Answer:an exoskeleton
Explanation:jus did it
Answer:
B
Explanation:
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate the amount of Ga ( s ) that can be deposited from a Ga ( III ) solution using a current of 0.220 A that flows for 40.0 min .
Answer: Amount of Gallium = 0.127g
Explanation:
Electrolysis equation is:
Ga3+ + 3e- ------> Ga
To calculate the charge
t = 40.0 min = 40.0 x 60 s = 2400 s
time, t = 2400s
Q = I*t =
= 0.22A x 2400s
= 528 C
1 mol of Ga requires 3 mol of electron
1 mol of electron = 1 Faraday =96485 C
So,1 mol of Ga requires 96485x 3= 289455 C
mol of Gallium = 528/289455 = 0.00182 mol
Molar mass of Ga = 69.72 g/mol
mass of Ga = number of moles x molar mass
= 0.00182mol * 69.72
g/mol
= 0.127g
or you can use this direct formula
m=(current*time/Faraday's)*(molar mass/no of electrons transferred)
keeping in mind Ga3+ + 3e- → Ga
n=3
m=(It/F)*(mew/n)
m =(0.22 x 2400/96485) x (69.72/3)
m=0.127 g
what is the atomic number of an oxygen atom with 8 protons and 10 neutrons in the nucleus.
A. 8
B.10
C.18
D. not enough information to calculate
Answer
A. 8
Explanation:
The Atomic number is equal to the number of protons.
I took the test and got it right so this is 100% correct. It is NOT 18 like some people say
The atomic number of an oxygen atom with 8 protons is 8, regardless of the number of neutrons.
Explanation:The atomic number of an atom is determined by the number of protons in its nucleus. This number also sorts elements into their correct position on the Periodic Table. Therefore, an oxygen atom with 8 protons will have an atomic number of 8, irrespective of the number of neutrons it has.
This is because neutrons do not influence the atomic number, only the atomic mass. So the correct answer to your question is: A. 8.
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The United States Mint uses electrolysis to copper plate zinc pennies by placing them in a Cu2+ solution and connecting the pennies and the copper electrode to a battery. Enter the half-reaction that takes place when pennies are plated with solid copper. Include phases.
Are the zinc pennies the cathode or the anode
Answer:
Cathode
Cu^2+(aq) + 2e ----> Cu(s)
Zinc is the cathode
Explanation:
The plating of copper is normally done by electrolysis. Electrolysis is generally defined as the chemical decomposition produced by passing an electric current through a liquid or solution containing ions.
There are two electrodes, the anode and the cathode. Recall that electrolysis is not a spontaneous process, hence energy from a battery is required to drive the reaction in the desired direction.
The metal to be plated is normally the cathode while the metal used to plate it is normally the anode. Since copper is to be plated on zinc, zinc must be the cathode while copper will be the anode.
The half-reaction that takes place when pennies are plated with solid copper is :
Cu^2+(aq) + 2e ----> Cu(s)
Copper plating is usually done by electrolysis. Electrolysis is commonly defined as the chemical decomposition produced by passing an electric current through a liquid or solution containing ions. The metal to be plated is usually the cathode and the metal used for plating is usually the anode. Copper is plated on zinc, so zinc must be the cathode and copper the anode.
Zinc is the cathode.
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Water causes many electrolytes to dissociate __________.
1. because of repulsive interactions between ions in the crystalline state.
2. because water molecules are dipoles and the dipoles orient in an energetically favorable manner to solvate the ions.
3. because the dispersion forces between ion and solvent are strong.
4. because it undergoes hydrogen bonding to large halide ions.
Answer:
2. because water molecules are dipoles and the dipoles orient in an energetically favorable manner to solvate the ions.
Explanation:
For the water to dissociate electrolytes, this one has to orientate its molecules in an energetically favorable way that allows them to interact with ions and dissociate electrolytes. This has to do with the way that intermolecular forces of a solute and a solvent, which is the water, interact to form a solution. The different intermolecular forces that interact in a solution are dipole-dipole force, ion-dipole interactions, Van Der Waals forces, and Hydrogen bonding.
What is a physical change?
A physical change is when there is an alteration to the material but does not affect at the molecular level. An example of a physical change would be cutting, crushing, freezing, and boiling a material object.
Water flowing at the rate of 13.85 kg/s is to be heated from 54.5 to 87.8°C in a heat exchanger by 54 to 430 kg/h of hot gas flowing counterflow and entering at 427°C (cpm = 1.005 kJ/kg · K). The overall Uo = 69.1 W/m^2.K. Calculate the exit-gas temperature and the heat-transfer area.
Answer:
=> 572.83 K (299.83°C).
=> 95.86 m^2.
Explanation:
Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.
Hence;
Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.
15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).
If we solve for x, we will get the value of x to be;
x = 572.83 K (2.99.83°C).
x is the temperature of the exit gas that is 572.83 K(299.83°C).
(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).
∆T = 93.87/ln 1.38.
∆T = 291.521K.
Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.
heat-transfer area = 1931146.394/69.1 × 291.521.
heat-transfer area= 95.86 m^2.
I want to convert atoms to moles. My friend tells my to multiply the number of atoms by 6.02 x 102. Is my friend
correct?
To convert atoms to moles, divide the number of atoms by Avogadro's number of 6.02 x 10²³ atoms per mole.
Explanation:Your friend is partially correct. In order to convert atoms to moles, you use the constant known as Avogadro's number (6.02 x 10²³ atoms per mole). However, it's important to note that you need to divide the number of atoms by Avogadro's number, not multiply it. Let's give an example:
Example: If we have 2.56 x 10²⁴ atoms of Uranium, we'd use Avogadro's number to convert this to moles like so: (2.56 x 10²⁴ atoms) / (6.02 x 10²³ atoms/mol) = 4.25 moles of Uranium
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No, your friend is not correct. To convert atoms to moles, divide the number of atoms by Avogadro's number, 6.022 × 10²³. This is because 1 mole of any substance contains 6.022 × 10²³ atoms.
To convert the number of atoms to moles, you should divide the number of atoms by Avogadro's number, which is 6.022 × 10²³. This relationship is based on the fact that 1 mole of any substance contains exactly 6.022 × 10²³ atoms, a constant known as Avogadro's number.
Step-by-Step Explanation:
Determine the number of atoms you have.Use the conversion factor: 1 mole = 6.022 × 10²³ atoms.Divide the number of atoms by 6.022 × 10²³ to find the number of moles.For example, if you have 1.2044 × 10²⁴ atoms of hydrogen:
Number of moles = 1.2044 × 10²⁴ atoms ÷ 6.022 × 10²³ atoms/mole.This equals 2 moles of hydrogen.Complete Question: -
I want to convert atoms to moles. My friend tells my to multiply the number of atoms by 6.02 x 10²³. Is my friend correct?
How many moles are in an 11mL solution of NaOh and KHP (C8H4O4)
Answer:
Calculate the molar concentration of the NaOH solution that you prepared Number of moles of KHP = Number of moles NaOH = 2.476 x 10 -3 moles Number of moles NaOH = Mb x Vb Mb = 2.476 x 10 -3 moles / 0.0250 L (equivalence point) = 0.0990 M 3
Explanation:
Butane (C4H10) burns completely with 110% of theoretical air entering at 74°F, 1 atm, 50% relative humidity. The dry air component can be modeled as 21% O2 and 79% N2 on a molar basis. The combustion products leave at 1 atm. For complete combustion of butane(C4H10) with theoretical amount of air, what is the number of moles of oxygen (O2) per mole of fuel?
To achieve complete combustion of one mole of butane (C₄H₁₀), 6.5 moles of oxygen (O₂) are necessary, as indicated by the balanced chemical equation for combustion.
To determine the number of moles of oxygen (O₂) required for the complete combustion of one mole of butane (C₄H₁₀), we must first write the balanced chemical equation for the combustion reaction:
C₄H₁₀ + O₂ → CO₂ + H₂O
After balancing the equation, we have:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
This balanced equation tells us that for every 2 moles of butane, we need 13 moles of oxygen for complete combustion, which means that for every mole of butane, we need 6.5 moles of oxygen.
Name the organic compound CH4
Answer:
Methane
Explanation:
Methane is a potent greenhouse gas with the formula CH₄. Hope this helps!
Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the following experimental values of virial coefficients: B = −152.5 cm3·mol−1 C = −5800 cm6·mol−2 (b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)]. (c) The steam tables (App. E).
Answer:
Explanation:
Given that:
the temperature [tex]T_1[/tex] = 250 °C= ( 250+ 273.15 ) K = 523.15 K
Pressure = 1800 kPa
a)
The truncated viral equation is expressed as:
[tex]\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}[/tex]
where; B = - [tex]152.5 \ cm^3 /mol[/tex] C = -5800 [tex]cm^6/mol^2[/tex]
R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹
Plugging all our values; we have
[tex]\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]
[tex]4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]
Multiplying through with V² ; we have
[tex]4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0[/tex]
[tex]4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0[/tex]
V = 2250.06 cm³ mol⁻¹
Z = [tex]\frac{PV}{RT}[/tex]
Z = [tex]\frac{1800*2250.06}{8.314*10^3*523.15}[/tex]
Z = 0.931
b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].
The generalized Pitzer correlation is :
[tex]T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345[/tex]
[tex]T__{\gamma}} = \frac{T}{T_c}[/tex]
[tex]T__{\gamma}} = \frac{523.15}{647.1}[/tex]
[tex]T__{\gamma}} = 0.808[/tex]
[tex]P__{\gamma}} = \frac{P}{P_c}[/tex]
[tex]P__{\gamma}} = \frac{1800}{22055}[/tex]
[tex]P__{\gamma}} = 0.0816[/tex]
[tex]B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}[/tex]
[tex]B_o = 0.083 - \frac{0.422}{0.808^{1.6}}[/tex]
[tex]B_o = 0.51[/tex]
[tex]B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}[/tex]
[tex]B_1 = -0.282[/tex]
The compressibility is calculated as:
[tex]Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}[/tex]
[tex]Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}[/tex]
Z = 0.9386
[tex]V= \frac{ZRT}{P}[/tex]
[tex]V= \frac{0.9386*8.314*10^3*523.15}{1800}[/tex]
V = 2268.01 cm³ mol⁻¹
c) From the steam tables (App. E).
At [tex]T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa[/tex]
V = 0.1249 m³/ kg
M (molecular weight) = 18.015 gm/mol
V = 0.1249 × 10³ × 18.015
V = 2250.07 cm³/mol⁻¹
R = 729.77 J/kg.K
Z = [tex]\frac{PV}{RT}[/tex]
Z = [tex]\frac{1800*10^3 *0.1249}{729.77*523.15}[/tex]
Z = 0.588
Final answer:
To determine Z and V for steam at 250°C and 1800 kPa, we can use the truncated virial equation with the given experimental values of the virial coefficients, or we can use the generalized Pitzer correlation to obtain the B value and then use the truncated virial equation. Alternatively, we can look up the values in the steam tables.
Explanation:
The question asks us to determine Z and V for steam at 250°C and 1800 kPa using three different methods: (a) The truncated virial equation with given experimental values of the virial coefficients, (b) The truncated virial equation with B value obtained using the generalized Pitzer correlation, and (c) The steam tables.
(a) To determine Z and V using the truncated virial equation with B and C values, we substitute the given temperature and pressure into the equation and solve for Z and V.
(b) To determine Z and V using the truncated virial equation with the B value obtained from the generalized Pitzer correlation, we substitute the given temperature and pressure into the equation and solve for Z and V.
(c) To determine Z and V using the steam tables, we look up the values for Z and V at the given temperature and pressure.
How much heat (in Joules) will be needed to vaporize 18.015 grams of liquid water at 100°C?
Answer:
40659.855 J
Explanation:
From the question given above, we obtained the following:
Mass (m) = 18.015g
Heat of vaporisation (ΔHv) = 2257 J/g
Heat (Q) =?
The heat required to vaporise the water can be calculated as follow:
Q = mΔHv
Q = 18.015 x 2257
Q = 40659.855 J
Therefore, the heat required to vaporise the water is 40659.855 J
Which of the following statements about bonding and hybridization is INCORRECT? (Select ALL incorrect statements) Group of answer choices Hybridization does not account for observed bond angles in molecules Single bonds are always pi bonds The length of a bond is determined by where the energy of the system is at its lowest point Multiple bonds always have a combination of sigma and pi bonds Pi bonds are always between unhybridized p orbitals
Answer:
-Hybridization does not account for observed bond angles in molecules.
-Single bonds are always pi bonds.
-The length of a bond is determined by where the energy of the system is at its lowest point
Explanation:
-The very first statement is incorrect because it does account for different bong angles as the hybrid orbitals are responsible for contributing for bond angles in a way that more the hybrid orbitals present the lesser the angles it forms.
-The second statement is incorrect because single bonds are considered as sigma bonds and not a pi bond.
-The third statement is incorrect because hybridization is responsible for deciding the bond length.
The incorrect statements about bonding and hybridization are that single bonds are always pi bonds and pi bonds are always between unhybridized p orbitals.
Explanation:The incorrect statements about bonding and hybridization are:
Single bonds are always pi bonds.Pi bonds are always between unhybridized p orbitals.Hybridization does not account for observed bond angles in molecules, so this statement is correct. The length of a bond is determined by where the energy of the system is at its lowest point, so this statement is also correct. Multiple bonds can have a combination of sigma and pi bonds, so this statement is correct as well.
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A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next
few minutes?
O Molecules in both the metal and the surrounding air will start moving at lower speeds.
O Molecules in both the metal and the surrounding air will start moving at higher speeds.
O The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
O The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Answer:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
Because the heat of the plate will be releases warming up the air making it move faster
An insulated tank having a total volume of 0.6 m3 is divided into two compartments. Initially one compartment contains 0.4 m3 of hydrogen (H2) at 127°C, 2 bar and the other contains nitrogen (N2) at 27°C, 4 bar. The gases are allowed to mix until an equilibrium state is attained. Assuming the ideal gas model with constant specific heats, determine
Answer:
See explaination
Explanation:
In order to have the detailed and step by step solution of the given problem, check or see the attached files.
A sample of helium gas occupies a volume of 152.0 mL at a pressure of 717.0 mm Hg and a temperature of 315.0 K. What will the volume be at a pressure of 463.0 mm Hg and a temperature of 777.0 K? The combined gas law equation is given below. P1V1 T1 = P2V2 T2 In this equation, P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.
Answer:
0.581 L or 581 mL
Explanation:
As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)
Write down the amounts you are given.
V1 = 0.152 L (I was taught to always convert milliliters to liters)
P1 = 717 mmHg
T1 = 315 K
V2 = ?
P2 = 463 mmHg
T2 = 777 K
The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.
(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)
0.346 = (463*V2) / (777)
0.346*777 = (463*V2) / (777)*777
268.842 = 463*V2
268.842/463 = (463*V2)/463
V2 = 0.581
Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.
What would the pressure be if 2.80atm of air is put into a 15.6L fixed volume cylinder and heated from 285K to 396K?
Answer:
3.89 atm
Explanation:
Given data
Initial pressure (P₁): 2.80 atmInitial temperature (T₁): 285 KInitial volume (V₁): 15.6 LFinal pressure (P₂): ?Final temperature (T₂): 396 KFinal volume (V₂): 15.6 L (=V₁)If we treat air as an ideal gas, we can calculate the final pressure using the Gay-Lussac's law.
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}\\P_2 = \frac{P_1 \times T_2 }{T_1} = \frac{2.80atm \times 396K }{285K}\\P_2 = 3.89 atm[/tex]
After the end of a normal inspiration, the volume of air in the lungs is about 2.8 L. Normally quiet inspiration is driven by a pressure difference of about 2 mm Hg. The air in the lungs is at 37C and after normal expiration it is at atmospheric pressure. Quiet inspiration is driven by the expansion of the chest cavity by contraction of the diaphragm, which expands the air in the lungs. How much is the air expanded to produce an decrease of 2 mmHg in pressure
Answer:
The Volume of the lungs that would produce 2 mmHg pressure decrease is
[tex]V_2 = 2.81 \ L[/tex]
Explanation:
From the question we are told that
The volume of air in the lungs is [tex]V = 2.8 \ L[/tex]
The pressure difference for quit normal inspiration is [tex]P = 2 \ mmHg[/tex]
The temperature of air in the lungs [tex]T = 37^oC[/tex]
The pressure after normal expiration is at [tex]T = 760 \ mmHg[/tex]
From ideal gas law we have that
[tex]PV= nRT[/tex]
Now since nRT is constant we have that
[tex]P_1 V_1 = P_2 V_2[/tex]
As the pressure decreased by 2 mmHg the volume becomes
[tex]V_2 = \frac{P_1 V_1}{P_2}[/tex]
[tex]V_2 = \frac{2.8 * 760}{758}[/tex]
[tex]V_2 = 2.81 \ L[/tex]
The change in the lung volume during the process of quiet inspiration can be calculated using Boyle's Law by determining the change in pressure and relating it to an equivalent change in volume. Once pressure values are converted to the same units, the Boyle's Law equation (P1V1 = P2V2) is used to solve for the final volume (V2) that represents the expanded lung volume.
Explanation:The changes in volume and pressure in the lungs during inspiration (breath in) can be described using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. In this scenario, a decrease in pressure by 2 mmHg drives an expansion of the lungs, and we are asked to determine this corresponding change in volume.
Firstly, the pressure change needs to be converted into standard pressure units - the SI unit is Pascal. 1 mm Hg equals 133.322 Pa, so a 2 mm Hg difference equals 266.644 Pascal (Pa).
Using Boyle's Law equation (P1V1=P2V2), where P1=the initial pressure, V1=the initial volume, P2=the final pressure (P1 - 2mm of Hg) and V2=the final volume that we are trying to find, we can solve for V2. P1 is atmospheric pressure, which is approximately 101,325 Pa, and V1 is the initial volume of air in lungs, which is 2.8 Litres.
Solving the equation gives the final volume, V2, after the pressure decrease and resulting lung expansion.
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Competition occurs when two or more organisms within an ecosystem seek the same resource. Which of the following is an example of a resource that organisms might compete for?
habitat
water
sunlight
food
Answer:
the answer is all of them
Explanation:
it is all of them because organisms in and ecosystem compete for anything and every thing that they need. Hope this helps!
A calorimeter weighing 123.7g has a quantity of 20C water added to it. The combined mass of the calorimeter + cold water is 198.3g.
61g of water is heated to 60C and is poured into the calorimeter. The temperature of the mixed cold and hot water and calorimeter is 38.5C. What is the calorimeter constant?
answer choices :
a ) 148.19 b) 91.29 c) 107.75 d) 161.58
Answer:
c) 107.75
Explanation:
Hot water lost = 61 g * 60C * (4.184 J g¯1 °C¯1) = 3,660
Cold water = 74.6g * 20C * (4.184 J g¯1 °C¯1) = 1,492
The difference is 3,660 - 1,492 = 2,168
Calorimeter Constant = Heat released by burning / Change in temperature
Calorimeter Constant = 2,168 / 40C * (1.987 J g¯1 °C¯1)
Calorimeter Constant = 107.75
Convert 26.02 x 1023 molecules of C2H8 to grams. Round your answer to the hundredths place.
Answer:
x= 138.24 g
Explanation:
We use the avogradro's number
6.023 x 10^23 molecules -> 1 mol C2H8
26.02 x 10^23 molecules -> x
x= (26.02 x 10^23 molecules * 1 mol C2H8 )/6.023 x 10^23 molecules
x= 4.32 mol C2H8
1 mol C2H8 -> 32 g
4.32 mol C2H8 -> x
x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8
x= 138.24 g
The correct answer is 156.69 * 10^46 grams.
How to convert molecules to grams?
To convert from molecules to grams, it is necessary to first convert the number of molecules of a substance by dividing by Avogadro’s number to find the number of moles, and then multiply the number of moles by the molar mass of this substance.Avogadro’s number is given as 6.022 x 10^23
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A chemist titrates 80.0mL of a 0.3184M pyridine C5H5N solution with 0.5397M HBr solution at 25°C . Calculate the pH at equivalence. The pKb of pyridine is 8.77.
Answer:pH = 2.96
Explanation:
C5H5N + HBr --------------> C5H5N+ + Br-
millimoles of pyridine = 80 x 0.3184 =25.472mM
25.472 millimoles of HBr must be added to reach equivalence point.
25.472 = V x 0.5397
V =25.472/0.5397= 47.197 mL HBr
total volume = 80 + 47.197= 127.196 mL
Concentration of [C5H5N+] = no of moles / volume=
25.472/ 127.196= 0.20M
so,
pOH = 1/2 [pKw + pKa + log C]
pKb = 8.77
pOH = 1/2 [14 + 8.77 + log 0.20]
pOH = 11.0355
pH = 14 - 11.0355
pH = 2.96
Final answer:
To calculate the pH at equivalence, we need to determine the concentration of pyridine and its conjugate acid. The pH at equivalence can be calculated using the Henderson-Hasselbalch equation, which relates the pH to the pKa and the concentration of the conjugate acid and base. The pKa value for the pyridinium ion can be determined by subtracting the pKb of pyridine from the pKw.
Explanation:
To calculate the pH at equivalence, we need to determine the concentration of pyridine and its conjugate acid. From the given information, we know that the initial volume of pyridine solution is 80.0 mL and its concentration is 0.3184 M. We also have the concentration of HBr solution, which is 0.5397 M. The reaction between pyridine and HBr is:
C5H5N (aq) + HBr (aq) → C5H5NH+Br- (aq)
This reaction forms the pyridinium ion (C5H5NH+) which is the conjugate acid of pyridine. At equivalence, the moles of pyridine and pyridinium ion are equal. Using the stoichiometry of the reaction, we can calculate the number of moles of pyridine:
Moles of pyridine = Volume of pyridine solution * Concentration of pyridine = 80.0 mL * 0.3184 M = 25.472 moles
Since the reaction is 1:1, the moles of pyridine also correspond to the moles of pyridinium ion. Therefore, the concentration of pyridinium ion is:
Concentration of pyridinium ion = Moles of pyridinium ion / Volume of pyridinium ion solution = 25.472 moles / 80.0 mL = 0.3184 M
Now, we can use the Henderson-Hasselbalch equation to calculate the pH at equivalence:
pH = pKa + log10 ([A-] / [HA])
Given that the pKb of pyridine is 8.77, we can determine the pKa of pyridinium ion:
pKa = 14.00 - pKb = 14.00 - 8.77 = 5.23
Substituting the values into the Henderson-Hasselbalch equation:
pH = 5.23 + log10 (0.3184 / 0.3184) = 5.23 + 0 = 5.23
Therefore, the pH at equivalence is 5.23.
Which of the following has nonvolatile bonds
Answer:
I can provide a proper answer since there are no bonds specified.
Explanation:
Water at 25 °C flows at 5 ft/s through a straight cylindrical tube made of benzoic acid, with a 1-inch inside diameter. If the tube is 10 ft long, estimate the mixing-cup average concentration of benzoic acid in the water leaving the tube. The Schmidt number for these conditions is Sc.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Final answer:
To estimate the mixing-cup average concentration of benzoic acid in water, the Schmidt number and fluid flow characteristics are vital. Understanding incompressible fluid flow through constrictions helps in analyzing scenarios like Venturi tubes.
Explanation:
Estimate the mixing-cup average concentration of benzoic acid in the water leaving the tube by considering the overall flow conditions in the cylindrical tube.
Sc (Schmidt number) represents the fluid flow characteristics such as the diffusion rate of momentum and mass transfer in the system.
Understanding concepts like incompressible fluid flow through constrictions can aid in analyzing scenarios like flow in a Venturi tube where diameters change.
What additional information is needed to solve this problem: If a sample of a gas 12.0 o C and 1.06 atm pressure is moved to a 2.30 L container at 24.9 0 C, what is the final pressure of the gas?
Answer:
The additional information required to solve this problem is the initial volume.
the final pressure P₂ of the gas is 1.108 atm
Explanation:
Given that :
A sample of gas at initial temperature [tex]T_1 = 12.0^0 \ C[/tex] = (12+273)K = 285 K
Pressure (P₁) = 1.06 atm
Initial Volume (V₁) = unknown ???
Final Volume (V₂) = 2.30 L
final temperature [tex]T_2 = 24.9^0 \ C[/tex] = (24.9 +273)K = 297.9 K
Find the final Pressure (P₂)
The relation between: Pressure, Volume and Temperature can be gotten from the ideal gas equation :
PV = nRT
The Ideal Gas Equation is also reduced to the General Gas Law or the combined Gas Law by assuming that n= 1 .
From ; PV = nRT
[tex]\frac{PV}{T} = R \ \ ( constant) \ if \ n=1[/tex]
∴ [tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = \frac{P_3V_3}{T_3}...= \frac{P_nV_n}{T_n} \ \ \ ( n \ constant)[/tex]
The additional information required to solve this problem is the initial volume.
This expression is a combination of Boyle's Law and Charles Law. From the combined Gas Law , it can be deduced that at constant volume, the pressure of a given mass(mole) of gas varies directly with absolute temperature.
∴ [tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex] if n & Volume (V) are constant .
[tex]P_2 = \frac{1.06*297.9}{285}[/tex]
P₂ = 1.108 atm
Thus, the final pressure P₂ of the gas is 1.108 atm
g A mixture of gases contains 6.46 g of N2O, 2.74 g of CO, and 5.40 g of O2. If the total pressure of the mixture is 4.33 atm, what is the partial pressure of each component? a) P(N2O) = 0.635 atm, P(CO) = 0.424 atm, and P(O2) = 3.27 atm. b) P(N2O) = 2.31 atm, P(CO) = 0.622 atm, and P(O2) = 1.40 atm. c) P(N2O) = 1.54 atm, P(CO) = 1.02 atm, and P(O2) = 1.77 atm. d) P(N2O) = 0.999 atm, P(CO) = 0.371 atm, and P(O2) = 2.96 atm. e) P(N2O) = 1.28 atm, P(CO) = 1.93 atm, and P(O2) = 1.12 atm.
Answer:
The correct answer is c) P(N2O) = 1.54 atm, P(CO) = 1.02 atm, and P(O2) = 1.77 atm
Explanation:
In order to calculate the partial pressures of the mixture components, we have to first calculate the number of moles:
For N₂O:
Molecular weight (MW): (14 g/mol x 2) + 16 g/mol= 44 g/mol
Number of moles of N₂O (n) = mass/Mw = 6.46g/44 g/mol= 0.1468 mol
For CO:
Molecular weight (MW): 12 g/mol + 16 g/mol= 28 g/mol
Number of moles of CO (n) = mass/Mw = 2.74 g/28 g/mol= 0.0978 mol
For O₂:
Molecular weight (MW): 16 g/mol x 2= 32 g/mol
Number of moles of O₂ (n) = mass/Mw = 5.40 g/32 g/mol= 0.1687 mol
Once calculated the number of moles of each component, we can calculate the total number of moles (nt):
nt = 0.1468 mol + 0.0978 mol + 0.1687 mol = 0.4133 moles
The partial pressure of a gas in a mixture can be calculated from the molar fraction of the gas (X) and the total pressure of the mixture (Pt=4.33 atm):
P(N₂O) = X(N₂O) x Pt
= (moles N₂O/nt) x Pt
= 0.1468 moles/0.4133 moles x 4.33 atm
= 1.538 atm
P(CO) = X(CO) x Pt
= (moles CO/nt) x Pt
= 0.0978 moles/0.4133 moles x 4.33 atm
= 1.0246 atm
P(O₂) = X(O₂) x Pt
= (moles O₂)/nt x Pt
= 0.1687 moles/0.4133 moles x 4.33 atm
= 1.767 atm
Evaluate each scenario described to determine the direction of heat flow.
ice cube to tap water
tap water to ice cube
Answer: tap water to ice cube
Explanation:
Answer: tap water to ice cube
Explanation:
A gas mixture contains CO, Ar and H2. What is the total pressure of the mixture, if
the mole fraction of H2 is 0.35 and the pressure of H2 is 0.58 atm?
Answer:
The total pressure of the mixture is 1.657 atm
Explanation:
Step 1: Data given
Mol fraction of H2 = 0.35
Pressure of H2 = 0.58 atm
Partial pressure gas = total pressure gas * mol fraction gas
Step 2: Calculate the total pressure
Partial pressure H2 = total pressure * mol fraction
0.58 atm = total pressure * 0.35
Total pressure = 0.58 atm / 0.35
Total pressure = 1.657 atm
The total pressure of the mixture is 1.657 atm
Considering the Dalton's partial pressure, the total pressure in the mixture of gases is 1.657 atm.
The pressure exerted by a particular gas in a mixture is known as its partial pressure.
So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:
[tex]P_{T} =P_{1} +P_{2} +... +P_{n}[/tex]
where n is the amount of gases in the mixture.
Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:
[tex]P_{A} =x_{A} P_{T}[/tex]
In this case, the partial pressure of gas H₂ can be expressed as:
[tex]P_{H_{2} } =x_{H_{2} } P_{T}[/tex]
You know:
[tex]P_{H_{2} }[/tex]= 0.58 atm[tex]x_{H_{2} }[/tex]= 0.35Replacing in the definition of partial pressure of gas H₂:
[tex]0.58 atm=0.35P_{T}[/tex]
Solving:
[tex]P_{T}=\frac{0.58 atm}{0.35}[/tex]
[tex]P_{T}[/tex]= 1.657 atm
In summary, the total pressure in the mixture of gases is 1.657 atm.
Learn more:
brainly.com/question/14239096?referrer=searchResults brainly.com/question/25181467?referrer=searchResults brainly.com/question/14119417What is the total amount of kinetic and potential energy of a substance?
Answer:
I THINK mechanical energy
How many milligrams of a 20mg sample of cesium-137 remain after 60 years
Approximately 0.237 milligrams of cesium-137 would remain after 60 years.
Explanation:The amount of a radioactive substance that remains after a certain amount of time can be calculated using the decay constant. For cesium-137, the decay constant is 0.0871 per year. To determine the amount remaining after 60 years, we can use the formula:
Amount remaining = initial amount * e^(-decay constant * time)
Substituting the values, we get:
Amount remaining = 20mg * e^(-0.0871 * 60) = 20mg * e^(-5.226) ≈ 0.237mg. Therefore, approximately 0.237 milligrams of cesium-137 would remain after 60 years.
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