Answer:
The volume you need to transfer from the stock solution is 0.145 l
Explanation:
Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:
number of moles in volume to transfer = number of moles in the final solution
Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:
Ci * Vi = Cf * Vf
where:
Ci = concentration of the stock solution.
Vi = volume of the stock solution to be transferred.
Cf = concentration of the final solution
Vf = volume of the final solution
Then, replacing with the data:
518 mM * Vi = 16.7 mM * 4.5 l
Vi = 16.7 mM * 4.5 l / 518 mM
Vi = 0.145 l or 145 ml
Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.
How many gram-moles of formic acid, CH2O2, are in 225 g of the compound? Numbe g-mol CH,0, How many pound-moles of formic acid are in 225 g of the compound? Number Ib-moles CH,O,
Explanation:
(a) Mass given = 225 g
Molecular mass of [tex]CH_{2}O_{2}[/tex] = [tex]12 + 2 + 16 \times 2[/tex] = 12 + 2 + 32 = 46 g/mol
As, it is known that number of moles is equal to the mass divided by molar mass.
Mathematically, No. of moles = [tex]\frac{mass}{\text{molecular mass}}[/tex]
Hence, calculate the number of moles of formic acid as follows.
No. of moles = [tex]\frac{225 g}{46 g/mol}[/tex]
= 4.89 mol or g mol
Hence, in 225 g of the compound there are 4.89 mol or g mol.
(b) It is known that 1 g = 0.0022 pound
So, pounds present in 4.89 g will be calculated as follows.
4.89 \times 0.0022 pound
= 0.0107 pound mole
Hence, in 225 g of the compound there are 0.0107 pound mole.
At 0.70 atmosphere pressure, a certain mass of a gas has a volume of 3.8 L. What will be the volume if the pressure is decreased to 0.409 atmosphere pressure?
Answer: The new volume of the gas will be 6.50 L
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.
The equation given by this law is:
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.
We are given:
[tex]P_1=0.70atm\\V_1=3.8L\\P_2=0.409atm\\V_2=?L[/tex]
Putting values in above equation, we get:
[tex]0.70atm\times 3.8L=0.409atm\times V_2\\\\V_2=6.50L[/tex]
Hence, the new volume of the gas will be 6.50 L
A woman informs her engineer husband that "hot water will freeze faster than cold water." He calls this statement nonsense but he does so in a polite sensitive manner. She answers by saying that she has actually timed the freezing process for ice trays in the home refrigerator and found that hot water does indeed freeze faster. As a friend, you are asked to settle the argument and preserve family harmony. Is there any logical explanation for the woman's observation
Answer:
Explanation:
It could be as a result of evaporation from the hot water causing it to have less mass than usual. Evaporation will allow the volume of the hot water to decrease and by so doing will cool faster than cold water which is at a higher mass. This is how I will explain my observations to them.
Final answer:
The concept of hot water freezing faster than cold water is known as the Mpemba effect and is influenced by various factors including evaporation and thermal energy transfer. Specific heat capacity plays a role in cooling processes, affecting the rate at which materials like metal are cooled when in contact with ice or water.
Explanation:
The observation that hot water freezes faster than cold water under certain conditions has been noted historically and is often referred to as the Mpemba effect. While it may seem counterintuitive, various experiments have shown that hot water can indeed freeze more quickly than cold water due to a number of factors, including evaporation (which reduces the mass of hot water), thermodynamic properties of water (such as specific heat capacity and the behavior of water molecules at different temperatures), and the environment in which the water is being frozen (like the surrounding air temperature and properties of the freezing surface).
Regarding the specific scenarios described, mixing hot and cold water or the implications of adding cold water to a hot liquid like tea or coffee, the result is affected by the thermal energy transfer between the substances. The outcome is influenced by the specific heat of the materials, the current temperatures, and the mass of the substances involved. For instance, pouring hot water into cold water would result in heat transfer from the hot water to both the cold water and its surroundings.
As for the question of cooling a hot piece of metal, the method of cooling (using ice versus cold water) will affect the rate at which the metal cools down, with the specific heat capacities playing a crucial role. Water generally has a higher specific heat capacity than ice, making it more efficient at absorbing heat.
One kilogram of saturated steam at 373 K and 1.01325 bar is contained in a rigid walled vessel. It has a volume of 1.673 m3. It is cooled to a temperature at which the specific volume of water vapour is 1.789 m. The amount of water vapour condensed in kilograms is (a) 0.0 (b) 0.065 (c) 0.1 (d) 1.0
Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
Initial volume [tex](v_{1})[/tex] = 1.673 [tex]m^{3}[/tex]
Final volume [tex](v_{2})[/tex] = 1.789 [tex]m^{3}[/tex]
As, the amount of water vapor condensed will be as follows.
[tex]\frac{(v_{2} - v_{1})}{v_{2}}[/tex]
= [tex]\frac{(1.789 m^{3} - 1.673 m^{3})}{1.789 m^{3}}[/tex]
= [tex]\frac{0.116 m^{3}}{1.789 m^{3}}[/tex]
= 0.065 kg
Hence, we can conclude that the amount of water vapour condensed in kilograms is 0.065 kg.
Two pipes with 1 meter diameters join to become one pipe with 2 meter diameter. Water was flowing in first pipe at 500 kg/s and oil was flowing in the second pipe with average velocity of 0.8 m/s. Oil mass density is 1500 kg/s. What will be the mass flow rate at the exit of the 2 meter diameter pipe?
Answer:
The mass flow rate at the exit of the 2 meter diameter pipe is 1442 kg/s
Explanation:
If the oil is flowing in the 1m diameter pipe at 0.8 m/s, we can calculate the flow as
[tex]Q=\bar{v}*A=\bar{v}*(\pi/4*D^{2} )=0.8m/s*0.785m^{2} =0.628m^{3}/s[/tex]
The mass flow is
[tex]M=\rho*Q=1500 kg/m3*0.628m3/s=942kg/s[/tex]
The mass flow rate at the exit of the 2 meter diameter pipe is
[tex]M=M_w+M_o=500 kg/s+942kg/s=1442kg/s[/tex]
When preparing 1-bromobutane from 1-butanol and HBr, why
mustit be washed with H2O, cold NaOH and NaCl?
Answer:
Answer has been given below
Explanation:
During reaction of 1-butanol and HBr, some unreacted amount of HBr might present in reaction mixture. Hence cold NaOH is used to neutralize unreacted HBr.Washing with [tex]H_{2}O[/tex] is done to remove the salt produced during neutralization i.e. NaBr from organic layer of 1-bromobutane.NaCl is added to remove water from organic layer. Because NaCl is more soluble in water than organic layer.Khoisan salts is the number 1 producer of salts in SA for both local and international markets. 500 kg of KCl is dissolved in sufficient water to make a saturated solution at 370 K. At 370 Kthe solubility of KCl is 42 mass %. The solution is cooled to 320 K and the solubility is 31,5 mass % It is assumed that no water is evaporated. 2.1. Determine the amount of water is added to the 500 kg of KCl to produce the required saturated solution at 370 K. (3) 2.2. Determine the mass of KCl crystals formed after the cooling process to a temperature of 320 K. (Use the formula method)
Answer:
690 kg of water.125 kg of KCl crystals are formed after cooling processExplanation:
1) A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent. The solubility of KCl at 370K is 42%mass -42kg of KCl in 100kg of water+KCl , thus, the amount of water added to 500kg of KCl is:
500 kg of KCl × [tex]\frac{100 kg Water+KCl}{42kgKCl}[/tex] = 1190 kg of water +KCl
1190 kg of water +KCl - 500 kg of KCl = 690 kg of water
2) The maximum amount that this solution could solubilize of KCl at 320K is:
1190 kg of water + KCl × [tex]\frac{31,5 kgKCl}{100kgWater+KCl}[/tex] = 375 kg
Thus, the mass of KCl crystals formed are:
500 kg of KCl - 375 Kg of KCl = 125 kg of KCl
I hope it helps!
Calculate the answer to the following expression to the correct number of significant figures.
4 X 10^8/2.0 X 10^-5
Answer : The answer will be [tex]\Rightarrow 2\times 10^{13}[/tex]
Explanation :
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
The rule apply for the multiplication and division is :
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
As we are given the expression :
[tex]\frac{4\times 10^8}{2.0\times 10^{-5}}[/tex]
[tex]\Rightarrow 2\times 10^{13}[/tex]
In the given expression, [tex]4\times 10^8[/tex] has 1 significant figure and [tex]2.0\times 10^{-5}[/tex] has 2 significant figures. From this we conclude that 1 is the least significant figures in this problem. So, the answer should be in 1 significant figures.
Thus, the answer will be [tex]\Rightarrow 2\times 10^{13}[/tex]
How are hydrogen bonds formed between water molecules? Use a drawing to illustrate.
Answer:
They hydrogen bonds exist at the covalents molecules, as water where the 0 is an atom electronegative. The hydrogen bonds are formed because the hydrogen gives its electron to oxygen (d-), that's why we say, there is a dipole at the molecule. As the dipole has been formed, the hydrogen who gave the electron to oxygen will try to get another electro from other molecules. That is how the hydrogen bonds are formed.
Explanation:
Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphere does the volume of a shell equal that of the interior sphere? Assume the shell thickness to be t = 1 nm.
Answer:
Radius of the interior sphere = 3.847 nm
Explanation:
The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:
[tex]V_s=V_o-V_i=V_i\\V_o=2*V_i\\[/tex]
If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):
[tex]r_o=r_i+e[/tex]
The first equation becomes
[tex]\frac{4 \pi}{3}*r_o^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\\frac{4 \pi}{3}*(r_i+e)^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\(r_i+e)^{3} = 2 *r_i^{3} \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0[/tex]
To find ri that satisfies this equation we have to find the roots of the polynomial.
Numerically, it could be calculated that ri=3.847 nm satisfies the equation.
So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.
267 moles of NADH are produced by the Citric Acid Cycle (CAC). How many moles of glucose needed to be broken down during glycolysis in order to produce enough pyruvate for the CAC to produce this many moles of NADH?
Answer:
45 moles
Explanation:
From glycolysis, 1 mole of glucose gives 2 moles of pyruvate which undergoes citric acid cycle.
1 mole of pyruvate undergoes citric acid cycle (After conversion to acetyl-CoA) gives 3 moles of NADH.
Also,
2 moles of pyruvate undergoes citric acid cycle (After conversion to acetyl-CoA) gives 6 moles of NADH.
Thus,
6 moles of NADH are produced from 2 moles of pyruvate or 1 mole of glucose.
1 mole of NADH is produced from 1/6 mole of glucose
267 moles of NADH are produced from [tex]\frac {1}{6}\times 267[/tex] moles of glucose.
Thus, moles of glucose needed to be broken ≅ 45 moles
Calculate the concentration of A bottle of wine contains 12.9% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm ethanol in wine in terms of mass percent and molality Mass percent Molality =
Answer:
The mass percentage of the solution is 10.46%.
The molality of the solution is 2.5403 mol/kg.
Explanation:
A bottle of wine contains 12.9% ethanol by volume.
This means that in 100 mL of solution 12.9 L of alcohol is present.
Volume of alcohol = v = 12.9 L
Mass of the ethanol = m
Density of the ethanol ,d= [tex]0.789 g/cm^3=0.789 g/mL[/tex]
[tex]1 cm^3=1 mL[/tex]
[tex]m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g[/tex]
Mass of water = M
Volume of water ,V= 100 mL - 12.9 mL = 87.1 mL
Density of water = D=1.00 g/mL
[tex]M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g[/tex]
Mass percent
[tex](w/w)\%=\frac{m}{m+M}\times 100[/tex]
[tex]\frac{10.1781 g}{10.1781 g+87.1 g}\times 100=10.46\%[/tex]
Molality :
[tex]m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}[/tex]
M = 87.1 g = 0.0871 kg (1 kg =1000 g)
[tex]=\frac{10.1781 g}{46 g/mol\times 0.0871 kg}[/tex]
[tex]m=2.5403 mol/kg[/tex]
To calculate the concentration of ethanol in wine, multiply the volume percent of ethanol by the density of ethanol. Then divide by the mass of the solution and multiply by 100 to get the mass percent.
Explanation:To calculate the concentration of ethanol in wine, we can use the mass percent formula. Mass percent is calculated by dividing the mass of the solute (ethanol) by the mass of the solution (wine), and multiplying by 100. The mass of ethanol can be found by multiplying the volume percent of ethanol (12.9%) by the density of ethanol (0.789 g/cm³). The density of wine is typically close to 1 g/cm³. So, to find the concentration of ethanol in terms of mass percent, we can follow these steps:
Assuming we have 100 mL of wine, we can calculate the volume of ethanol in the wine by multiplying 100 mL by 12.9% (0.129).Next, we can convert the volume of ethanol to mass by multiplying it by the density of ethanol. This gives us 0.129 mL * 0.789 g/cm³ = 0.101961 g.Since the total mass of the wine is equal to the mass of ethanol (since ethanol is the only solute), the concentration of ethanol in terms of mass percent is 0.101961 g / 100 mL * 100% = 0.101961%.Learn more about Concentration of Ethanol in Wine here:https://brainly.com/question/14641045
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An astronaut uses a laboratory balance and weighs an object on earth and again on the moon Which statement below about the weight and mass of the object is true? A) The mass and weight will be identical on the earth and the moon. B) The mass will be the same on earth and moon but the weight will be less on the moon C) Both the mass and weight will be different on earth and moon. D) The weight will be the same on earth and moon but the mass will be less on the moon.
Answer:
B) The mass will be the same on earth and moon but the weight will be less on the moon
Explanation:
Mass of a substance is a intrinsic property. It is same throughout the universe. On the other hand,
Weight = Mass × Gravitational acceleration
Weight depends on the gravitational acceleration of the planet on which the substance is present.
Thus, the astronaut weight on Moon will be approximately six times less than the weight on Earth as gravitational acceleration on Moon is six times less than the gravitational acceleration on Earth the but number of the atoms in body of the astronaut has not changed and thus, mass is same same at two places.
The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi
The factor 0.01 corresponds to the prefix D. 'centi'.
In the metric system, prefixes are used to denote different multiples or fractions of a base unit. Here's a breakdown of the prefixes relevant to this question:
milli-: This prefix represents the factor $10⁻³ $ or 0.001.
deci-: This prefix represents the factor $10⁻¹ $ or 0.1.
deka-: This prefix represents the factor $10¹ or 10.
centi-: This prefix represents the factor $10⁻² $ or 0.01.
A water main made out of galvanized iron has an inner pipe diameter of 0.1 meters, an inlet pressure of 1000 kPa, outlet pressure of 500 kPa, water flow rate of 5 m3/min, and 10 90 degree elbows. What length galvanized iron line should be used to reach the desired outlet pressure?
Explanation:
The given data is as follows.
Diameter = 0.1 m, [tex]P_{1}[/tex] = 1000 kPa
[tex]P_{2}[/tex] = 500 kPa
Change in pressure [tex]\Delta P[/tex] = 1000 kPa - 500 kPa = 500 kPa
Since, 1000 Pa = 1 kPa. So, 500 kPa will be equal to [tex]500 \times 10^{3}[/tex].
Q = 5 [tex]m^{3}/min[/tex] = [tex]\frac{5}{60} m^{3}/sec[/tex] = 0.0833 [tex]m^{3}/sec[/tex]
It is known that Q = [tex]A \times V[/tex]
where, A = cross sectional area
V = speed of the fluid in that section
Hence, calculate V as follows.
V = [tex]\frac{Q}{A}[/tex]
= [tex]\frac{Q \times 4}{\pi \times d^{2}}[/tex]
= [tex]\frac{0.0833 \times 4}{3.14 \times (0.1)^{2}}[/tex]
= 10.61 m/sec
Also it is known that Reynold's number is as follows.
Re = [tex]\frac{\rho \times V \times d}{\mu}[/tex]
= [tex]\frac{1000 \times 10.61 \times 0.1}{10^{-3}}[/tex]
= 1061032.954
As, it is given that the flow is turbulent so we cannot use the Hagen-Poiseuille equation as follows. Therefore, by using Blasius equation for turbulent flow as follows.
[tex]\Delta P = \frac{0.241 \times \rho^{0.75} \times \mu^{0.25} \times L}{D^{4.75}} \times Q^{1.75}[/tex]
[tex]500 \times 10^{3} = \frac{0.241 \times (1000)^{0.75} \times (10^{-3})^{0.25} \times L}{(0.1)^{4.75}} \times (0.0833)^{1.75}[/tex]
[tex]500 \times 10^{3} = 5535.36 \times L[/tex]
L = [tex]\frac{500 \times 10^{3}}{5535.36}[/tex]
= 90.328 m
Thus, we can conclude that 90.328 m length galvanized iron line should be used to reach the desired outlet pressure.
Two moles of a monatomic ideal gas are contained at a pressure of 1 atm and a temperature of 300 K; 34,166 J of heat are transferred to the gas, as a result of which the gas expands and does 1216 J of work against its surroundings. The process is reversible. Calculate the final temperature of the gas.
Answer:
Final temperature is 302 K
Explanation:
You can now initial volume with ideal gas law, thus:
V = [tex]\frac{n.R.T}{P}[/tex]
Where:
n are moles: 2 moles
R is gas constant: 0,082 [tex]\frac{atm.L}{mol.K}[/tex]
T is temperature: 300 K
P is pressure: 1 atm
V is volume, with these values: 49,2 L
The work in the expansion of the gas, W, is: 1216 J - 34166 J = -32950 J
This work is:
W = P (Vf- Vi)
Where P is constant pressure, 1 atm
And Vf and Vi are final and initial volume in the expansion
-32950 J = -1 atm (Vf-49,2L) × [tex]\frac{101325 J}{1 atm.L}[/tex]
Solving: Vf = 49,52 L
Thus, final temperature could be obtained from ideal gas law, again:
T = [tex]\frac{P.V}{n.R}[/tex]
Where:
n are moles: 2 moles
R is gas constant: 0,082 [tex]\frac{atm.L}{mol.K}[/tex]
P is pressure: 1 atm
V is volume: 49,52 L
T is final temperature: 302 K
I hope it helps!
A 25 liter gas cylinder contains 10 mole of carbon dioxide. Calculate the pressure, in atm, of the gas at 325K.
Answer: The pressure of carbon dioxide gas is 11 atm
Explanation:
To calculate the pressure of gas, we use the equation given by ideal gas equation:
PV = nRT
where,
P = pressure of the gas = ?
V = Volume of gas = 25 L
n = number of moles of gas = 10 mole
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = 325 K
Putting values in above equation, we get:
[tex]P\times 25L=10mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 325K\\\\P=11atm[/tex]
Hence, the pressure of carbon dioxide gas is 11 atm
Question: Consider the molecule cyclohexene, which
containsonly hydrogen and carbon atoms.
When a sample is burned in oxygen, 4.822g of CO(2) and1.650g
of H(2)O are obtained.
What is its simplest formula?
Answer:
The simplest formula is C₃H₅.
Explanation:
The moles of CO₂ are the moles of Carbon because oxygen is in excess. Thus, the double of moles of H₂O are the moles oh hydrogen (Double because H₂O has two hydrogens for each mole).
4,822g CO₂ × (1mol / 44,01 g) = 0,1096 moles of C
1,650g H₂O × (1mol / 18,01 g) × 2= 0,1832 moles of H
The next step is divide these numbers between the smaller one, thus:
0,1096 moles / 0,1096 moles = 1
0,1832 moles / 0,1096 moles = 1,672
The last step is to find a number to multiple that gives integers:
1 × 3 = 3 Carbon
1,672 × 3 = 5 hydrogen
Thus, the simplest formula is C₃H₅.
I hope it helps!
You have ice cubes in your freezer at home. The ice cubes melt into water. This type of change is called a... O a. nuclear change O b.physical change O c. chemical change
Calculate the mass of 12 mL of a substance whose density is 1.1 g/mL. Express your answer in grams using the correct number of significant figures. Do not enter your answer using scientific notation.
Final answer:
To find the mass of the substance, multiply the given density (1.1 g/mL) by the given volume (12 mL) to get a mass of 13.2 grams, with the answer rounded to two significant figures.
Explanation:
To calculate the mass of a substance, you can use the formula mass = density × volume. Given a density of 1.1 g/mL and a volume of 12 mL, you simply multiply these two values together to find the mass.
Mass = 1.1 g/mL × 12 mL = 13.2 g
The density has two significant figures and the volume has two as well, so our final answer will be reported with two significant figures, hence 13.2 grams is the mass.
The chemical formula for sucrose (also known as table sugar) is C12H22011. What is the molar mass of sucrose? O a. 144 g/mole O b. 166 g/mole O c. 22 g/mole O d. 29 g/mole O e. 342 g/mole
1. Make a 100 mL volume of 100 mM acetic acid (HAC) HAC (glacial, 100%, 17.4 M, 1.05 g/ml, MW 60.05) ddh,0 ML add to 100 mL Measure 250 ml beaker. Add mL of ddH20 using a 100-ml graduated cylinder and transfer this volume to a ul of 100 mM HAC to the water. Mix with a stir bar.
Final answer:
To prepare a 100 mM acetic acid solution, calculate the moles needed, measure the glacial acetic acid based on its density and volume, and dilute with distilled water to the final volume. Mix well for homogeneity.
Explanation:
The question involves preparing a solution of acetic acid and relates to the subjects of molarity and solution preparation in chemistry. The solution is to be made to a specified concentration using a direct dilution from a more concentrated stock.
Steps to Prepare 100 mM Acetic Acid Solution
First, we calculate the number of moles of acetic acid required for 100 mL of a 100 mM solution using the molar mass of acetic acid (MW 60.05). Second, we use the density of glacial acetic acid (1.05 g/mL) and the volume needed (from the calculated moles and molarity) to determine how much of the glacial acetic acid to measure out. Finally, we add this measured amount of acid to a beaker and dilute to the mark with distilled water (ddH2O) to achieve the desired final concentration. Mixing with a stir bar ensures a homogeneous solution.
To calculate the mole fraction of acetic acid in a solution, we need the number of moles of acetic acid and the number of moles of water in the solution. This can be accomplished by converting the mass of each component to moles using their respective molar masses.
Calculate the number of mol of helium in a 2.01-L balloon at 29°C and 2.71 atm of pressure. Be sure to answer all parts. Imol
Answer:
0.220 mol He
Explanation:
Hello,
Based on the ideal gas equation of state:
[tex]PV=nRT\\n=\frac{PV}{RT} \\n=\frac{2.71 atm * 2.01 L}{0.082\frac{atm*L}{mol*K}*302K } \\n=0.220 mol[/tex]
Best regards.
Pyridine is a conjugate base which reacts with H to form pyridine hydrochloride. The hydrochloride dissociates to yield H' with a pKof 5.36. Describe the preparation of a pyridine buffer at pH 5.2 starting with 0.1M pyridine and 1.OM HCI. You may start with one liter of the 0.1 M pyridine.
Answer:
To one liter of the 0.1 M pyridine you need to add 41 mL of 1,0M HCl to obtain a buffer at 5,2
Explanation:
The reaction is:
pyridine-H⁺ ⇄ pyridine + H⁺ pka = 5,36; k = [tex]10^{-5,36}[/tex]
Using Henderson-Hasselbalch formula:
5,2 = 5,36 + log[tex]\frac{[Py-H^+]}{[Py]}[/tex]
0,692 = [tex]\frac{[Py-H^+]}{[Py]}[/tex] (1)
As total intial moles are 0,1:
0,1 = Py-H⁺ moles + Py moles (2)
Replacing (2) in (1) final moles of both Py-H⁺ and Py are:
Py: 0,059 moles
Thus:
Py-H⁺: 0,041 moles
Moles in reaction are:
Py: 0,1-x moles
H⁺: Y-x moles Y are initial moles of H⁺
Py-H⁺: x moles
Knowing x = 0,041 moles, pyridine volume is 1L and HCl molarity is 1 mol/L and [H⁺] = [tex]10^{-5,2}[/tex]
[tex]10^{-5,2}[/tex] = [tex]\frac{Y-0,041moles}{Y+1L}[/tex]
Y = 0,04100605 moles≡ 41 mL of 1,0M HCl
I hope it helps!
A house has an area of 234 m2. What is its area in each unit?
A)km2
B)dm2
C)cm2
Answer:
a) [tex]0.000324 km^2[/tex]
b) [tex]32400 dm^2[/tex]
c) [tex]3.24x10^6 cm^2[/tex]
Explanation:
To do the different conversions we need to know the follow:
[tex]1 km->1000 m so 1 km^2-> 1000^2 m^2= 1x10^6 m^2[/tex]
[tex]324 m^2*(1 km^2/1x10^6 m^2) = 0.000324 km^2[/tex]
[tex]1m-> 10 dm so 1m^2->10^2 dm^2=100 dm^2[/tex]
[tex]324 m^2*(100 dm^2/1 m^2)=32400 dm^2[/tex]
[tex]1m-> 100 cm so 1m^2->100^2 cm^2=10000 cm^2[/tex]
[tex]324 m^2*(10000 cm^2/1 m^2)=3.24x10^6 cm^2[/tex]
Answer :
(A) [tex]234m^2=2.34\times 10^8km^2[/tex]
(B) [tex]234m^2=2.34\times 10^4dm^2[/tex]
(C) [tex]234m^2=2.34\times 10^6cm^2[/tex]
Explanation :
The conversion used for area from [tex]m^2[/tex] to [tex]km^2[/tex] is:
[tex]1m^2=10^6km^2[/tex]
The conversion used for area from [tex]m^2[/tex] to [tex]dm^2[/tex] is:
[tex]1m^2=100dm^2[/tex]
The conversion used for area from [tex]m^2[/tex] to [tex]cm^2[/tex] is:
[tex]1m^2=10000cm^2[/tex]
Part (A):
As, [tex]1m^2=10^6km^2[/tex]
So, [tex]234m^2=\frac{234m^2}{1m^2}\times 10^6km^2=2.34\times 10^8km^2[/tex]
Part (B):
As, [tex]1m^2=100dm^2[/tex]
So, [tex]234m^2=\frac{234m^2}{1m^2}\times 100dm^2=2.34\times 10^4dm^2[/tex]
Part (C):
As, [tex]1m^2=10000cm^2[/tex]
So, [tex]234m^2=\frac{234m^2}{1m^2}\times 10000cm^2=2.34\times 10^6cm^2[/tex]
A worker is exposed to the following airborne concentrations of ethylbenzene: 8:00 – 10:00 125 ppm 10:15 – 12:15 88 ppm 12:45 – 3:15 112 ppm 3:30 – 5:00 70 ppm What is the 8-hour TWA exposure for the employee (calculate the exposure)? Does it exceed the OSHA PEL?
Answer:
The 8-hour TWA exposure for the employee is 101 ppm and it exceeds the OSHA PEL of 100 ppm for ethylbenzene.
Explanation:
The TWA for 8 hours is calculated by the sum of airbone concentrations multiplied by the time it has been exposed to that period. The total is divided by 8 which refers to the 8 hours total the employee has been exposed.
TWA = (125x2+88x2+112x2.5+70x1.5)/8.
The OSHA PEL is a known number for every compound and it can be find in PEL tables. In the case of ethylbenezene, it is 100 ppm.
A mixture of methanol and methyl acetate contains 15.0 weight percent methanol. Determine the number of gmols of methanol in 110.0 kilograms of the mixture.
Answer:
There are 550.5 moles of methanol in 110.0 kilograms of the mixture.
Explanation:
A solution 15% weight of methanol means there is 15g of methanol per 100g of the mixture or 0.1kg of the mixture. Also, the molar mass of methanol (CH3OH) is:
[tex]m_{C} + 4xm_{H} + m_{O} = 12.0g/mol + 4x1.0g/mol + 16.0g/mol = 32.0g/mol[/tex]
Thus, dividing 15g by molar mass
[tex]15.0g / 32.0\frac{g}{mol} = 0.5moles[/tex] we find there is 0.5 moles of methanol per 0.1Kg of the mixture. Calculating the number of mols of methanol in 110.0 kilograms of the mixture:
[tex]\frac{110.1Kgx0.5moles}{0.1Kg} = 550.5 moles[/tex]
Therefore, there is 550.5 moles of methanol in 110.0 kilograms of the mixture.
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l) In an industrial synthesis of urea, a chemist combines 132.0 kg of ammonia with 211.4 kg of carbon dioxide and obtains 172.7 kg of urea. determine the limiting reactant
Answer:
NH₃ is the limiting reagent
Explanation:
Mass of ammonia = 132.0kg
Mass of carbon dioxide = 211.4kg
Mass of urea = 172.7kg
Unknown:
Limiting reagent = ?
Solution
First, we write the balanced stoichiometeric equation:
2NH₃ + CO₂ → CH₄N₂O + H₂O
The reactant that is present in short supply determines the amount of product that is formed in a reaction. This reactant is called the limiting reagent.
To establish the limiting reagent, we need to go find out what is happening at the start of the reaction:
Convert the masses of the reactants to moles.
Number of moles of NH₃ = [tex]\frac{mass}{molar mass}[/tex]
Molar mass of NH₃ = 14 + (3x1) = 17g/mol
Number of moles of NH₃ = [tex]\frac{132}{17}[/tex] = 7.765mole
Number of moles of CO₂ = [tex]\frac{mass}{molar mass}[/tex]
Molar mass of CO₂ = 12 + (2 x 16) = 44g/mol
Number of moles of CO₂ = [tex]\frac{211.4}{44}[/tex] = 4.805mole
From the reaction equation:
2 moles of NH₃ reacted with 1 mole of CO₂
so 7.765 mole of NH₃ will require [tex]\frac{7.765}{2}[/tex]mole, 3.883 of CO₂
But we are given 4.805mole of CO₂.
Therefore, CO₂ gas is in excess and NH₃ is the limiting reagent.
The limiting reactant for the synthesis of CH₄N₂O given the data is ammonia, NH₃
Balanced equation2NH₃ + CO₂ —> CH₄N₂O + H₂O
Molar mass of NH₃ = 17 g/mole
Mass of NH₃ from the balanced equation = 2 × 17 = 34 g = 0.034 Kg
Molar mass of CO₂ = 44 g/mole
Mass of CO₂ from the balanced equation = 1 × 44 = 44 g = 0.044 Kg
SUMMARY
From the balanced equation above,
0.034 Kg of NH₃ reacted with 0.044 Kg of CO₂
How to determine the limiting reactantFrom the balanced equation above,
0.034 Kg of NH₃ reacted with 0.044 Kg of CO₂
Therefore,
132 Kg of NH₃ will react with = (132 × 0.044) / 0.034 = 170.8 of Kg of CO₂
From the calculation made above, we can see that only 170.8 of Kg of CO₂ out of 172.7 Kg given is needed to react completely with 132 Kg of NH₃.
Thus, NH₃ is the limiting reactant
Learn more about stoichiometry:
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The density of a certain type of plastic is 0.81 g/cm3. If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick, what is its mass?
mass = _____ x 10_____g
g
Enter your answer in scientific notation.
Answer: The mass of plastic sheet is [tex]8.1\times 10^4g[/tex]
Explanation:
The plastic sheet is in the form of cuboid. To calculate the volume of cuboid, we use the equation:
[tex]V=lbh[/tex]
where,
V = volume of cuboid
l = length of cuboid = 10.0 m = 1000 cm (Conversion factor: 1 m = 100 cm)
b = breadth of cuboid = 1.0 m = 100 cm
h = height of cuboid = 1 cm
Putting values in above equation, we get:
[tex]V=1000\times 100\times 1=10^5cm^3[/tex]
To calculate mass of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
We are given:
Density of plastic sheet = [tex]0.81g/cm^3[/tex]
Volume of plastic sheet = [tex]10^5cm^3[/tex]
Putting values in equation 1, we get:
[tex]0.81g/cm^3=\frac{\text{Mass of plastic sheet}}{10^5cm^3}\\\\\text{Mass of plastic sheet}=8.1\times 10^4g[/tex]
Hence, the mass of plastic sheet is [tex]8.1\times 10^4g[/tex]
The mass of the plastic sheet is found by multiplying the density (0.81 g/cm³) with its volume (100,000 cm³), resulting in a mass of 81,000 g, or 8.1 × 10⁴ g in scientific notation.
To calculate the mass of a plastic sheet with a given density, you need to multiply its density by its volume. The density of the plastic is given as 0.81 g/cm³. To find the volume, convert all dimensions into the same unit, in this case, centimeters, because the density is given in g/cm³. The sheet is 10.0 m long (1000 cm), 1.0 m wide (100 cm), and 1 cm thick. Therefore, the volume is 1000 cm × 100 cm × 1 cm = 100,000 cm³.
Now, calculate the mass using the formula: mass = density × volume, so mass = 0.81 g/cm³ × 100,000 cm³. The mass of the plastic sheet is thus 81,000 g. To express this in scientific notation, we write it as 8.1 × 10⁴ g.
A runner runs 4339 ft in 7.45 minutes. What is the runner's average speed in miles per hou 6.62 mi/hr O 0.110 mi/hr 6.618 mi/h e 582 mi/hr
Answer: The average speed of the runner is 6.618 miles/hr
Explanation:
Average speed is defined as the ratio of total distance traveled to the total time taken.
To calculate the average speed of the runner, we use the equation:
[tex]\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}[/tex]
We are given:
Distance traveled = 4339 ft
Time taken = 7.45 mins
Putting values in above equation, we get:
[tex]\text{Average speed of runner}=\frac{4330ft}{7.45min}=582.42ft/min[/tex]
To convert the speed into miles per hour, we use the conversion factors:
1 mile = 5280 ft
1 hr = 60 mins
Converting the speed into miles per hour, we get:
[tex]\Rightarrow \frac{528.42ft}{min}\times (\frac{1miles}{5280ft})\times (\frac{60min}{1hr})\\\\\Rightarrow 6.618mil/hr[/tex]
Hence, the average speed of the runner is 6.618 miles/hr