Answer:
[H+] = 0.0562 M
Explanation:
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H⁺ concentration in a solution with a pH of 1.25 is 0.0562.
What is pH?pH may be defined as a measure of acidity of a substance.
More precisely, pH is defined as the negative logarithm of the hydrogen ion concentration.
pH = - log [H⁺]
The range of pH extends from zero to 14.
A pH value of 7 is neutral, because pure water has a pH value of exactly 7. Values lower than 7 are acidic; values greater than 7 are basic or alkaline.
pH measurements can be made using litmus paper or pH paper known to change colors around a certain pH value.
A universal indicator is a mixture of indicator solutions intended to provide a color change over a pH range of 2 to 10.
pH measurements can also be done with pH meters that give the exact value of pH.
Given,
pH = 1.25
We know that, pH = - log [H⁺]
[H⁺] = antilog [ -pH]
[H⁺] = antilog [ -1.25]
= 0.0562 M
Therefore, H⁺ concentration in a solution with a pH of 1.25 is 0.0562.
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What information about water is needed to calculate the enthalpy change for converting 1 mol H2O(g) at 100 °C to H2O(l) at 80 °C? (a) Heat of fusion, (b) heat of vaporization, (c) heat of vaporization and specific heat of H2O(g), (d) heat of vaporization and specific heat of H2O(l), (e) heat of fusion and specific heat of H2O(l).
The information needed includes the heat of vaporization and specific heat of H₂O(l).
Explanation:The information needed to calculate the enthalpy change for converting 1 mol H₂O(g) at 100 °C to H₂O(l) at 80 °C is the (d) heat of vaporization and specific heat of H₂O(l).
The heat of vaporization is required because it accounts for the energy needed to convert water from a liquid to vapor state. The specific heat of H₂O(l) is needed to determine the amount of heat required to change the temperature of liquid water from 100 °C to 80 °C.
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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Which of the following is used to represent each element?
formula
equation
symbol
O
coefficient
A 475 ml sample of a gas was collected at room temperature of 23.5 °C and a pressure of
756 mm Hg. Calculate the volume of the gas if the conditions were altered to 10.0 °C and a
pressure of 722 mm Hg.
Answer:
The volume when the conditions were altered is 0.5109 L or 510.9 mL
Explanation:
Using the general gas equation,
P1 V1 / T1 = P2 V2 / T2
where;
P1 = 756 mmHg
V1 = 475 ml = 0.475 L
T1 = 23.5°C = 23.5 + 273K = 275.5 K
P2 = 722 mm Hg
T2 = 10°C = 10 + 273 K = 283 K
V2 = ?
Rearranging to make V2 the subject of the formula, we obtain:
V2 = P1 V1 T2 / P2 T1
V2 = 756 * 0.475 * 283 / 722 * 275.5
V2 = 101, 625.3 / 198911
V2 = 0.5109 L or 510.9 mL
What volume would 32.0 g of NO2 gas occupy at 3.12 atm and 18.0 Celsius
Answer:
Volume of 32.0 g [tex]NO_{2}[/tex] at 3.12 atm and 18.0 [tex]^{0}\textrm{C}[/tex] is 5.33 L.
Explanation:
Molar mass of [tex]NO_{2}[/tex] = 46.0055 g/mol
So, 32.0 g of [tex]NO_{2}[/tex] = [tex]\frac{32.0}{46.0055}[/tex] mol of [tex]NO_{2}[/tex] = 0.696 mol of [tex]NO_{2}[/tex]
Let's assume [tex]NO_{2}[/tex] behaves ideally.
Then [tex]P_{NO_{2}}V_{NO_{2}}=n_{NO_{2}}RT[/tex] , where P, V, n, R and T represents pressure, volume, number of moles, gas constant and temperature (in kelvin scale) respectively.
Here [tex]P_{NO_{2}}[/tex] = 3.12 atm, [tex]n_{NO_{2}}[/tex] = 0.696 mol, R = 0.0821 L.atm/(mol.K) and T = (273+18.0)K = 291 K
So, [tex]V_{NO_{2}}=\frac{n_{NO_{2}}RT}{P_{NO_{2}}}[/tex]
or, [tex]V_{NO_{2}}=\frac{(0.696mol)\times (0.0821\frac{L.atm}{mol.K})\times (291K)}{3.12atm}[/tex]
or, [tex]V_{NO_{2}}=5.33L[/tex]
So, volume of 32.0 g [tex]NO_{2}[/tex] at 3.12 atm and 18.0 [tex]^{0}\textrm{C}[/tex] is 5.33 L.
The Volume of NO₂ is 542.8L
What is Ideal Gas Law ?Ideal Gas Law relates the macroscopic property of gas , It states that the product of pressure and volume of 1 mole of a gas is equal to the product of absolute temperature and universal gas constant.
For n mole of gas,
PV = nRT
In the question it is given that mass of NO₂ is 32 g
P = 3.12 atm
T = 18°Celsius = 291 K
Volume = ?
n = m/M = 32/46 = 0.7
3.12 * V = 0.7*0.0821*291
V = 5.36 L
Therefore the Volume of NO₂ is 5.36L
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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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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
Determine the number of bonding electrons and the number of nonbonding electrons in the structure of SI2. Enter the number of bonding electrons followed by the number of nonbonding electrons, separated by a comma, in the dot structure of this molecule.
The molecule SI2 does not exist but if the student meant SiI2, it would have 8 bonding electrons and 14 nonbonding electrons.
Explanation:The molecule SI2 does not exist. It may have been a typo. If the student meant Si2, this molecule also doesn't exist because silicon, a member of group 14 on the periodic table, needs 4 electrons to achieve a full outer shell, and thus generally forms 4 bonds. If we are talking about SiI2, or silicon diiodide, it would have 8 bonding electrons and 14 nonbonding electrons. This is because all four of the bonding electrons from silicon and four from iodine atoms make up the bonding electrons while the remaining 7 electrons from each iodine atom make up the nonbonding electrons.
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A student adds solid KCl to water in a flask. The flask is sealed with a stopper and thoroughly shaken until no more solid KCl dissolves. Some solid KCl is still visible in the flask. The solution in the flask is A) saturated and is at equilibrium with the solid KCl B) saturated and is not at equilibrium with the solid KCl C) unsaturated and is at equilibrium with the solid KCl D) unsaturated and is not at equilibrium with the solid KCl
Answer:
Option (A) saturated and is at equilibrium with the solid KCl
Explanation:
A saturated solution is a solution which can not dissolve more solute in the solution.
From the question given above, we can see that the solution is saturated as it can not further dissolve any more KCl as some KCl is still visible in the flask.
Equilibrium is attained in a chemical reaction when there is no observable change in the reaction system with time. Now, observing the question given we can see that there is no change in flask as some KCl is still visible even after thorough shaking. This simply implies that the solution is in equilibrium with the KCl solid as no further dissolution occurs.
9) 20 g of calcium chloride per 100 g of water at 90°C will be...
10 points
Сас,
KNO,
K₂ Cigar
Solubility (g of salt in 100 g H,0)
Pb(NO3)2
KCR
20
A
KCIO,
0
10
Ce (80)
20 30 40 50 60 70 80 90 100
Temperature (°C)
O
saturated
O
unsaturated
O
supersaturated
According to solubility, 20 g of calcium chloride per 100 g of water at 90°C will be saturated solution.
What is solubility?Solubility is defined as the ability of a substance which is basically solute to form a solution with another substance. There is an extent to which a substance is soluble in a particular solvent. This is generally measured as the concentration of a solute present in a saturated solution.
The solubility mainly depends on the composition of solute and solvent ,its pH and presence of other dissolved substance in it . It is also dependent on temperature and pressure which is maintained through the process.Concept of solubility is not valid for chemical reactions which are irreversible. The dependency of solubility on various factors is due to interactions between the particles, molecule or ions.
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2. Carbon monoxide is a poisonous carbon oxygen compound (as opposed to carbon
dioxide). Provide the complete MO diagram including the atomic orbitals for both
carbon and oxygen as well as the MOs for CO assuming the bonding is the same as for
C2 in the reference. Also indicate the bond order for CO from the MO diagram. (14
points)
Answer:
Carbon monoxide has a bond order of 3
Explanation:
A molecular orbital diagram ( MO diagram ), is a qualitative descriptive tool tha explains chemical bonding in molecules in terms of the molecular orbital theory in general and particularly adopts the linear combination of atomic orbitals (LCAO) method in its approach. A fundamental principle of these theories is that as atoms bond to each other to form molecules, a certain number of atomic orbitals combine to form an equal number of molecular orbitals , although the electrons involved may be redistributed among the orbitals. This tool is very well suited for simple diatomic molecules such as hydrogen molecule , oxygen molecule, and carbon monoxide but becomes more complex when discussing even comparatively simple polyatomic molecules, such as methane. MO diagrams usually explains why some molecules exist and others do not exist. It is used to predict bond strength of molecules, as well as the electronic transitions that can take place.
The MO diagram for carbon monoxide is shown in the image attached. It can clearly be seen that carbon monoxide has a bond order of three. The combination of carbon and oxygen atoms to form molecular orbitals of appropriate energy is also seen in the image.
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!
How do neutrons stars generate the energy they need to survive
The neutron stars are the smallest but heaviest density supergiant star. The surfaces of the neutron stars undergo thermonuclear fusion to produce energy to survive.
What is thermonuclear fusion?Thermonuclear fusion or just simply nuclear fusion is a nuclear reaction that involves the combining (fusion) of the atomic nuclei together to produce different nuclei in the presence of high temperatures.
Unlike nuclear fission, it needs high temperature and energy to produce atomic nuclei. It is a process generally seen in the universe including the Sun, stars, galaxies, etc.
Neutron stars also use their energy to undergo thermonuclear fusion and produce heavy nuclei from lighter nuclei. The hydrogen atom combines fuses to form helium atoms resulting in high energy production.
Therefore, neutron stars generate energy by thermonuclear fusion.
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What happens if you break a magnet in half?
a) Each half will be a new magnet, with both a north and south pole.
b) One half will have a north pole only and one half will have a south pole only.
c) Neither half will have a pole.
d) Neither half will be able to attract or repel
Answer:
neither half will have a pole
Which pair of elements can be used to determine the age of a fossil that is over one billion years old?
Answer:
Potassium and Argon (⁴⁰K-⁴⁰Ar)
Explanation:
Potassium-Argon dating is a method used in geology and archeology to date rocks or volcanic ash, generally the oldest.
It is based on the principle of radioactive decay, it is a process that has a half-life of 1248 million years, during which time the gas is concentrated in the rock crystals. Taking advantage of this known rhythm and half-life, the method lends itself to dating samples ranging from 100.000 years to several billion years.
Answer:
Its D to sum it up.
Explanation:
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
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 the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 ⋅ 10-8. What is the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 x 10-8.
a. 7.7
b. 0
c. 7.14
d. 14.28
e. 9.05
f. 6.86
Answer:
c. 7.14
Explanation:
The buffer solution is formed by a weak acid ( hypochlorous acid, HClO) and its conjugate base (hypochlorite ClO⁻, coming from sodium hypochlorite NaClO). We can calculate the pH using the Henderson-Hasselbach equation.
pH = pKa + log [base]/[acid]
pH = -log 3.8 × 10⁻⁸ + log 0.111/0.211
pH = 7.14
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
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.
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.
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.
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.
Calculate the rate of emission of SO2 in g/s that results in a centerline (y = 0) concentration at ground level of 1.786 x 10-3 g/m3 one kilometer from the stack. The time of measurements was 1 P.M. on a clear summer afternoon. The wind speed was 2.0 m/s measured at a height of 10 m. The effective stack height is 96 m. No inversion is present.
Answer:
790 g/s
Explanation:
The first thing to do is to make sure to calculate the standard deviation of the y-axis and the z-axis and the values for both axis are; 215 m and 450 m for y-axis and z-axis respectively.
The next thing to do now is to find the rate of emission of SO2 in units of g/s using the relation by using the formula below;
Emission rate of pollutant, E = (x × π × by × bz × g) ÷ [ e^(-1/2) × (y/by)^2] × [e^(-1/2)× (h/bz)^2].
Where g = wind direction, y and h are the distance in metres.
Therefore, slotting in the values into the Emission rate of pollutant equation above, we have;
Emission rate of pollutant,E =[ 1.412 × 10^-3] × π × 215 × 450 × 1.8] ÷ [e^(-1/2)(0/215)^2] × e^ (-1/2)(94/450)^2.
Emission rate of pollutant,E = 790 g/s.
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.
Some people must eat a low-sodium diet with no more than 2,000 mg of sodium per day. By eating 1 cracker, 1 pretzel, and 1 cookie, a person would ingest 149 mg of sodium. If a person ate 8 pretzels and 8 cookies, he or she would ingest 936 mg of sodium. By eating 6 crackers and 7 pretzels, the person would take in 535 mg of sodium. Find the amount of sodium in each.
Answer:
The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
Explanation:
Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.
So, the following three equations can be written as per given information:
x+y+z = 149 ........(1)
8y+8z = 936 ........(2)
6x+7y = 535 .........(3)
From equation- (2), y+z = [tex]\frac{936}{8}[/tex] = 117
By substituting the value of (y+z) in equation- (1) we get,
x = 149-(y+z) = 149-117 = 32
By substituting the value of x into equation- (3) we get,
y = [tex]\frac{535-(6\times 32)}{7}[/tex] = 49
By substituting the value of y into equation- (2) we get,
z = (117-49) = 68
So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
Final answer:
To find the amount of sodium in each item, set up a system of equations and solve to get x = 9, y = 55, and z = 85.
Explanation:
To find the amount of sodium in each item, we can set up a system of equations.
Let x represent the amount of sodium in 1 cracker, y represent the amount of sodium in 1 pretzel, and z represent the amount of sodium in 1 cookie.
From the given information, we can create the following equations:
x + y + z = 1498y + 8z = 9366x + 7y = 535Solving this system of equations, we can find the values of x, y, and z. The amount of sodium in 1 cracker is 9 mg, 1 pretzel is 55 mg, and 1 cookie is 85 mg.
19.24 The mechanism for acetal hydrolysis has been heavily investigated. In one study, which explored rates as well as stereochemical aspects,[5] compound 1 was treated with aqueous acid to afford compound 2. Draw the structure of 2, clearly showing the configuration of any chiral center(s). c19s124
Answer:
Explanation:
check the attached file below for the diagram and better expalnation.
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.
Learn more about Titration here:https://brainly.com/question/31271061
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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]