [tex]1\text{ psi}=144lbf/ft^2[/tex]
So, [tex]220\text{ psi}=\frac{220\text{ psi}}{1\text{ psi}}\times 144lbf/ft^2=31680lbf/ft^2[/tex]
The 220 psi is equal to [tex]31680lbf/ft^2[/tex]
(c) The conversion used from torr to pascal is:
[tex]1\text{ torr}=133.322Pa[/tex]
As we are given that 120 torr. Now we have to convert into Pa.
As, [tex]1\text{ torr}=133.322Pa[/tex]
So, [tex]120\text{ torr}=\frac{120\text{ torr}}{1\text{ torr}}\times 133.322Pa=15998.6Pa[/tex]
The 120 torr is equal to 15998.6 Pa.
(d) 1.0 atm to inches of glycerin
Formula used : [tex]P=\rho gh[/tex]
where,
P = pressure of glycerin = 1.0 atm = 101325 Pa
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
[tex]\rho[/tex] = density of glycerin = [tex]1260kg/m^3[/tex]
Now put all the given values in above formula, we get:
[tex]101325Pa=(1260kg/m^3)\times (9.8m/s^2)\times h[/tex]
[tex]h=8.206\frac{Pa.m^2s^2}{kg}=8.206m=323.07\text{ inches}[/tex]
Conversion used :
[tex]1Pa=\frac{kg}{ms^2}\\\\1m=39.37inches[/tex]
The 1.0 atm to inches of glycerin is equal to 323.07 inches.
(e) The conversion used from [tex]psi[/tex] to [tex]lbf/ft^2[/tex] is:
[tex]1\text{ mmHg}=0.359lbf/ft^2[/tex]
As we are given that 1050 mmHg. Now we have to convert into [tex]lbf/ft^2[/tex].
As, [tex]1\text{ mmHg}=0.359lbf/ft^2[/tex]
So, [tex]1050\text{ mmHg}=\frac{1050\text{ mmHg}}{1\text{ mmHg}}\times 0.359lbf/ft^2=376.95lbf/ft^2[/tex]
The 1050 mm Hg is equal to [tex]376.95lbf/ft^2[/tex]
To convert 60 feet of water to Pa, multiply by the conversion factor 2.98907 kPa. To convert 220 psi to lbf/ft^2, multiply by the conversion factor 144 lbf/ft^2. To convert 120 torr to Pa, multiply by the conversion factor 133.322 Pa. To convert 1.0 atm to inches of glycerin, we need additional information. To convert 1050 mm Hg to lbf/ft^2, multiply by the conversion factor 0.01934 lbf/ft^2.
Explanation:a) To convert 60 feet of water to Pa, we can use the conversion factor 1 ftH2O = 2.98907 kPa. First, convert feet to meters by multiplying by 0.3048, then multiply by the conversion factor to get the answer in Pa.b) To convert 220 psi to lbf/ft^2, we can use the conversion factor 1 psi = 144 lbf/ft^2. Multiply psi by the conversion factor to get the answer in lbf/ft^2.c) To convert 120 torr to Pa, we can use the conversion factor 1 torr = 133.322 Pa. Multiply torr by the conversion factor to get the answer in Pa.d) To convert 1.0 atm to inches of glycerin, we need to know the relationship between atm and inches of glycerin. Without this information, we cannot perform the conversion.e) To convert 1050 mm Hg to lbf/ft^2, we can use the conversion factor 1 mm Hg = 0.01934 lbf/ft^2. Multiply mm Hg by the conversion factor to get the answer in lbf/ft^2.33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b. What is the mole fraction of fructose in this solution? c. What is the solution's average molar mass? d. What is the specific molar volume of the solution?
Explanation:
Mass of fructose = 33.56 g
Mass of water = 18.88 g
Total mass of the solution = Mass of fructose + Mass of water = M
M = 33.56 g + 18.88 g =52.44 g
Volume of the solution = V = 40.00 mL
Density =[tex]\frac{Mass}{Volume}[/tex]
a) Density of the solution:
[tex]\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL[/tex]
b) Molar mass of fructose = 180.16 g/mol
Moles of fructose = [tex]n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol[/tex]
Molar mass of water = 18.02 g/mol
Moles of water= [tex]n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol[/tex]
Mole fraction of fructose in this solution:[tex]\chi_1[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}[/tex]
[tex]\chi_1=0.1510[/tex]
Mole fraction of water = [tex]\chi_2=1-\chi_1=0.8490[/tex]
c) Average molar mass of of the solution:
=[tex]\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol[/tex]
[tex]=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol[/tex]
d) Mass of 1 mole of solution = 42.50 g/mol
Density of the solution = 1.311 g/mL
d) Specific molar volume of the solution:
[tex]\frac{\text{Average molar mass}}{\text{Density of the mass}}[/tex]
[tex]=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol[/tex]
Mass transfer rate in convection is ..... mass transfer in conduction
a) more than
b) less than
c) equal to
d) no relation to
Answer:
(a) more than
Explanation:
Conduction:
Transfer of heat due to direct contact between two mediums at different temperatures, without having any of the bodies traveling. Therefore, conduction heat transfer occurs by the transfer of momentum (molecular) from always the same group of molecules in one medium to another group of molecules in another medium.
Example: Heat transfer INSIDE a solid.
Convention:
Transfer of heat or mass due to at least one traveling medium, where the transfer of momentum is not bounded anymore to the same groups of molecules. Molecules moving to transfer their momentum and keep flowing to the next group, also allowing other molecules behind to do the same. Example: heat transfer by the wind.
Hence, the Mass transfer rate in convection is more than mass transfer in conduction
Final answer:
The mass transfer rate in convection is generally more than (option a) that in conduction due to the macroscopic movement of mass that facilitates faster heat transfer.
Explanation:
The question asks to compare the mass transfer rate in convection to the mass transfer rate in conduction. The correct comparison is that the mass transfer rate in convection is generally more than the mass transfer rate in conduction. This is because conduction is the process by which heat is directly transmitted through a substance when there is a difference of temperature, without movement of the material. In contrast, convection involves the movement of mass (typically a fluid such as air or water) which contributes to a faster transfer of heat. An empirical equation for the rate of heat transfer by forced convection can be represented by Q/t = hA(T₂ – T₁), showing that this rate is dependent on the temperature difference (T₂ – T₁), contact surface area (A), and the convective heat transfer coefficient (h).
Select 2-possible ointment bases. Select one or more:
a. Hypromellose and CaCl2
b. xanthan gum
c. petroleum jelly
d. Polyethylene glycol 4000/600 mixture
Answer:
The correct option is: c. petroleum jelly, d. Polyethylene glycol 4000/600 mixture
Explanation:
Topical medications are used for the treatment of ailments and include ointments, gels, lotions creams etc. that can applied directly on the surface of the body i.e. skin.
An ointment base medication gets rapidly absorbed into the skin. Some of the examples of ointment bases include water-soluble bases: polyethylene glycol, hydrocarbon bases: petroleum jelly, paraffin wax.
Answer:
yo mama
Explanation:
in the fali, the relative humidity is high in early morning and as the temperature increases, the relative humidity decreases because (3 pts) A. saturation pressure of water increases B. water condenses C. water evaporates D. saturation pressure of water decreases
Answer:
The correct option is: A. saturation pressure of water increases
Explanation:
Relative humidity, at a given temperature, is defined as ratio of partial pressure of the water vapor preset in the air mixture and equilibrium vapor pressure above pure water.
Relative humidity of a system is dependent upon the pressure and temperature. When the amount of water vapor in a system is constant, the higher relative humidity is higher at low temperature and lower at high temperature. This is because at high temperatures, the air capacity and the saturation pressure increases.
Therefore, in the morning when the temperature is low in fali, the relative humidity is high. As the temperature increases, the relative humidity in fali decreases.
out of 2.00 50 mL of water is added to 30 mL of a 0.50 M salt solution. What is the new concentration? (To write your answer using scientific notation use 1.0E-1 instead of 1.0 x 10-1)
Answer:
The concentration of the new solution will be 0.31 M
Explanation:
The number of moles per liter of the solution is 0.50 mol. Then, in 50 ml there will be:
50 ml · (0.50 mol / 1000 ml) = 0.025 mol.
If we add 30 ml and assuming that the solution is an ideal solution, the final volume will be 80 ml.
Then, 0.025 mol will be present in 80 ml solution. In 1 l there will be:
1000 ml · (0.025 mol / 80 ml) = 0.31 mol.
The concentration of the new solution will be 0.31 M.
You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A quick analysis reveals that the pH of the water is 8.40 when it should be 7.20. The pool is 7.00 m wide, 18.0 m long, and has an average depth of 1.50m What is the minimum (in the absence of any buffering capacity) volume (mL) of 12.0 wt% H2SO4 (SG 1.080) that should be added to return the pool to the desired pH? mL
Answer:
3,78 mL of 12,0wt% H₂SO₄
Explanation:
The equilibrium in water is:
H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)
The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = [tex]10^{-7,2}[/tex]
Thus, you need to add:
[H⁺] = [tex]10^{-7,2} - 10^{-8}[/tex] = 5,31x10⁻⁸ M
The total volume of the pool is:
7,00 m × 18,0 m ×1,50 m = 189 m³ ≡ 189000 L
Thus, moles of H⁺ you need to add are:
5,31x10⁻⁸ M × 189000 L = 1,00x10⁻² moles of H⁺
These moles comes from
H₂SO₄ → 2H⁺ +SO₄²⁻
Thus:
1,00x10⁻² moles of H⁺ × [tex]\frac{1 H_{2}SO_{4} moles}{2H^+ mole}[/tex] = 5,00x10⁻³ moles of H₂SO₄
These moles comes from:
5,00x10⁻³ moles of H₂SO₄ × [tex]\frac{98,1g}{1mol}[/tex] × [tex]\frac{100 gsolution}{12 g H_{2}SO_{4} }[/tex] × [tex]\frac{1mL}{1,080 g}[/tex] = 3,78 mL of 12,0wt% H₂SO₄
I hope it helps!
For IR radiation with û = 1,130 cm 1, v=__THz
Answer: The frequency of the radiation is 33.9 THz
Explanation:
We are given:
Wave number of the radiation, [tex]\bar{\nu}=1130cm^{-1}[/tex]
Wave number is defined as the number of wavelengths per unit length.
Mathematically,
[tex]\bar{\nu}=\frac{1}{\lambda}[/tex]
where,
[tex]\bar{\nu}[/tex] = wave number = [tex]1130cm^{-1}[/tex]
[tex]\lambda[/tex] = wavelength of the radiation = ?
Putting values in above equation, we get:
[tex]1130cm^{-1}=\farc{1}{\lambda}\\\\\lambda=\frac{1}{1130cm^{-1}}=8.850\times 10^{-4}cm[/tex]
Converting this into meters, we use the conversion factor:
1 m = 100 cm
So, [tex]8.850\times 10^{-4}cm=8.850\times 10^{-4}\times 10^{-2}=8.850\times 10^{-6}m[/tex]
The relation between frequency and wavelength is given as:[tex]\nu=\frac{c}{\lambda}[/tex]
where,
c = the speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\nu[/tex] = frequency of the radiation = ?
Putting values in above equation, we get:
[tex]\nu=\frac{3\times 10^8m/s}{8.850\times 10^{-4}m}[/tex]
[tex]\nu=0.339\times 10^{14}Hz[/tex]
Converting this into tera Hertz, we use the conversion factor:
[tex]1THz=1\times 10^{12}Hz[/tex]
So, [tex]0.339\times 10^{14}Hz\times \frac{1THz}{1\times 10^{12}Hz}=33.9THz[/tex]
Hence, the frequency of the radiation is 33.9 THz
The quantity of energy released or absorbed as heat during a chemical reaction is called the 28. a. temperature. c. entropy d. free energy enthalpy of reaction.
Answer: enthalpy of reaction.
Explanation:
Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like [tex]^0C[/tex] and [tex]K[/tex]
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
Free energy is the amount of energy that can be converted into useful work.
Enthalpy of the reaction is the difference between the energy of products and the energy of reactants. it is either the heat released or absorbed during the reaction. It is either positive or negative.
Heather just drank 40.0 grams of water (H20). How many moles of water did she just drink? O a. 2.22 moles b. 45 moles O c. 40.0 moles O d. 0.45 moles O e.720 moles
Answer:
This question begins with something, you should know: molar mass from water is aproximately 18 g/m, so if 18 grams of water are contained in 1 mole, the 40 grams occuped 2.22 moles. As you see, opcion a is the best!
Explanation:
The airspeed velocity of an unladen swallow is 10.00 meters per second. What is it in miles per hour? You must use appropriate significant figures to receive full credit (3 pts) 1 in = 2.54 cm, 100 cm = 1m, 12 in =1ft, 5280 ft = 1 mile, 60 s = 1 min, 60 min = 1 hour
Answer : The speed in miles per hour is 22 mile/hr.
Explanation :
The conversion used from meters to miles is:
[tex]1m=100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}[/tex]
The conversion used from second to hour is:
[tex]1s=\frac{1}{60}min\times \frac{1hr}{60min}[/tex]
The conversion used from meter per second to mile per hour is:
[tex]1\frac{m}{s}=\frac{100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}}{\frac{1}{60}min\times \frac{1hr}{60min}}[/tex]
[tex]1m/s=2.2mile/hr[/tex]
As we are given the speed 10.00 meter per second. Now we have to determine the speed in miles per hour.
As, [tex]1m/s=2.2mile/hr[/tex]
So, [tex]10.00m/s=\frac{10.00m/s}{1m/s}\times 2.2mile/hr=22mile/hr[/tex]
Therefore, the speed in miles per hour is 22 mile/hr.
Which of the following statements about material bonding is correct? C O a. Ionic bonds are formed by the sharing of valence electrons among two or more atoms b. Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other C c. Metallic elements with metallic bonds have atoms that donate valence electrons to other atoms, thus filling the outer energy shells of these other atoms d. Covalent bonds are formed by atoms that donate their valence electrons to form a "sea" of electrons surrounding the atoms
Answer: Option (b) is the correct answer.
Explanation:
In material bonding, there occurs Vander waal foces between the molecules in which their is either an induced or permanent dipole moment that attract molecules towards each other.
And, due to these forces the molecules are held together.
On the other hand, in a ionic bond there will always be transfer of electrons from one atom to another. This is because on atom which loses its valence electrons acquires a positive charge and another atom which gains the electrons acquires a negative charge.
Hence, these opposite charges strongly gets attracted towards each other forming a strong bond.
Whereas in a covalent bond, there will be sharing of electrons between the combining atoms.
In a metallic bond, there occurs a sea of electrons which is uniformly distributed throughout the solid substance or material.
Thus, we can conclude that the statement, Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other, about material bonding is correct.
Final answer:
The correct statement is that Van der Waals bonds are formed by Van der Waals forces due to induced or permanent dipoles, unlike ionic or covalent bonds.
Explanation:
The correct statement about material bonding is option b: Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other. Unlike ionic and covalent bonds, Van der Waals bonds involve weaker, more temporary attractions. In contrast, ionic bonds occur when electrons are transferred from one atom to another, usually between metals and nonmetals, creating a positive and a negative ion that attract each other.
Finally, covalent bonds involve the sharing of electron pairs between atoms, and in metallic bonds, valence electrons are not donated to other atoms; instead, they form a 'sea of electrons' that surrounds the metal ions.
15 g of anhydrous calcium chloride is dissolved in 185 mL of water. What is molarity of the prepared solution? 81.1 M O0.73 M 0.081 M O0.073 M 1.07 M
Answer:
When you prepared the solution, you will find that the molarity is 0.73M
Explanation:
First of all you should get by the periodic table, molar mass in the anhydrouds calcium.
CaCl2 · 0H20 = 110.98 g/m
Now we have to find out how many mols are 15 g.
So 15g / 110.98 g/m = 0.135 moles
This moles are in 185 ml of water. Molarity as you should know are moles of solute in 1 L of solution (either 1000 ml)
185 ml ______ 0.135 moles
1000 ml _____ x x = (1000*0.135) /185 = 0.730M
According to the following thermochemical equation, if 951.1 g of NO 2 is produced, how much heat is released at constant pressure? 2NO(g) + O2(g) 2NO 2(g); DH -114.4 kJ
Answer:
The amount of released is 1,182 kJ.
Explanation:
When heat is released at constant pressure, this change in energy is known as enthalpy (ΔH°) of the reaction. Enthalpy is an extensive property, so it depends on the amount of reacting material. Let's take a look at the provided equation:
2 NO(g) + O₂ ⇄ 2 NO₂(g) ΔH° = -114.4 kJ
Since this equation is balanced with 2 moles of NO₂(g), we can say that 114.4 kJ are released every 2 moles of NO₂(g) produced. By convention, when enthalpies are negative, it means that energy is released and the reaction is exothermic. Conversely, positive enthalpies mean energy is absorbed and the reaction is endothermic.
We can calculate the amount of energy released taking into account the previous relationship (-114.4 kJ/2 moles of NO₂(g)), the mass of NO₂(g) produced (951.1g) and its molar mass (46.00g/mol). The calculations would be:
[tex]951.1g.\frac{1molNO_{2} }{46.00g} .\frac{-114.4kJ}{2molesNO_{2} } =-1,182kJ[/tex]
The amount of heat released at constant pressure when 951.1 g of [tex]NO_2[/tex] is produced is approximately -544.5 kJ.
The molar mass of [tex]NO_2[/tex] can be calculated from its molecular weight:
Nitrogen (N) has an atomic mass of approximately 14.01 g/mol,
Oxygen (O) has an atomic mass of approximately 16.00 g/mol.
Thus, the molar mass of [tex]NO_2[/tex] is:
[tex]\[ \text{Molar mass of NO}_2 = 14.01 \text{ g/mol} + 2 \times 16.00 \text{ g/mol} = 46.01 \text{ g/mol} \][/tex]
Now, we can calculate the number of moles of [tex]NO_2[/tex] produced:
[tex]\[ \text{moles of NO}_2 = \frac{\text{mass of NO}_2}{\text{molar mass of NO}_2} = \frac{951.1 \text{ g}}{46.01 \text{ g/mol}} \approx 20.67 \text{ mol} \][/tex]
The thermochemical equation given is:
[tex]\[ 2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g); \quad \Delta H = -114.4 \text{ kJ} \][/tex]
This equation tells us that 2 moles of NO2 are produced for every -114.4 kJ of heat released at constant pressure. Therefore, the heat released for the production of 20.67 moles of NO2 is:
[tex]\[ \text{Heat released} = \text{moles of NO}_2 \times \frac{-114.4 \text{ kJ}}{2 \text{ mol}} \][/tex]
[tex]\[ \text{Heat released} = 20.67 \text{ mol} \times \frac{-114.4 \text{ kJ}}{2 \text{ mol}} \approx -544.5 \text{ kJ} \][/tex]
A sample of lake water was analyzed to determine the amount of metals found in the lake. The standard deviation of the sampling method was found to be ±6.0%±6.0% . The standard deviation of the analytical method used to determine the amount of metals in the sample was determined to be ±2.4%±2.4% . What is the overall standard deviation?
Answer:
The overall standard deviation, s = 6.46 %
Given:
Sampling variance, [tex]s_{b} = \pm 6.0% = 0.06[/tex]
Analytical variance, [tex]s_{a} = \pm 2.4% = 0.024[/tex]
Solution:
Variance additive is given by:
[tex]s^{2} = s_{a}^{2} + s_{b}^{2}[/tex] (1)
where
s = overall variance
Also, we know that:
Standard Deviation, [tex]\sigma = \sqrt{variance}[/tex]
Therefore the standard deviation of the sampling, analytical and overall sampling is given by taking the square root of eqn (1) on both the sides:
[tex]s = \sqrt{s_{a}^{2} + s_{b}^{2}}[/tex]
[tex]s = \sqrt{0.024^{2} + 0.06^{2}} = 0.0646[/tex]
s = 6.46 %
The overall standard deviation considering both the sampling method (±6.0%) and analytical method (±2.4%) is calculated by squaring each to get variances, summing these, and taking the square root of the sum, resulting in an overall standard deviation of ±6.464%.
To calculate the overall standard deviation when given two separate standard deviations from different sources (sampling and analytical method), you should first square each standard deviation to find the variances. Then, add the variances together, and finally, take the square root of the total to find the overall standard deviation.
The standard deviation of the sampling method is ±6.0%, and this value squared gives 36.0%. The analytical method's standard deviation is ±2.4%, which squared gives 5.76%. Adding these two variances together, we get 36.0% + 5.76% = 41.76%. The overall standard deviation is then the square root of 41.76%, which is approximately ±6.464%.
Therefore, the combined or overall standard deviation considering both the sampling and analytical methods is ±6.464%.
10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of the carbon dioxide if the temperature and pressure change to 75 oC and 15 kPa.
Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and [tex]7.2m^3[/tex] respectively.
Explanation :
First we have to calculate the new or final volume of carbon dioxide gas.
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 10 kPa
[tex]P_2[/tex] = final pressure of gas = 15 kPa
[tex]V_1[/tex] = initial volume of gas = [tex]10m^3[/tex]
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]50^oC=273+50=323K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]75^oC=273+75=348K[/tex]
Now put all the given values in the above equation, we get:
[tex]\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}[/tex]
[tex]V_2=7.2m^3[/tex]
The new volume of carbon dioxide gas is [tex]7.2m^3[/tex]
Now we have to calculate the new density of carbon dioxide gas.
[tex]PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}[/tex]
Formula for new density will be:
[tex]\rho_2=\frac{P_2M}{RT_2}[/tex]
where,
[tex]P_2[/tex] = new pressure of gas = 15 kPa
[tex]T_2[/tex] = new temperature of gas = [tex]75^oC=273+75=348K[/tex]
M = molar mass of carbon dioxide gas = 44 g/mole
R = gas constant = 8.314 L.kPa/mol.K
[tex]\rho[/tex] = new density
Now put all the given values in the above equation, we get:
[tex]\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}[/tex]
[tex]\rho_2=0.2281g/L[/tex]
The new density of carbon dioxide gas is 0.2281 g/L
Final Temperature in Heating Applesauce. A mixture of 454 kg of applesauce at 10°C is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet tem perature of the applesauce. (Hint: In Appendix A.4, a heat capacity for applesauce is given at 32.8C. Assume that this is constant and use this as the average com)
Answer:
[tex]T_f = 76.46°C[/tex]
Explanation:
Given data:
Mass of mixture = 454 kg
Initial temperature is 10°C
Heat added is Q = 121300 kJ
Heat capacity (Applesuace) at 32.8°C is 4.02kJ/kg K
From heat equation we have
[tex]Q = mCp(T_f -T_i)[/tex]
[tex]\frac{Q}{mCp} = (T_f -T_i)[/tex]
[tex]T_f = T_i + \frac{Q}{mCp}[/tex]
Putting all value to get required final temperature value
[tex]T_f = \frac{121300}{454\times 4.02} + 10[/tex]
[tex]T_f = 76.46°C[/tex]
How many cubic centimeters are there in 200 milliliters?
Answer:
200 mL = 200 cm³
Explanation:
The relationship between cm³ and mL is 1:1.
1 cm³ = 1 mL
Thus, 200 mL is converted to cm³ as follows:
(200 mL)(1 cm³/1 mL) = 200 cm³
The combustion of fuel in your car engine requires oxygen gas, which is supplied as air (21% oxygen molecules) into the engine. Consider a car that is using 100% ethanol, C2H5OH, as fuel. If your engine intakes 4.73 L of air per minute at 1.00 atm and 25ºC, what is the maximum volume of ethanol (0.789 g/mL) that can be burned per minute? Hint: You can ignore the "per minute" information because both the ethanol and air are being quantified per minute. Enter your answer to three significant figures in units of mL.
Answer:
the maximum volume that can be burned per minute is: 0,895 mL of ethanol.
Explanation:
The combustion of ethanol is:
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
With gas law:
PV/RT = n
Where P is pressure (1,00 atm)
V is volume (4,73 L of air per minute)
R is gas constant (0,082 atmL/molK)
T is temperatue(25°C≡298,15K)
And n are moles, replacing:
n = 0,193 moles of air per minute.
These moles of air contain:
0,193 moles air ×[tex]\frac{21 molesO_2}{100 molesAIR}[/tex] = 0,0406 moles O₂
Thus, the maximum volume that can be burned per minute is:
0,046 moles O₂[tex]\frac{1molC_{2}H_{5}OH}{3molesO_2} \frac{46,07 g}{1mol} \frac{1mL}{0,789g}[/tex] = 0,895 mL of ethanol per minute
I hope it helps!
If the concentration of an analyte in a solution prepared by digesting a 2.05 g solid sample and making up to 100 mL, is found to be 21.1 mg L-1. What was the mass percentage of the analyte in the original sample?
Answer:
Mass percentage of analyte = 0.10%
Explanation:
The mass of analyte in the solution is calculated as follows:
n = CV = (21.1 mgL⁻¹)(100 mL)(1L/1000mL) = 2.11 mg
The mass percentage of the analyte is calculated as follows:
(g analyte)/(g sample) x 100%
(2.11 mg)(1g/1000g) / (2.05g) x 100% = 0.10%
Select the cations responsible for water hardness.1) Ca2+ 2) CO3−2 3) CaCO3 4) Mg2+
Answer:
The correct options are: 1. Ca²⁺ and 4. Mg²⁺
Explanation:
Hard water is the water with high mineral content. Temporary hardness and permanent hardness are the two types of hardness of water.
Temporary hardness is due to the presence of dissolved bicarbonate minerals. These minerals present in the water, dissociate to give multivalent calcium cations (Ca²⁺) and magnesium cations (Mg²⁺).
Therefore, the presence of metal cations such as calcium cations (Ca²⁺) and magnesium cations (Mg²⁺) makes the water hard.
The compressibility factor of any gas at a specific value of reduced temperature, T Tc, and reduced pressure, P Pc does not have approximately the same value for all species. Select one: True False
Answer:
False
Explanation:
According to principle of corresponding state, at reduced states or corresponding state, behavior of all gases are similar.
This principle is proposed by van der Waals.
In other words, all gases at same reduced temperature, reduced pressure and reduced volume deviate from ideal gas behavior to the same degree or have same compressibility factor.
Reduced quantities are defined as:
Reduced pressure [tex]P_R = \frac{P}{P_c}[/tex]
Reduced temperature [tex]T_R = \frac{T}{T_c}[/tex]
Reduced volume [tex]V_R = \frac{V}{V_c}[/tex]
Where,
P_c = Critical pressure
V_c = Critical volume
T_c = Critical temperature
The compressibility factor (Z_c) at critical temperature is given by,
[tex]Z_c=\frac{P_c V_c}{n_c k_B T_c}[/tex]
Crtitical parameters (critical temperaures, critical pressure and critical volume) can be expressed in terms of van der Waals parameters a and b.
Principle of corresponding state can also be stated as gases at the same reduced pressure and reduced temperature have same reduced volume.
Hence, compressibility factor at reduced state will be same for all gase. so the given statement is false.