Answer:
See explanation
Explanation:
Sea water = a mixture of different substances in water. If the water evaporates, some of those substances remain as salts. It's not a compound because it's a mixture of compounds and substances. An example of a compound is water ( because it has hydrogen atoms and an oxygen atom)
Calcium = Calcium is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because we can find it on the left side of the periodic table, in group 2. It's an alkaline earth metal, what makes that calcium is a reactive metal.
Argon = Argon is found on the periodic table, so it can't be a mixture or compound. It is a non-metallic element. We can find it in group 18 on the periodic table. Argon is a noble gas, so non-metallic.
Water = a compound because its only made of 2 atoms : oxygen and hydrogen. Reasons why water is a compound and not a mixture are:
- The ormation of a compound is a chemical change which is followed by absorption of energy or evolution of energy, in case of water, electricity is required.
- Mixtures can be separated by physical separation techniques ,Water can not be separated into it its elements by physical separation techniques. But by the absorption of chemical energy.
Air = Mixture because it can be separated into different atoms, molecules,.. like oxygen, nitrogen etc. by the physical separation techniques (fractional distillation).
Carbon Monoxide = Carbon Monoxide is not found on the periodic table so it cannot be an element. It's made of 2 elements, this means caron monoxide is a compound. It's not a mixture since the elements cannot be separated by physical separating techniques.
Iron = Iron is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because:
-High melting point
- Some metals form a dull oxide layer, this explains the shiny luster surface
- Electrical conductivity and thermal conductivity
Sodium chloride = NaCl cannot be found on the periodic table, so it isn't an element. It's a compound because it only has 2 atoms (elements). Those elements cannot be seperated by physical separating techniques, but would require electricity. So it's a compound.
Diamond = is a solid form of the element Carbon. It's an allotrope of carbon. They have the same physical state but in distinct form. Technically diamond is a non-metallic element. Since it's seen as carbon.
Brass = Brass is a mixture of the elements of copper and zinc. Those elements can be separated by physical techniques.
Copper = Copper is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because:
-High melting point
- Some metals form a dull oxide layer, this explains the shiny luster surface
- Electrical conductivity and thermal conductivity
Dilute sulphuric acid = This is a mixture. Sulphuric acid is a compound but to dilute it's added in water, what is another compound. So it's a mixture of different compounds.
Sulphur = Can be found in the periodic table so it cannot be a mixture or compound. It's part of the metalloids, therefore, it can be concluded that sulfur is a non-metal. It belongs to the oxygen family.
Oil = Oil is a mixture of hydrocarbon compounds which varies in lengths.
Nitrogen = Nitrogen is found on the periodic table, so it can't be a mixture or compound. It can be found as a gas so it is a non-metallic element.
Ammonia = a compound of nitrogen and hydrogen with the formula NH3. Those elements cannot be separated with physical separating techniques.
The Safe Drinking Water Act (SDWA) sets a limit for mercury-a toxin to the central nervous system-at 0.002 mg/L. Water suppliers must periodically test their water to ensure that mercury levels do not exceed 0.002 mg/L. Suppose water becomes contaminated with mercury at twice the legal limit (0.004 mg/L). Part A How much of this water would have to be consumed to ingest 0.150 g of mercury
In order to ingest 0.150g of mercury from water contaminated with mercury at twice the legal limit (0.004 mg/L), one would need to consume 37500 liters of this contaminated water.
Explanation:Your question is looking to find out how much water contaminated with mercury at twice the legal limit would need to be consumed to ingest 0.150g of mercury. First, we need to convert the g of mercury you want to find out to the same unit as the water contamination level, which is mg/L. So, 0.150 g = 150 mg.
Then, we will divide this amount by the contamination level, which is 0.004 mg/L. Therefore, 150 divided by 0.004 = 37500 L. So, one would have to consume 37500 liters of water at that level of contamination to ingest 0.150 g of mercury.
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One would have to consume 37,500 litres of water to ingest 0.150 grams of mercury at a concentration of 0.004 mg/L.
To solve this problem, we need to calculate the volume of water that contains 0.150 grams of mercury when the concentration of mercury is twice the legal limit, which is 0.004 mg/L.
First, we convert the mass of mercury from grams to milligrams, since the concentration is given in milligrams per litre.
1 gram = 1000 milligrams
So, 0.150 grams of mercury is equivalent to:
[tex]\[ 0.150 \text{ g} \times 1000 \text{ mg/g} = 150 \text{ mg} \][/tex]
Now, we know that the water is contaminated at a concentration of 0.004 mg/L, which means there are 0.004 milligrams of mercury in every litre of water.
To find out how many litres of water contain 150 mg of mercury, we set up the following proportion:
[tex]\[ \frac{0.004 \text{ mg}}{1 \text{ L}} = \frac{150 \text{ mg}}{x \text{ L}} \][/tex]
Solving for ( x ) (the volume of water in litres), we get:
[tex]\[ x = \frac{150 \text{ mg}}{0.004 \text{ mg/L}} \]\\[/tex]
[tex]\[ x = 37500 \text{ L} \] \\[/tex]
[tex]\[ x = \frac{150 \text{ mg}}{0.004 \text{ mg/L}} \] \\[/tex]
[tex]\[ x = 37500 \text{ L} \][/tex]
A reaction between substances Y and Z is
representedstochiometrically by
Y2 + Z2 ---> 2YZ
The rate constant obeys the Arrhenius equation. At 435.
K
the rate constant is k = 2.95 e-03 L/mol-s and A = 3.00 e+10
L/mol-s
What is the activation energy (KJ/mol) for
thisreaction?
Answer : The value of activation energy for this reaction is 108.318 kJ/mol
Explanation :
The Arrhenius equation is written as:
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
Taking logarithm on both the sides, we get:
[tex]\ln k=-\frac{Ea}{RT}+\ln A[/tex] ............(1)
where,
k = rate constant = [tex]2.95\times 10^{-3}L/mol.s[/tex]
Ea = activation energy = ?
T = temperature = 435 K
R = gas constant = 8.314 J/K.mole
A = pre-exponential factor = [tex]3.00\times 10^{+10}L/mol.s[/tex]
Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:
[tex]\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)[/tex]
[tex]Ea=108318.365J/mol=108.318kJ/mol[/tex]
Therefore, the value of activation energy for this reaction is 108.318 kJ/mol
Given a water solution that contains 1.704 kg of HNO:/kg H:O and has a specific gravity of 1.382 at 20 °C, express the composition in the following ways: (a) Weight percent HNO (b) Pounds HNO3 per cubic foot of solution at 20 °c (c) Molarity (gmoles of HNOs per liter of solution at 20 °C)
Answer:
a) 63,0%
b) 54,4 Pounds HNO₃ per cubic foot of solution
c) 13,8 M
Explanation:
a) Weight percent is the ratio solute:solution times 100:
[tex]\frac{1,704 kg HNO_3}{2,704 kg Solution}[/tex] = 63,0%
b) Pounds HNO₃ per cubic foot of solution at 20 °c
Pounds HNO₃:
1,704 kg [tex]\frac{2,20462 pounds}{1 kg}[/tex] = 3,7567 pounds
Cubic foot:
2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex]x[tex]\frac{1 cubic foot }{28,3168 L}[/tex] = 0,069 ft³
Thus:[tex]\frac{3,7567 pounds}{0,069ft^3} =[/tex] = 54,4 Pounds HNO₃ per cubic foot of solution
c) Moles of HNO₃:
1704 g HNO₃ [tex]\frac{1 mol HNO_3}{63,01 g }[/tex] = 27,04 moles
Liters of solution:
2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex] = 1,96 L of solution
Molarity:
[tex]\frac{27,04 mol}{1,96 L}[/tex] = 13,8 M
Final answer:
The composition of the solution can be expressed as 63.02 wt% HNO3, 54.28 lb HNO3/ft3, and a molarity of 14.2 M at 20 °C.
Explanation:
To express the composition of a water solution with 1.704 kg of HNO3 per kg of H2O and a specific gravity of 1.382 at 20 °C, we can calculate the following:
Weight Percent HNO3
Weight percent (wt%) is calculated as the mass of the solute divided by the total mass of the solution, multiplied by 100. For 1.704 kg HNO3 in 1 kg of water, the total mass of the solution is 1.704 kg + 1 kg = 2.704 kg. The weight percent HNO3 is then ((1.704 kg) / (2.704 kg)) × 100 = 63.02 wt%.
Pounds HNO3 per Cubic Foot of Solution at 20 °C
Firstly, we need to convert the specific gravity to density in g/mL: 1.382 (Specific Gravity) × 1.000 g/mL (density of water) = 1.382 g/mL. To find the density in lb/ft3, we multiply by 62.43 lb/ft3 which is the conversion factor from g/mL to lb/ft3. Then, multiply the density of the solution by the weight percent of HNO3 to find pounds HNO3 per cubic foot of solution: (1.382 g/mL) × (62.43 lb/ft3) × (63.02%) = 54.28 lb HNO3/ft3.
Molarity of HNO3 at 20 °C
Using the density of the concentrated HNO3 solution, 1.42 g/mL, and the given formula for molarity which is Molarity = [(%) (d)/ (Mw)] × 10, we substitute the values: Molarity = [(63.02) (1.42 g/mL) / (63.01)] × 10 = 14.2 M.
How much heat to change 1 mole of ice at -25 °C to steam at +125 °C? • heat to warm ice - heat to melt ice (no temperature change) - heat to warm water • heat to boil water (no temperature change) - heat to warm steam
Answer : The amount of heat changes is, 56.463 KJ
Solution :
The conversions involved in this process are :
[tex](1):H_2O(s)(-25^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(125^oC)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change or heat changes = ?
n = number of moles of water = 1 mole
[tex]c_{p,s}[/tex] = specific heat of solid water = [tex]2.09J/g^oC[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]
[tex]c_{p,g}[/tex] = specific heat of liquid water = [tex]1.84J/g^oC[/tex]
m = mass of water
[tex]\text{Mass of water}=\text{Moles of water}\times \text{Molar mass of water}=1mole\times 18g/mole=18g[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[18g\times 4.18J/gK\times (0-(-25))^oC]+1mole\times 6010J/mole+[18g\times 2.09J/gK\times (100-0)^oC]+1mole\times 40670J/mole+[18g\times 1.84J/gK\times (125-100)^oC][/tex]
[tex]\Delta H=56463J=56.463KJ[/tex] (1 KJ = 1000 J)
Therefore, the amount of heat changes is, 56.463 KJ
what is the purpose of a pipe rack in chemical process?
Answer:
Pipe racks are construction in chemical and other industries plants, that support the pipe line, electric cables and instrument cable.
Explanation:
The pipe racks also used to support mechanical equipment as valve and vessels. You can transfer material between equipment and sorage or utility areas. Pipe racks aren´t only non-building constructions that have similiraties to the Steel buildings but also have additional loads style. The requirements found in the building codes apply and dictate some of the design requirements.
Some industry references exist to help the designer apply the intent of the code and follow expected engineering practices.
Pipe racks have design criteria: In most of the United States, the governing building code is the International Building Code.
Includes:
*Dead Loads
*Live Loads
*Thermal Loads
*Earthquake Loads
*Wind Loads
*Rain Loads
*Snow Loads
*Ice Loads
*Load Combinations
Also have Design Considerations:
*Layout
*Seismic
*Seismic System Selection
*Period Calculations
*Analisys Procedure
*Selection
*Equivalent Lateral Force Method Analysis
*Modal Response Spectra Analysis
*Drift
*Seismic Detailing Requirements
*Wind
*Pressures and Forces
*Coatings
*Fire Protection
*Torsion on Support Beams
A 0.1375 g sample of solid magnesium is burned in
aconstant-volume bomb calorimeter that has a heat capacity of3024
J/ oC. The temperature increases by 1.126oC.
Calculate the heat given off by burning Mg, inkJ/g and in
kJ/mol.
Answer:
24.76 kJ/g
844.8 kJ/mol
Explanation:
The heat produced by the burning of the solid magnesium is equal to the heat absorbed by the calorimeter, which causes its temperature to increase.
First, we calculate the heat absorbed by the calorimeter, where C is the heat capacity and Δt is the temperature change.
Q = CΔt = (3024 J·°C⁻¹)(1.126 °C) = 3405 J
This is the same amount of heat that was produced by burning the magnesium
Now we can calculate the heat produced per gram of magnesium:
(3405 J)(1 kJ/1000 J) / (0.1375 g) = 24.76 kJ/g
We can convert grams to moles using the atomic weight of Mg (24.305 g/mol).
(24.76 kJ/g)(24.305 g/mol) = 844.8 kJ/mol
a reasonable pka for a weak base is:
10.3
0.9
5.8
7.4
13.1
Answer:
0.9
Explanation:
The pka represents the force by which the molecules need to dissociate for the acids ,
Hence , lower the pka stronger will be the acid and that therefore will dissociate completely and vice versa , for a weak acid higher the pka .
And in case of a base , it will be completely reversed , lower pKa , weaker base ,
and higher pKa , stronger base .
From the data of the question ,
0.9 is the lowest value of the pKa , hence , weakest base .
Normally carbon forms how many and what type of bonds?
Answer:
Carbon generally forms four covalent chemical bonds.
Explanation:
Carbon is a chemical element that belongs to group 14 of the periodic table and has atomic number 6. It is a member of the p-block and is nonmetallic in nature.
The ground-state electronic configuration of carbon atom is 1s²2s²2p². Thus, it has 4 valence electrons and is said to be tetravalent.
Therefore, carbon generally forms four covalent chemical bonds.
Picture this in your mind as you read it: You weigh an empty, clean, dry beaker on the balance (scale). It weighs 32.15 grams. You are instructed to measure out 2.0 grams of solid X and put it in the beaker you just weighed. Using weighing paper, you weigh out 2.0 grams of solid X and carefully transfer it into the beaker. You now weigh the beaker WITH the solid in it and get a mass of 34.40 grams. What is the weight of the solid, using the correct number of significant figures? 2.0 grams, just as you weighed it before putting it in the beaker. 2.00 grams, using the correct number of significant figures. The mass is 34.40 g -32.15 g -2.3 grams The mass is 34.40 g -32.15 g -2.25 grams
Answer:
2.25 g
Explanation:
The mass of the solid X must be the total mass (beaker + solid X) less than the mass of the beaker. Then:
mass of the solid X = 34.40 - 32.15
mass of the solid X = 2.25 g
The difference of 0.25 g must occur for several problems: an incorrect weight in the balance, the configuration of the balance, the solid can be hydrophilic and absorbs water, and others.
One particular super red giant star has a surface temperature of 2,822 K. What is the wavelength (in nm) of the spectral maximum in the emission of light by this star?
Answer: [tex]1027nm[/tex]
Explanation:
Using Weins displacement law:
[tex]\lambda_{max}=\frac{b}{T}[/tex]
where [tex]\lambda_{max}[/tex] = wavelength
b = constant =[tex]2898\micro mK[/tex]
T = Temperature in Kelvin = 2822 K
Putting the values we get:
[tex]\lambda _{max}=\frac{2898\micro mK}{2822K}=1.027\micro m[/tex]
[tex]1\micro m=10^3nm[/tex]
[tex]1.027\micro m=\frac{10^3}{1}\times 1.027=1.027\times 10^3nm[/tex]
Thus wavelength of the spectral maximum in the emission of light by this star is [tex]1027nm[/tex]
A solution of HNO3HNO3 is standardized by reaction with pure sodium carbonate. 2H++Na2CO3⟶2Na++H2O+CO2 2H++Na2CO3⟶2Na++H2O+CO2 A volume of 27.71±0.05 mL27.71±0.05 mL of HNO3HNO3 solution was required for complete reaction with 0.9585±0.0007 g0.9585±0.0007 g of Na2CO3Na2CO3 , (FM 105.988±0.001 g/mol105.988±0.001 g/mol ). Find the molarity of the HNO3HNO3 solution and its absolute uncertainty.
Answer:
(0,653±0,002) M of HNO₃
Explanation:
The reaction of standarization of HNO₃ with Na₂CO₃ is:
2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻
To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:
0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =
9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃
Molarity is moles divide liters, thus, molarity of HNO₃ is:
1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃
The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:
ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =
0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M
Thus, molarity of HNO₃ solution and its absolute uncertainty is:
(0,653±0,002) M of HNO₃
I hope it helps!
a 50-g sample contains 32.1 g of oxygen 2 g of hydrogen and 15.9 g of carbon
Answer:
i need more to solve this
Explanation:
Options for full question:
(A) % oxygen is 32.1%
(B) % hydrogen is 4%
(C) total percent composition of all elements is approximately 100%
(D) % carbon is 15.9%
Answer:
Options B and C
Explanation:
Information Given;
Mass of sample - 50g
Mass of Oxygen - 32.1g
Mass of Hydrogen - 2g
Mass of Carbon - 15.9g
The percentage composition of an element in a compound is the mass percentage of the element present in the compound. It tells the mass percentage of each element present in a compound.
The formular is given as;
Percentage composition = (Mass of element / Mass of compound) * 100
For oxygen;
Percentage Composition = (32.1 / 50) * 100 = 0.642 * 100= 64.2%
For Hydrogen;
Percentage Composition = (2 / 50) * 100 = 0.04 * 100= 4%
For Carbon;
Percentage Composition = (15.9 / 50) * 100 = 0.318 * 100= 31.8%
Total percentage composition of all elements = 4% + 31.8% + 64.2% = 100%
Based on this,
Option A is incorrect
Option B is Correct.
Option C is Correct.
Option D is incorrrect
Acetone must be kept in a closed, pressure tank due to evaporation. Acetone evaporates at a rate of 44 grams per minute left bracket StartFraction g Over min EndFraction right bracket g min. If the tank holds 5454 pound-mass [lbm] of acetone and is allowed to remain open to the atmosphere, how many days [d] will it take for half of the acetone to evaporate?
The vapor pressure of acetone at 25.0°C under 1 atm of pressure is approximately 2.05 atm.
Explanation:The molar entropy of vaporization of acetone under 1 atm of pressure can be calculated using the Clausius-Clapeyron equation. The equation is given as:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where P1 and P2 are the initial and final vapor pressures, ΔHvap is the molar enthalpy of vaporization, R is the ideal gas constant, T1 and T2 are the initial and final temperatures respectively. By rearranging the equation, we can solve for the final vapor pressure:
P2 = P1 * exp(ΔHvap/R * (1/T1 - 1/T2))
Substituting the given values, we have:
P1 = 1 atm
T1 = 56°C = 329 K
T2 = 25°C = 298 K
ΔHvap = 31.3 kJ/mol = 31,300 J/mol
R = 8.314 J/(mol·K)
Plugging these values into the equation:
P2 = 1 atm * exp(31300 J/mol / (8.314 J/(mol·K)) * (1/329 K - 1/298 K))
Simplifying the equation gives:
P2 ≈ 2.05 atm
Therefore, the vapor pressure of acetone at 25.0°C is approximately 2.05 atm.
A gas effuses 1.55 times faster than propane (C3H8)at the
same
temperature and pressure.
What is the mass of the gas?
Answer: The mass of the gas is 18.3 g/mol.
Explanation:
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
[tex]\frac{Rate_{X}}{Rate_{C_3H_8}}=1.55[/tex]
[tex]\frac{Rate_{X}}{Rate_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{X}}}[/tex]
[tex]1.55=\sqrt{\frac{44}{M_{X}}[/tex]
Squaring both sides and solving for [tex]M_{X}[/tex]
[tex]M_{X}=18.3g/mol[/tex]
Hence, the molar mas of unknown gas is 18.3 g/mol.
Answer: The molar mass of the unknown gas is 18.3 g/mol
Explanation:
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
We are given:
[tex]\text{Rate}_{\text{(unknown gas)}}=1.55\times \text{Rate}_{C_3H_8}[/tex]
We know that:
Molar mass of propane = 44 g/mol
Taking the ratio of the rate of effusion of the gases, we get:
[tex]\frac{\text{Rate}_{\text{(unknwon gas)}}}{\text{Rate}_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{\text{(unknown gas)}}}}[/tex]
Putting values in above equation, we get:
[tex]\frac{1.55\times \text{Rate}_{C_3H_8}}{\text{Rate}_{C_3H_8}}=\sqrt{\frac{44}{M_{\text{(unknown gas)}}}}[/tex]
[tex]1.55=\sqrt{\frac{44}{M_{\text{(unknown gas)}}}}\\\\1.55^2=\frac{44}{M_{\text{unknwon gas}}}\\\\M_{\text{unknwon gas}}=\frac{44}{2.4025}\\\\M_{\text{unknwon gas}}=18.3g/mol[/tex]
Hence, the molar mass of the unknown gas is 18.3 g/mol
Dagastinium, on the other hand, is a rare material that has a temperature dependent constant pressure heat capacity of cP = 0.1 x T J/(mol.K) in the range from 273 to 303 K. What is the entropy change of 2 moles of Dagastinium between 273 and 303 K?
Answer : The change in entropy is 6 J/K
Explanation :
To calculate the change in entropy we use the formula:
[tex]\Delta S=\int \frac{dQ}{T}[/tex]
and,
[tex]Q=nC_pdT[/tex]
[tex]\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p}dT}{T}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
n = number of moles = 2 moles
[tex]T_f[/tex] = final temperature = 303 K
[tex]T_i[/tex] = initial temperature = 273 K
[tex]C_{p}[/tex] = heat capacity at constant pressure = [tex]0.1\times T(J/K.mol)[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=2\int\limits^{303}_{273}{\frac{(0.1\times TdT}{T}[/tex]
[tex]\Delta S=2\times 0.1\int\limits^{303}_{273}dT[/tex]
[tex]\Delta S=2\times 0.1\times [T]^{303}_{273}[/tex]
[tex]\Delta S=2\times 0.1\times (T_f-T_i)[/tex]
[tex]\Delta S=2\times 0.1\times (303-273)[/tex]
[tex]\Delta S=6J/K[/tex]
Therefore, the change in entropy is 6 J/K
integrated rate law for second order unimolecular irreversible
Answer:
The rate law for second order unimolecular irreversible reaction is
[tex]\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }[/tex]
Explanation:
A second order unimolecular irreversible reaction is
2A → B
Thus the rate of the reaction is
[tex]v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}[/tex]
rearranging the ecuation
[tex]-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}[/tex]
Integrating between times 0 to t and between the concentrations of [tex][A]_{0}[/tex] to [A].
[tex]\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}[/tex]
Solving the integral
[tex]\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }[/tex]
Using the integrated rate law for second-order reactions, we can determine the concentration of butadiene gas after a certain time period by applying the rate constant and initial concentration into the equation 1/[A] = kt + 1/[A]0 where k is the rate constant, t is the time and [A]0 is the initial concentration.
Explanation:The integrated rate law for second-order reactions in kinetics helps us understand how the concentration of a reactant changes over time in a unimolecular and irreversible reaction. In the example provided, using the second-order integrated rate law equation, we want to find the concentration of butadiene (C4H6) after 10 minutes given its initial concentration and reaction rate constant.
The second-order integrated rate equation can be written as:
1/[A] = kt + 1/[A]0
With a y-intercept of 1/[A]0 and slope of the rate constant k.
To solve for the concentration after 10 minutes, we can plug in the given values:
Initial concentration of butadiene [A]0 = 0.200 MRate constant k = 5.76 × 10-2 L mol-1 min-1Time t = 10.0 minThen we calculate [A] after 10.0 min using the equation:
1/[A] = (5.76 × 10-2 L mol-1 min-1 × 10.0 min) + (1/0.200 M)
The above calculation will yield the final concentration [A] of butadiene after 10 minutes.
What is the theoretical Van’t Hoff factor for the following substances: HCL & ethanol (CH3CH2OH)
Answer:
i=2 for HCl and i=1 for ethanol
Explanation:
Hello,
Since the hydrochloric acid is composed by sodium and chloride ions which are completely dissociated, two types of ions are present, that's why i becomes 2.
On the other hand, as long as the ethanol doesn't present dissociation in aqueous solution, the Van't Hoff factor becomes 1.
Best regards.
A 200 ml sample of 0.1015 Miric acid is mbred with 2300 ml of water. What is the molar concentration of nitric acid in the final solution 0 406 M 127M 325 x 10 M 5.08 x 10 M 8.12 x 10 M water Backspace lu 'o Pin
Answer: Molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]
Explanation:
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]HNO_3[/tex] solution = 0.1015 M
[tex]V_1[/tex] = volume of stock [tex]HNO_3[/tex]solution = 200 ml
[tex]M_2[/tex] = molarity of dilute [tex]HNO_3[/tex] solution = ?
[tex]V_2[/tex] = volume of dilute [tex]HNO_3[/tex] solution = (2300 +200 )ml = 2500 ml
Putting in the values we get:
[tex]0.1015M\times 200=M_2\times 2500[/tex]
[tex]M_2=8.12\times 10^{-3}M[/tex]
Thus the molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]
The heat of vaporization of water is: O a. the amount of heat/energy required to convert 1 gram of water at 0°C to 1 gram of steam at 100°C. O b. the amount of heat/energy required to convert 1 gram of ice at 0°C to 1 gram of liquid water at 0°C. O c. the amount of heat/energy required to convert 1 gram of liquid water at 100°C to 1 gram of steam at 100°C
Answer: Option (c) is the correct answer.
Explanation:
Boiling point is defined as the point at which vapor pressure of a liquid becomes equal to the atmospheric pressure.
Boiling point of water is [tex]100^{o}C[/tex].
Whereas when we heat one mole of a liquid at its boiling point without any change in temperature then the heat required to bring out change from liquid to vapor state is known as heat of vaporization.
Thus, we can conclude that the heat of vaporization of water is the amount of heat/energy required to convert 1 gram of liquid water at [tex]100^{o}C[/tex] to 1 gram of steam at [tex]100^{o}C[/tex].
What is the sum of the numbers of neutrons and electrons in the ion 208 Pb2+?
Answer:
For ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = 206
Explanation:
Lead, chemical symbol Pb, is a chemical element which belongs to the group 14 of the periodic table. The atomic number of lead is 82 and it is a member of the p-block.
The isotope of lead with the mass number 208, has 126 neutrons.
Since, atomic number = number of protons = number of electrons for neutral atom
Therefore, for ²⁰⁸Pb: number of electrons= 82
So, for ²⁰⁸Pb²⁺ cation: number of electrons= 82 - 2 = 80
Therefore, for ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = number of electrons + number of neutrons = 80 electrons + 126 neutrons = 206.
A protein has a binding site for a single ligand with a A.G-16.7 KJ/mol at 298 K. What is Keg for this reaction?
Explanation:
The given data is as follows.
[tex]\Delta_{r} G[/tex] = -16.7 kJ/mol = [tex]-16.7 \times 10^{3}[/tex], T = 298 K
R = 8.314 J/mol K, [tex]K_{eq}[/tex] = ?
Relation between [tex]\Delta_{r} G[/tex] and [tex]K_{eq}[/tex] is as follows.
[tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]
Hence, putting the values into the above equation as follows.
[tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]
[tex]-16.7 \times 10^{3} J/mol[/tex] = [tex]-8.314 J/mol K \times 298 K ln K_{eq}[/tex]
[tex]ln K_{eq}[/tex] = [tex]\frac{-16.7 \times 10^{3} J/mol}{-8.314 J/mol K \times 298 K}[/tex]
= 6.740
[tex]K_{eq}[/tex] = antilog (6.740)
= 846
Thus, we can conclude that [tex]K_{eq}[/tex] for given values is 846.
Assuming steam to be an ideal gas, calculate its specific volume and density at a pressure of 90 lb/in2 and a temperature of 650F Compare your answer using data from the steam tables (appendix of your textbook). Problem 2 (25 points). Find the mass of air in a closed chamber measuring 35ft by 20ft by 10ft, when the pressure is 17 lb/in? and the temperature is 75F, assuming air to be an ideal gas.
Answer:
1) Sv = 0.4584 m³/Kg...assuming steam as an ideal gas
% deviation from the values in the steam tables
⇒ % dev = 45 %
2) mass air = 272.617 Kg; assuming air to be an ideal gas
Explanation:
ideal gas:
PV = RTn
molar volume:
⇒ V/n = RT/P
∴ P = 90 psi * ( 0.06895 bar/psi ) = 6.2055 bar
∴ T = 650 F = 343.33 °C = 616.33 K
∴ R = 0.08314 bar.L/mol.K
⇒ V/n = (( 0.08314 )*(616.33 K )) / 6.2055 bar
⇒ V/n = 8.2574 L/mol * ( m³/1000L ) = 8.2574 E-3 m³/mol
specific volume ( Sv ):
∴ Mw = 18.01528 g/mol
⇒ Sv = 8.2574 E-3 m³/mol * ( mol / 18.01528 g ) * ( 1000 g/Kg )
⇒ Sv = 0.4584 m³/Kg
steam table:
∴ P = 6.2055 bar ≅ 6 bar → Sv = 0.3157 m³/Kg
⇒ % deviation = (( 0.4584 - 0.3157 ) / 0.3157) * 100
⇒ % dev = 45.2 %; significant value, assuming steam to be a ideal gas
2) mass air, assuming ideal gas:
∴ V = 20ft * 35ft * 10ft = 7000ft³ * ( 28.3168 L/ft³ ) = 198217.6 L
∴ P = 17 psi * ( 0.06895 bar/psi ) = 1.172 bar
∴ T = 75 °F = 23.89 °C = 296.89 K
∴ R = 0.08314 bar.L/K.mol
⇒ n air = PV/RT = (( 1.172 )*( 198217.6 )) / (( 0.08314 )*( 296.89 ))
⇒ n air = 9411.616 mol air
∴ Mw air = 28.966 g/mol
⇒ mass air = 9411.616 mol * ( 28.966 g/mol ) = 272616.892 g = 272.617 Kg
What is the wavenumber of the radiation emitted when a hydrogen
atom makes a transition corresponding to a change in energy of
1.634 x 10-18 J?
Answer: Wavenumber of the radiation emitted is [tex]0.08\times 10^{8}m^{-1}[/tex]
Explanation:
The relationship between wavelength and energy of the wave follows the equation:
[tex]E=\frac{hc}{\lambda}[/tex]
where,
E = energy of the radiation = [tex]1.634\times 10^{-18}J[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength of radiation = ?
Putting values in above equation, we get:
[tex]1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m[/tex]
[tex]\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}[/tex]
Thus wavenumber of the radiation emitted is [tex]0.08\times 10^{8}m^{-1}[/tex]
A pure crystalline substance has no movement at absolute zero temperature. O 1st Law of Thermodynamics O 3rd Law of Thermodynamics O Law of Absolute Zero 2nd Law of Thermodynamics
3rd Law of Thermodynamics
The correct answer is the third law of thermodynamics.
"The entropy of a perfect crystal is zero when the temperature of the crystal is equal to absolute zero (0 K).”
The temperature scale they refer to is the zero in kelvin degrees, this is what we call absolute zero.
If the entropy is zero, all physic processes stop, and the entropy of the system is minimum and constant.
Final answer:
The concept described refers to the Third Law of Thermodynamics, which states that the entropy of a perfect crystalline substance is zero at absolute zero temperature. This law is often used to compute standard entropy values and predict entropy changes during phase transitions and chemical reactions. Option b.
Explanation:
The statement describing a pure crystalline substance at absolute zero temperature having no movement refers to the Third Law of Thermodynamics. This law states that the entropy of a perfectly ordered, crystalline substance at absolute zero temperature (0 K) is zero. Entropy, often denoted by S, is a measure of the disorder or randomness in a system, and it increases with temperature as molecular motion increases.
At absolute zero, all molecular motion ceases, meaning a perfectly crystalline substance has only a single microstate available to it (W = 1). Since there is just one possible arrangement for the particles, the entropy is zero as per the Boltzmann equation. This is part of the Third Law of Thermodynamics, which can be used to calculate entropy changes for phase transitions and chemical reactions under standard conditions.
A sample of mercury has a mass of 607.0 lb and a volume of 0.717 ft. What is its specific gravity? Number SG= What are the units of specific gravity? Ib/ft none g/cm2 kg/m
Answer:
Specific gravity of mercury is 13.56 and it is an unit-less quantity.
Explanation:
Mass of the mercury = m = 607.0 lb = 275330.344 g
1 lb = 453.592 g
Volume of the mercury = v = [tex]0.717 ft^3=20,303.18 mL[/tex]
[tex]1 ft^3 = 28316.847 mL[/tex]
Density of the mercury = d=[tex]\frac{m}{v}=\frac{275330.344 g}{20,303.18 mL}[/tex]
d = 13.56 g/mL
Specific gravity of substance = Density of substance ÷ Density of water
[tex]S.G=\frac{d}{1 g/mL}[/tex]
Specific gravity of mercury :
[tex]S.G=\frac{13.56 g/mL}{1 g/mL}=13.56[/tex] (unit-less quantity)
What is the Damkohler number? b) What is the significance of a system with a low Damkohler number?
Explanation:
(a) A number which is dimensionless and provides us an estimate of the degree of conversion which can be achieved in CSTR, that is, continuous stirred tank reactor is known as Damkohler number.
This number is denoted as Da.
Mathematically, Da = [tex]\frac{\text{reaction rate}}{\text{convection rate}}[/tex]
Da = [tex]\frac{-rA \times V}{Fa_{o}}[/tex]
Now, for first order system, Da = [tex]\frac{-rA \times V}{Fa_{o}}[/tex]
= [tex]\frac{k \times CA_{o} \times V}{v \times CA_{o}}[/tex] = Tk
where, rA = rate of reaction
V = volume of reactor
[tex]Fa_{o}[/tex] = molar flow rate of component A
k = rate constant
[tex]CA_{o}[/tex] = initial concentration of A
v = volumetric flow rate of A
T = residence time
(b) Since, from a given Damkohler number we can figure out the possible conversion of CSTR, that is, continuous stirred tank reactor.
So, if we have a low Damkohler number then the system will give us a less conversion formula. As the conversion is as follows.
X = [tex]\frac{Da}{Da + 1}[/tex]
Hence, we can conclude that [tex]Da \leq 0.1[/tex] will give less than 10% conversion as calculated by using above formula.
Final answer:
The Damkohler number is a dimensionless number used in chemical engineering to characterize the importance of reaction rates relative to a system's residence time. A system with a low Damkohler number indicates that the chemical reaction is much faster compared to the transport of material. This means that the reaction can be considered essentially instantaneous, and the reactants are fully converted into products within the system before they have a chance to be transported out.
Explanation:
The Damkohler number is a dimensionless number used in chemical engineering to characterize the importance of reaction rates relative to a system's residence time. It is defined as the ratio of the characteristic time for a chemical reaction to occur to the characteristic time for a system to transport material through itself.
A system with a low Damkohler number indicates that the chemical reaction is much faster compared to the transport of material. In practical terms, this means that the reaction can be considered essentially instantaneous, and the reactants are fully converted into products within the system before they have a chance to be transported out.
For example, if we have a packed bed reactor (a reactor where reactants flow through a bed of solid catalyst particles), a low Damkohler number implies that the reaction is so fast that the reactants are fully converted into products even before they can flow through the bed, resulting in high conversion levels.
The heat capacity of chloroform (trichloromethane,CHCl3)
in the range 240K to 330K is given
byCpm/(JK-1mol-1) = 91.47
+7.5x10-2(T/K). In a particular experiment,
1.0molCHCl3 is heated from 273K to 300K. Calculate the
changein molar entropy of the sample.
Answer : The change in molar entropy of the sample is 10.651 J/K.mol
Explanation :
To calculate the change in molar entropy we use the formula:
[tex]\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p,m}dT}{T}[/tex]
where,
[tex]\Delta S[/tex] = change in molar entropy
n = number of moles = 1.0 mol
[tex]T_f[/tex] = final temperature = 300 K
[tex]T_i[/tex] = initial temperature = 273 K
[tex]C_{p,m}[/tex] = heat capacity of chloroform = [tex]91.47+7.5\times 10^{-2}(T/K)[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=1.0\int\limits^{300}_{273}{\frac{(91.47+7.5\times 10^{-2}(T/K))dT}{T}[/tex]
[tex]\Delta S=1.0\times [91.47\ln T+7.5\times 10^{-2}T]^{300}_{273}[/tex]
[tex]\Delta S=1.0\times 91.47\ln (\frac{T_f}{T_i})+7.5\times 10^{-2}(T_f-T_i)[/tex]
[tex]\Delta S=1.0\times 91.47\ln (\frac{300}{273})+7.5\times 10^{-2}(300-273)[/tex]
[tex]\Delta S=8.626+2.025[/tex]
[tex]\Delta S=10.651J/K.mol[/tex]
Therefore, the change in molar entropy of the sample is 10.651 J/K.mol
What is the formula for aluminum nitrite ?
Answer: The chemical formula for aluminium nitrite is [tex]Al(NO_2)_3[/tex]
Explanation:
The given compound is formed by the combination of aluminium and nitrite ions. This is an ionic compound.
Aluminium is the 13th element of periodic table having electronic configuration of [tex][Ne]3s^23p^1[/tex].
To form [tex]Al^{3+}[/tex] ion, this element will loose 3 electrons.
Nitrite ion is a polyatomic ion having chemical formula of [tex]NO_2^{-}[/tex]
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
So, the chemical formula for aluminium nitrite is [tex]Al(NO_2)_3[/tex]
The formula for aluminum nitrite is Al(NO3)3.
The formula for aluminum nitrite is Al(NO2)3.
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Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a distance of 1 nm. Please calculate the force of attraction between the anion and cation. The force of attraction is given by: (9 x 109 V/C) (Z)(2)(e2) FA valence of the ions, e = charge of an electron 1.602 x 10-19 C Where Z1 and Z2 r distance between ions 1N 1 (V C/m)
Answer:
Force of attraction = 35.96 [tex]\times 10^{27}[/tex]N
Explanation:
Given: charge on anion = -2
Charge on cation = +2
Distance = 1 nm = [tex]10^{-9}[/tex] m
To calculate: Force of attraction.
Solution: The force of attraction is calculated by using equation,
[tex]F = \dfrac{k \times q_1 q_2}{ \r^2}[/tex] ---(1)
where, q represents the charge and the subscripts 1 and 2 represents cation and anion.
k = [tex]8.99 \times 10^9 \ Nm^{2}C^{-2}[/tex]
F = force of attraction
r = distance between ions.
Substituting all the values in the equation (1) the equation becomes
[tex]F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }[/tex]
Force of attraction = 35.96 [tex]\times 10^{27}[/tex]N
If 12.5 mL of 0.200 M NaOH was used to reach the equivalence point for in the titration of 25.0 mL of unknown weak acid, HA, the concentration of HA is a. 0.050M b. 0.100M C.0.200M d.0.400M
Answer:
b. 0.100M
Explanation:
The balanced chemical reaction is: NaOH + HA ⇒ H₂O + NaA
The NaOH and HA react in a 1:1 molar ratio, so at the equivalence point, the amount of NaOH added equals the amount of HA that was present in the solution.
The amount of NaOH that was added can be calculated and set equal to the amount of HA that must have been present to react with it.
n = CV = (0.200 mol/L)(12.5 mL) = 2.50 mmol NaOH = 2.50 mmol HA
Thus, there were 2.50 mmol of HA in 25.0 mL. The concentration can be calculated as follow:
C = n/V = (2.50 mmol)/(25.0mL) = 0.100 M