Answer:
The correct answer is (c) 1 × 10⁻⁹ M
Explanation:
Hydroxide ion = OH⁻
Hydrogen ion = H⁺
The autoionization equilibrium of water at 25ºC has a water constant Kw which is expressed as follows:
Kw = [H⁺] x [OH⁻]= 1 x 10⁻¹⁴
If we know the concentration of hydroxide ion ([OH⁻]) we can calculate the hydrogen ion concentration ([H⁺]) as follows:
[H⁺] = Kw/[OH⁻]= (1 x 10⁻¹⁴)/1 x 10⁻⁵ M= 1 x 10⁻⁹ M
The hydrogen ion concentration of a solution with a hydroxide ion concentration of 1 × 10-5 M is 1 × 10-9 M.
Explanation:In chemistry, the product of the concentration of hydrogen ions [H+] and the concentration of hydroxide ions [OH-] in a solution always equals 1 × 10–14 M² at a temperature of 25°C (this is known as the ion product of water).
Since the concentration of hydroxide ions [OH-] in your solution is given as 1 × 10-5 M, you can calculate the concentration of hydrogen ions [H+] by dividing the ion product of water, 1 × 10-14 M², by the concentration of hydroxide ions [OH-]. Therefore, the hydrogen ion concentration [H+] in your solution would be 1 × 10-14 M² / 1 × 10-5 M = 1 × 10-9 M.
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Consider the titration of a 20.0 mL sample of 0.500 M HCN (Ka =6.17x10-10) with 0.250 M KOH. a. (6pt) What is the initial pH? b. (7pt) What is the pH at 6.00 mL base added? c. (8pt) What is the pH when 40.00 mL of KOH is added to reach the equivalence point? d. (8pt) What is the pH if 42.00 mL of KOH is titrated with the solution?
Answer:
a. pH = 4.75
b. pH = 9.20
c. pH = 8.42
d. pH = 13.53
Explanation:
This is a titration between a strong base, the KOH and a weak acid, HCN.
The initial pH is the pH, when you did not add the base yet, so it is the pH of the HCN
HCN + H2O ⇄ H₃O⁺ + CN⁻
Initial 0.5 - -
Eq. 0.5-x x x
Ka = x² / (0.5-x) = 6.17ₓ10⁻¹⁰
Ka is really small, so we can say that 0.5-x = 0.5. Then,
x² = 6.17ₓ10⁻¹⁰ . 0.5
x = √(6.17ₓ10⁻¹⁰ . 0.5) = 1.75×10⁻⁵ → [H₃O⁺]
pH = - log [H₃O⁺] → - log 1.75×10⁻⁵ = 4.75
b. First of all, we determine the moles of base, we are adding.
0.250 mol/L . 0.006 L = 0.0015 moles
In conclussion we have 0.0015 moles of OH⁻
Now, we determine the moles of our acid.
0.500 mol/L . 0.020L = 0.01 moles
The 0.0015 moles of OH⁻ will be neutralized with the acid, so:
HCN + OH⁻ → H₂O + CN⁻
0.01 0.0015 0.0085
The hydroxides are neutralized with the proton from the weak acid, so we have 0.0085 moles of cyanide and 0.0085 moles of HCN. (0.01-0.0015)
Our new volume is 20 mL and 6mL that we added, so, 26mL
This is a buffer with the weak acid, and its conjugate base.
Our concentrations are 0.0085 moles / 0.026 L = 0.327 M
We apply Henderson-Hasselbach
pH = pKa + log (base/acid) → pH = 9.20 + log (0.327/0.327)
pH = pKa
c. When we add 40 mL, our volume is 20mL +40mL = 60 mL
These are the moles, we add:
0.040 L . 0.250 mol/L = 0.01 moles of KOH (moles of OH⁻)
HCN + OH⁻ → H₂O + CN⁻
0.01 0.01 0.01
All the hydroxides have neutralized all the moles from the HCN, so we only have in solution, cyanhide. This is the equivalence point.
0.01 moles / 0.060 L = 0.16 M → [CN⁻]
pH at this point will be
CN⁻ + H₂O ⇄ HCN + OH⁻ Kb = 1.62ₓ10⁻⁵ (Kw/Ka)
In. 0.16 - -
Eq. 0.16-x x x
Kb = x² / (0.16-x)
We can also assume that 0.16-x = 0.16. Then:
[OH⁻] = √(Kb . 0.16) → √(1.62ₓ10⁻⁵ . 0.16) = 2.59×10⁻⁶
- log [OH⁻] = pOH → - log 2.59×10⁻⁶ = 5.58
pH = 14 - pOH → 14 - 5.58 = 8.42
This is a basic pH, because the titration is between a weak acid and a strong base.
d. When we add 42 mL of base, our volume is 20mL + 42 mL = 62 mL
We add 0.5 mol/L . 0.062L = 0.031 moles
These are the moles of OH⁻ , so as we have neutralized all the acid with 40 mL, with 42 mL of base, we only have base in solution.
0.031 moles - 0.01 moles = 0.021 moles of OH⁻
[OH⁻] = 0.021 moles / 0.062L = 0.34M
- log [OH⁻] = pOH → - log 0.34 = 0.47
pH = 14-pH → 14 - 0.47 = 13.53
This answer details the titration of HCN with KOH, explaining ways to calculate the pH values at different stages of the titration: the initial state, partway through, at the equivalence point when all acid has been neutralized, and after the equivalence point.
Explanation:We are considering the titration of a solution of HCN with KOH. First, we calculate the initial pH using the formula pH=-log10[H+], where the H+ concentration can be determined from the Ka expression for HCN.
For question b, when 6.00 mL base KOH is added, we need to determine which ion is in excess, and then use its concentration to calculate the pH. In this case, there are still excess HCN ions leading to a buffer solution. The Henderson-Hasselbalch equation can be used to determine the pH.
For question c, at the equivalence point, 40.00 mL of base KOH has been added. All HCN has been titrated to form CN-, so the pH is determined by the OH- ions hydrolyzed from CN-. We use the Kb of CN- (which can be calculated using Kw/Ka) and the total volume to calculate [OH-], and then pH = 14 - pOH.
Finally, for question d, after 42.00 mL of KOH, OH- is in excess. We can calculate the [OH-] and then find the pH using pH = 14 - pOH.
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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:
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
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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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.
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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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.
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
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:
Name the organic compound CH4
Answer:
Methane
Explanation:
Methane is a potent greenhouse gas with the formula CH₄. Hope this helps!
A tank contains helium gas at 191 mm HG nitrogen gas at 0.261 ATM and neon at 522 torr what is the total pressure in MM Hg
Answer:
911.36mmhg
Explanation:
1 torr is almost equivalent to 1mmhg.but to convert from atm to mmhg,multiply by 760
Final answer:
The total pressure in the tank, after converting all pressures to the same unit (mm Hg), is the sum of the partial pressures of helium, nitrogen, and neon, resulting in a value of 911.36 mm Hg.
Explanation:
To calculate the total pressure in a mixture of gases, you can use Dalton's Law of partial pressures which states that the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.
Since the pressure values are given in different units, we first need to convert them all to millimeters of mercury (mm Hg). The pressure of helium gas is already given in mm Hg. The nitrogen gas pressure needs to be converted from atmospheres (ATM) to mm Hg by multiplying by 760 mm Hg/ATM, and the neon gas pressure in torr simply needs to be directly converted to mm Hg because 1 torr equals 1 mm Hg.
Now we can sum these values to find the total pressure in the tank.
The conversion and summation are as follows:
Helium: 191 mm Hg
Nitrogen: 0.261 ATM x 760 mm Hg/ATM = 198.36 mm Hg
Neon: 522 torr = 522 mm Hg
Adding these together, the total pressure in the tank is:
191 mm Hg + 198.36 mm Hg + 522 mm Hg = 911.36 mm Hg
Therefore, the total pressure in the tank in mm Hg is 911.36 mm Hg.
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.
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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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.
At constant pressure, 50.0 mL of 0.100 M KOH and 45.0 mL of 0.100 M HNO3 are mixed in a styrofoam cup.The two solutions are initially at 25.00 °C and the final temperature is 25.65 °C. Calculate the ΔHneutralization (kJ/mol) in term of moles of HNO3 since it’s the limiting reagent. Assume a specific heat capacity of 4.184 J/(°C•g) and that the density of the solution is the same as water (0.997 g/mL).
Answer:
≈ -57.2 kJ/mol
Explanation:
Total volume of the solution = (50.0 + 45.0) mL = 95.0 mL.
Density of the solution = 0.997 g/mL.
Mass of the solution = (volume of the solution)*(density of the solution)
= (95.0 mL)*(0.997 g/mL)
= 94.715 g
Heat gained by the solution, qsoln = (mass of the solution)*(specific heat capacity)*(change in temperature)
= (94.715 g)*(4.184 J/ºC.g)*(25.65 – 25.00)ºC
= 257.5869 JAccording to the principle of thermochemistry,
qsoln + qrxn = 0
where qrxn denotes the heat change during the neutralization reaction.
Therefore,
(257.5869 J) + qrxn = 0
======> qrxn = -257.5869 J
HNO3 is the limiting reactant.
Moles of limiting reactant = (volume of the limiting reactant in L)*(molarity of limiting reactant)
= (45.0 mL)*(0.100 M)
= (45.0 mL)*(1 L)/(1000 mL)*(0.100 M)
= 0.0045 mole.
ΔHneutralization = qrxn/(moles of HNO3)
= (-257.5869 J)/(0.0045 mol)
= -57241.5 J/mol
= (-57241.5 J/mol)*(1 kJ)/(1000 J)
= -57.2415 kJ/mol
≈ -57.2 kJ/mol
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
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.
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brainly.com/question/14239096?referrer=searchResults brainly.com/question/25181467?referrer=searchResults brainly.com/question/14119417Which of the following is used to represent each element?
formula
equation
symbol
O
coefficient
4.
What volume of 0.120 M HNO3(aq) is needed to
completely neutralize 150.0 milliliters of 0.100 M
NaOH(aq)?
A. 62.5 mL
B. 125 ml
C.
180. mL
D. 360. mL
Answer:
B) 125 mL
Explanation:
M1V1=M2V2
(0.120M)(x)=(150.0 mL)(0.100M)
x= 125 mL
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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]
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 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
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.
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
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.
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.
Given a balanced chemical equation below: 3Cu(s) + 2H3PO4 --- > Cu3(PO4)2 + 3H2 How many moles of copper are needed to react with 5 moles of phosphoric acid?
Answer:
7.5 moles
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
3Cu + 2H3PO4 —> Cu3(PO4)2 + 3H2
From the balanced equation above,
3 moles of Cu reacted with 2 moles of H3PO4.
Therefore, Xmol of Cu will react with 5 moles of H3PO4 i.e
Xmol of Cu = (3 x 5)/2
Xmol of Cu = 7.5 moles
Therefore, 7.5 moles of Cu are needed to react with 5 moles of H3PO4.
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!
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
learn more about conversion below,
https://brainly.com/question/24513742
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