Answer:
The two changing properties in Charles’ law are temperature and volume.
Explanation:
Charles’ law state the presence of direct relationship between temperature and volume of the system of gas molecules at constant pressure condition. In this law, the expansion of gas has been explained with the increase of temperature.
As the temperature is increased or the system of gas molecules are heated, the gas molecules tend to expand their volume to maintain the pressure same.
So the temperature and volume are directly proportional at constant pressure. Thus the two changing properties of Charles’ law is temperature and volume. The mathematical representation of Charles’ law is
V∝T (at constant Pressure)
[tex]V=kT[/tex]
Here k is the non-zero constant and V and T are volume and temperature respectively.
Determine the change in volume that takes place when a 3.25-L sample of N2(g) is heated from 250.0 K to 406.8 K. Enter your answer in the box provided. L
Answer: The change in volume is 2.05 L
Explanation:
To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where,
[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.
[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.
We are given:
[tex]V_1=3.25L\\T_1=250.0K\\V_2=?\\T_2=406.8K[/tex]
Putting values in above equation, we get:
[tex]\frac{3.25L}{250.0K}=\frac{V_2}{406.8K}\\\\V_2=5.30L[/tex]
Change in volume = [tex]V_2-V_1=(5.30-3.25)L=2.05L[/tex]
Hence, the change in volume is 2.05 L
If heat flows into a system and the system does work on the surroundings, what will be the signs on q and w? Select the correct answer below O positive q, positive w O positive q, negative w O negative q, positive w O negative q, negative w
Answer:
q = Positive
w = Negative
Explanation:
As per first law of thermodymanics,
ΔE = q + w
Where,
ΔE = Change in internal energy
q = Heat absorbed or heat released by the system
w = Work done
Sign conventions are used for heat transfer and work done during a thermodynamics process.
Sign convention for Heat transfer
q is positive when heat is added to the system or heat absorbed by the system this is because energy of the system is increased.q is negative when heat is withdrawn from the system or heat released by the system.Sign convention for Work done
w is positive if work is done on the system or work is done by the surroundingsw is negative if work is done on the system or work is done on the surrounding.In the given question, work is done on the surroundings so, w is negative.
Heat flows into a system or in other word heat is added to the system,
So q is positive.
A piece of an unknown metal has a volume of 19.8 ml and a mass of 210.0 grams. The density of the metal is g/mL A piece of the same metal with a mass of 86.0 grams would have a volume of mL.
Answer: The density of the metal is 10.60 g/mL and the volume occupied by 86.0 grams is 8.11 mL
Explanation:
To calculate the density of unknown metal, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex] ......(1)
Volume of unknown metal = 19.8 mL
Mass of unknown metal = 210.0 g
Putting values in equation 1, we get:
[tex]\text{Density of unknown metal}=\frac{210.0g}{19.8mL}\\\\\text{Density of unknown metal}=10.60g/mL[/tex]
The density of the metal remains the same.
Now, calculating the volume of unknown metal, using equation 1, we get:
Density of unknown metal = 11.45 /mL
Mass of unknown metal = 86.0 g
Putting values in above equation, we get:
[tex]10.60g/mL=\frac{86.0g}{\text{Volume of unknown metal}}\\\\\text{Volume of unknown metal}=8.11mL[/tex]
Hence, the density of the metal is 10.60 g/mL and the volume occupied by 86.0 grams is 8.11 mL
Waste water treatment plant. A municipal water treatment plant for a small community. Waste water 32,000 m3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks and microbes break down the organic materials. An entering feed has a BOD of 200 mg O2/liter and the effluent has a negligible BOD. Find the rate of reaction in the treatment tanks.
Answer:
The reaction rate k is 0.0012563 (1/hour).
Explanation:
We considered the reactions occurring in the plant as first order, and represented by this equation:
[tex]y = L (1- e^{-kt})[/tex]
where y is the BOD at time t, L is the initial value of BOD and k is the reaction rate.
If we replaced with the values
y = 2 mg O2/l (1% of the initial value)
L = 200 mg 02/l
t = 8 hr
We can calculate k
[tex]y=L(1-e^{-kt})\\\\k=-(1/t)*ln(1-y/L) = -(1/8)*ln(1-2/200)=-(1/8)*(-0.01)=0.0012563 \, hour^{-1}[/tex]
The reaction rate k is 0.0012563 1/hour.
Final answer:
The rate of reaction for a municipal wastewater treatment plant can be found by calculating the total oxygen consumed to lower the BOD, using the volume of wastewater and the residence time. The initial BOD is 200 mg O2/liter and is reduced to a negligible level, indicating a complete reaction over the course of the 8-hour residence time.
Explanation:
Finding the Rate of Reaction in Wastewater Treatment Tanks
The student is tasked with finding the rate of reaction for a municipal wastewater treatment plant where wastewater is treated with aerobic bacteria that decompose organic material. The process significantly lowers the biochemical oxygen demand (BOD) from an entering feed of 200 mg O2/liter to a negligible level in the effluent. The key to solving for the rate of reaction is utilizing the given information: the wastewater volume (32,000 m3/day) and the mean residence time (8 hours)
To find the rate of reaction, we need to calculate the amount of oxygen consumed by the microorganisms to reduce the BOD to negligible levels. The initial BOD is given as 200 mg O2/liter, and since it's reduced to a negligible level, we can assume that virtually all the initial oxygen demand is consumed by the reaction. The volume converted to liters (since BOD is in mg/liter) and residence time in days must be considered to calculate the daily rate of reaction:
Volume flow rate in liters/day: 32,000 m3/day * 1,000 liters/m3Total oxygen demand per day: Volume flow rate * BODSince the residence time is 8 hours (or 1/3 of a day), we must divide the total oxygen demand by 3 to obtain the daily rate of oxygen consumed in the tanks.Once the daily rate is found, the rate of reaction will be the amount of oxygen consumed per day divided by the volume of wastewater treated in one day (in liters), providing the rate in mg O2/liter/day.
What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?
Now if instead of pure water a buffer is used: HPO4-2/H2PO4- pKa = 7.2 Assume the initial pH of this buffer is 7 (like the pure water example). -First you must you must use the Henderson-Hasselbalch equation to determine the ratio of HPO4-2/H2PO4- , which is 0.063M to .1M. Using the same amount of HCl added (.001M), determine the change in pH that occurs to the buffer when the HCl is added.
(i already answered the first part, I just need the second part. Show and explain your work please!)
Answer:
pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0
Explanation:
It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:
pH = pka + log₁₀
Where A⁻ is conjugate base and HA is conjugate acid
The equilibrium of phosphate buffer is:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,2
Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:
7,0 = 7,2 + log₁₀ [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]
Ratio obtained is:
0,63 = [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]
As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M
As the amount added of HCl is 0,001 M the concentrations in equilibrium are:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺
0,1 M +x 0,063M -x 0,001M -x -because the addition of H⁺ displaces the equilibrium to the left-
Knowing the equation of equilibrium is:
[tex]K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}[/tex]
Replacing:
6,20x10⁻⁸ = [tex]\frac{[0,063-x][0,001-x]}{[0,1+x]}[/tex]
You will obtain:
x² -0,064 x + 6,29938x10⁻⁵ = 0
Thus:
x = 0,063 → No physical sense
x = 0,00099990
Thus, [H⁺] in equilibrium is:
0,001 M - 0,00099990 = 1x10⁻⁷
Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =
-log₁₀ [1x10⁻⁷] = 7,0
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!
I hope it helps!
A student labeled 5 tubes 0, 1, 2, 3, 4, 5. Tube 0 contained undiluted protein lysate. The student used the tube 0 lysate to prepared 10-fold serial dilutions in the tubes labeled 1-5. What is the dilution of each of the tubes?
Answer:
Tube 1: 10⁻¹, tube 2: 10⁻², tube 3: 10⁻³, tube 4: 10⁻⁴, tube 5: 10⁻⁵.
Explanation:
Serial dilutions are dilutions that the concentration decreases by the same quantity in each successive step. It means that the undiluted will be used for the first step, then the first will be used for the second, and successively. So, for a 10-fold, the concentration must decrease 1/10 in each step, it means that the dilution will be 1/10 in the first one (because it's 1 in tube 0).
In tube 1, the dilution is 1/10 = 0.1 = 10⁻¹;
In tube 2, the dilution will decrease more 1/10, so it will be 1/100x10 = 1/100 = 0.01 = 10⁻²;
In tube 3, it will be 1/1000x1/10 = 1/1000 = 0.001 = 10⁻³
In tube 4, it will be 10⁻⁴, and
In tube 5, it will be 10⁻⁵.
How many vibrational modes does ethanol have? Explain why.
Answer:
The ethanol has 21 vibrational modes.
Explanation:
A molecule can show 3 types of motions: one external called translational and two internal called rotational and vibrational.
In order to calculate the vibrational modes of a molecule we need to know the degrees of freedom of this molecule, it means the number of variables that are involved in the movement of this particle.
If we know that atoms are three dimensional we will know that they have 3 coordinates expressed as 3N. But the atoms are bonded together so they can move not only in translational but also rotational and vibrational. So, the rotational move can be described in 3 axes and the other vibrational move can be described as
3N-5 for linear molecules
3N-6 For nonlinear molecules like ethanol
So using the formula for nonlinear molecules where N is the amount of atoms in the chemical formula, so ethanol has 9 atoms
3(9)-6= 21
Thus, ethanol has 21 vibrational modes.
Final answer:
Ethanol has 21 vibrational modes, which can be calculated using the formula 3N-6 for non-linear molecules, where N equals the number of atoms in the molecule. The vibrational modes consist of various bond stretching and bending movements, and these can only occur at certain quantized frequencies.
Explanation:
The number of vibrational modes in a molecule depends on the number of atoms it contains and the types of bonds present. In the case of ethanol (C2H5OH), which is a relatively complex molecule, the actual number of vibrational modes can be calculated using the formula 3N-6 for non-linear molecules, where N is the number of atoms. Considering ethanol has 9 atoms, we can calculate its vibrational modes as 3(9)-6, resulting in 21 vibrational modes.
The reason for the 3N-6 rule is that each atom contributes three degrees of freedom (x, y, and z axes), but we subtract the six degrees corresponding to the translational and rotational movements of the molecule as a whole. The vibrational modes in a molecule are the various ways that the molecule's bonds can stretch, bend, and twist. These modes consist of various stretching and bending movements of the carbon-hydrogen, carbon-carbon, and carbon-oxygen bonds within the ethanol molecule.
It's important to note that molecular vibrations are quantized, meaning a molecule can only access these modes at specific frequencies. When a molecule like ethanol is exposed to infrared radiation that matches the frequency of one of its vibrational modes, it can absorb that radiation, causing a transition to a higher energy vibrational state.
The formula for acetic acid is CH3COOH. Seana wants to know how many oxygen atoms are in 0.12 moles of acetic acid. O a. 1x 10^25 atoms O b. 60 atoms Oc. 14x 10^23 atoms O d. 4.0 x 10^-24 atoms O e. 7.2 x 10^22 atoms
Answer:
c. 1.4 x 10²³ oxygen atoms
Explanation:
The number of oxygen atoms in one molecule of CH₃COOH is 2.
Avogadro's constant relates the number of molecules in one mole:
6.022 × 10²³ mol⁻¹
Thus, the number of oxygen atoms in one molecule of acetic acid can be converted to the number of oxygen atoms in one mole of acetic acid:
(2 oxygen atoms / molecule)(6.022 × 10²³ molecule / mol) = 1.204 x 10²⁴ atoms per mole
Finally, the number of oxygen atoms in 0.12 moles of acetic acid are calculated:
(1.204 x 10²⁴ atoms / mol)(0.12 mol) = 1.4 x 10²³ atoms/mol
The volume of a sphere is given by V-(4/3)r where r is the radius. The density of magnesium is 1.74 g/cm What is the mass of a magnesium sphere with a radius of 0.80 cm? • V= x2 m= TEMPERAURE CONVERTIONS show formulas used Convert 38.0 °F 10 °C = 3.33 (38-32)x(5/6)=3333°C Convert 23.5 °C 10 °F - 245 (22.5%6/5)+32=72.5°F SPD.
Answer:
The answer to your question is:
mass = 3.74 g
Explanation:
Data
V = (4/3) πr³
density = 1.74 g/cm³
radius = r = 0.80 cm
Process
V = (4/3) π(0.8)³ Substitution
V = 2.1446 cm³
mass = density x volume
mass = 1.74 x 2.1446 Substitution
mass = 3.74 g
I don't understand if the second section is also a question.
Calculate the mole fraction of NaCl in a solution prepared by dissolving 117 g NaCl in 1.11 kg H2O.
1.12 × 10-2
6.29 × 10-2
1.57 × 10-2
9.91 × 10-1
3.15 × 10-2
The mole fraction of NaCl in the solution is 3.15 × 10-2.
Explanation:The mole fraction of NaCl in a solution prepared by dissolving 117 g NaCl in 1.11 kg H2O can be calculated by dividing the moles of NaCl by the total moles of solute and solvent. First, we convert the mass of NaCl to moles using its molar mass. Then, we calculate the moles of water by dividing its mass by its molar mass. Finally, we divide the moles of NaCl by the sum of moles of NaCl and moles of water to get the mole fraction of NaCl in the solution.
Given:
Mass of NaCl = 117 gMass of H2O = 1110 g (1.11 kg)Molar mass of NaCl = 58.44 g/molMolar mass of H2O = 18.02 g/molStep 1: Convert mass of NaCl to moles:
Moles of NaCl = (mass of NaCl / molar mass of NaCl) = (117 g / 58.44 g/mol) = 2 moles
Step 2: Convert mass of H2O to moles:
Moles of H2O = (mass of H2O / molar mass of H2O) = (1110 g / 18.02 g/mol) = 61.55 moles
Step 3: Calculate mole fraction of NaCl:
Mole fraction of NaCl = (moles of NaCl / (moles of NaCl + moles of H2O)) = (2 moles / (2 moles + 61.55 moles)) = 3.15 × 10-2
The units of density are kg/m2. If the density of a liquid is 760.0 kg/m' what is the specific volume? a) 1.316 x 10 m2/kg b) 1.316 x 10 m3/kg c) without the molecular weight of the liquid it is impossible to determine the specific volume d) none of the above are correct
Answer:
The correct answer is: 1.316 . 10⁻³ m³/kg.
Explanation:
The density (ρ) of a substance is the ratio of its mass (m) to its volume (V). At constant temperature and pressure, its value is constant and it is an intrinsic property of materials. The units of density are kg/m³.
[tex]\rho = \frac{m}{V}[/tex]
The specific volume (ν) of a substance is the ratio of its mass to its volume. We can see that it is the reciprocal of density and an intrinsic property of matter as well. Therefore, the units of specific volume are m³/kg.
[tex]\nu = \frac{V}{m}=\frac{1}{\rho }[/tex]
Given we know the density of the liquid, we can use this relationship to find out its specific volume:
[tex]\nu =\frac{1}{\rho }=\frac{1}{760.0kg/m^{3} } =1.316 .10^{-3} m^{3} /kg[/tex]
A specific brand of gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-gpiece. How many kilograms of dietary fat are in a box containing 1.00 lb of candy?
Express your answer numerically in kilograms.
Answer: 0.14 kg
Explanation:
Gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-g piece
That is 1 piece of candy weighs 22.7 g and contains 7.00 g of dietary fat
Converting the mass in pounds to kg
1 lb = 0.45 kg = 450 grams (1kg=1000g)
Number of pieces = [tex]\frac{450}{22.7g}=20pieces[/tex]
1 piece contains = 7 g of dietary fat
Thus 30 pieces would contain =[tex]\frac{7}{1}\times 20=140g[/tex] of dietary fat
1 g = 0.001 kg
Thus 140 grams =[tex]\frac{0.001}{1}\times 140=0.14kg[/tex]
Thus 0.14 kg of dietary fat are in a box containing 1.00 lb of candy.
Assigned as ir in an internal combustion engine is at 440'F and 150 psía, with a volume of 10 in3. what is the mass of air b,) ou may assume ideal gas
Answer: The mass of air is 1.18 g
Explanation:
To calculate the number of moles, we use the equation given by ideal gas equation:
[tex]PV=nRT[/tex]
Or,
[tex]PV=\frac{m}{M}RT[/tex]
where,
P = pressure of the gas = 150 psia = 10.2 atm (Conversion factor: 1 psia = 0.068 atm)
V = Volume of gas = [tex]10in^3=0.164L[/tex] (Conversion factor: [tex]1in^3=0.0164L[/tex] )
m = mass of air = ?
M = Average molar mass of air = 28.97 g/mol
R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]440^oF=499.817K[/tex] (Conversion factor: [tex](T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}[/tex] )
Putting values in above equation, we get:
[tex]10.2atm\times 0.164L=\frac{m}{28.97g/mol}\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 499.817K\\\\m=1.18g[/tex]
Hence, the mass of air is 1.18 g
Two scientists work together on an experiment, but they have different hypotheses. When the scientists look at the experimental results, they interpret the data in different ways and come to different conclusions. Which of the following does this situation best illustrate?
Answer:
When experiments are carried out or research is done in a certain field of knowledge the scientist at first hypothesise certain knowledge or make theoretical hypotheses about that field of knowledge.
And then conduct the experiment or research to derive certain conclusions and get answers which they can apply on the hypothesis made or on the previous knowledge they have and thereby confirm or negate the hypothesis.
When two scientists are working on similar experiment and they tend to differ in the conclusions drawn by the result, as they get from experiment then it is called as confirmation bias among the scientists.
What is the solubility of helium (in units of grams per liter) in water at 25 C, when the He gas over the solution has a partial pressure of 0.284 atm? ky for He at 25 °C is 3.26x10 molL atm
Answer:
9.2584 mol/L is the solubility of helium in water at 25 °C.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{He}=K_H\times p_{H_2O}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]3.26\times 10 mol/L atm[/tex]
[tex]p_{He}[/tex] = partial pressure of carbonated drink = 0.284 atm
Putting values in above equation, we get :
[tex]C_{He}=3.26\times 10 mol/L atm\times 0.284 atm=9.2584 mol/L[/tex]
9.2584 mol/L is the solubility of helium in water at 25 °C.
A chemist prepares a solution of magnesium fluoride MgF2 by measuring out 0.00598μmol of magnesium fluoride into a 50.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's magnesium fluoride solution. Round your answer to 2 significant digits.
Answer:
0,12 μmol/L of MgF₂
Explanation:
Preparation of solutions is a common work in chemist's life.
In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in μmol/L
You have 0,00598 μmol but not Liters.
To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:
50,0 mL (1L/1000mL) = 0,05 L of water.
Thus, concentration in μmol/L is:
0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-
I hope it helps!
The concentration of the chemist's magnesium fluoride solution is 0.12 μmol/L when rounded to two significant digits, calculated by dividing the moles of solute by the volume of the solution in liters.
Explanation:To calculate the concentration of the magnesium fluoride solution in μmol/L, we use the formula C = (n/V), where 'C' is the concentration, 'n' is the number of moles of solute, and 'V' is the volume of solution in liters.
First, we convert the volume from milliliters to liters: 50 mL = 0.050 L.
The number of moles (n) of MgF2 given is 0.00598 μmol, which is already in the correct units for our calculation.
Therefore, the concentration (C) of the solution is:
C = n/V
C = 0.00598 μmol / 0.050 L
C = 0.1196 μmol/L
When rounded to two significant digits, the concentration of the magnesium fluoride solution is 0.12 μmol/L.
Assume a pill has a dosage of 350 mg of medication. How much medication is this in grams?
Answer:
0.350 grams
Explanation:
The given mass of the pill that has to be taken as a dosage for the medication = 350 mg
The medication has to be determined in grams which means that milligram has to be converted to grams.
The conversion of mg to g is shown below:
[tex]1\ milligram=10^{-3}\ gram[/tex]
So,
[tex]350\ milligram=350\times 10^{-3}\ gram[/tex]
[tex]350\ milligram=0.350\ gram[/tex]
The medication required in grams = 0.350 grams
Final answer:
To convert 350 mg to grams, divide by 1000, resulting in 0.35 grams of medication.
Explanation:
To convert the dosage of medication from milligrams to grams, you need to know the conversion factor between these two units.
There are 1000 milligrams in one gram.
Therefore, you can find the amount in grams by dividing the milligram dosage by 1000.
For the pill with a dosage of 350 mg of medication, the conversion to grams would be:
Total given grams/1000
= 350 mg ÷ 1000
= 0.35 grams
A conjugate acid is the species that a. remains after a base has given up a proton. b. is formed by the addition of a proton to a base. c. is formed by the addition of a proton to an acid. d. remains after an acid has given up a proton.
Answer:
b.
Explanation:
A conjugate acid-base pair are related to each other through the addition or loss of a proton. Acids react by giving up a proton, and the resulting product of that reaction is the conjugate base. Thus, the conjugate base will have one proton less than the acid. Similarly, bases react by receiving protons, so the conjugate acid will have one proton more than the conjugate base that reacted.
In the following example, HA is the acid while A⁻ is the conjugate base:
HA + H₂O ⇒ H₃O⁺ + A⁻
Answer option a. and c. are not correct since bases to not give up protons and acids to not receive protons. Base on the above information, a conjugate base reacts to produce the conjugate acid and vice versa. Thus, to produce the conjugate acid, the conjugate base received a proton. The correct answer is b.
A conjugate acid is formed by the addition of a proton to a base, and has the potential to donate that proton back, reverting to its original base form.
Explanation:A conjugate acid is the species that is formed by the addition of a proton to a base. When we describe acid-base reactions in terms of conjugate acids and conjugate bases, we are applying the Brønsted-Lowry Acid-Base Theory. In this context, a conjugate acid/base pair is defined by the gain or loss of a single hydrogen ion. For example, when a base accepts a proton (H+), it forms its conjugate acid, and this conjugate acid has the potential to donate that proton back, essentially reverting to its original base form.
Conversely, when an acid donates a proton, it forms a conjugate base, which can accept a proton in a reverse reaction, thus potentially reforming the original acid. These relationships are fundamental to understanding acid-base chemistry and reactions where proton transfer occurs.
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What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A) 2.18x10 (B) 7.84*10J (C) 8.62x10 J (D) 5.34x10 3 Answer A
The kinetic energy of an electron can be calculated utilizing the equation KE = hf - BE, but this specific equation cannot be solved without additional information such as the binding energy, frequency of the light radiation, or Plank's constant. Energy can also be calculated using Planck's equation,E = hf, or for a specific orbital using 13.6 eV / n², where n refers to the level of the orbital.
Explanation:The kinetic energy acquired by an electron in a hydrogen atom after absorbing light radiation can be found by utilizing the equation KE = hf - BE (kinetic energy equals energy of radiation minus binding energy). In this case, if the electron absorbs a light radiation of energy 1.08x101 J, it's crucial to determine the binding energy first.
However, given the information available, further clarification is needed since binding energy, frequency of the radiation, or Plank's constant (h) are not specified in the question.
Similar energy calculations involve using Planck's equation E = hf, where E is energy, h is Planck’s constant, and f is the frequency of radiation. Furthermore, energy can also be calculated for a specific orbital of a hydrogen atom using the equation 13.6 eV / n², where n refers to the level of the orbital.
Remember, when using these equations you may need to convert your units appropriately to reach the correct answer.
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Your boss tells you that water at 68 F is flowing in a cooling tower loop at a rate of 600 gpm, in an 8-inch inside diameter pipe. Your boss asks you whether this flow is the laminar or turbulent flow regime. Calculate the Reynolds number and state whether the flow is laminar or turbulent.
Explanation:
The volumetric flow rate of water will be as follows.
q = [tex]600 gpm \times \frac{0.000063 m^{3} s^{-1}}{1 gpm}[/tex]
= 0.0378 [tex]m^{3}/s [/tex]
Diameter = [tex]8 in \times \frac{0.0254 m}{1 in}[/tex]
= 0.2032 m
Relation between area and diameter is as follows.
A = [tex]\frac{\pi}{4} \times D^{2}[/tex]
= [tex]\frac{3.14}{4} \times (0.2032 m)^{2}[/tex]
= 0.785 x 0.2032 x 0.2032
= 0.0324 [tex]m^{2}[/tex]
Also, q = A × V
or, V = [tex]\frac{q}{A}[/tex]
= [tex]\frac{0.0378 m^{3}/s}{0.0324 m^{2}}[/tex]
= 1.166 m/s
As, viscosity of water = 1 cP = [tex]10^{-3}[/tex] Pa-s
Density of water = 1000 [tex]kg/m^{3}[/tex]
Therefore, we will calculate Reynolds number as follows.
Reynolds number = [tex]\frac{D \times V \times density}{viscosity}[/tex]
= [tex]\frac{0.2032 m \times 1.166 m/s \times 1000}{10^{-3}}[/tex]
= 236931.2
Hence, the flow will be turbulent in nature.
Thus, we can conclude that the Reynolds number is 236931.2 and flow is turbulent.
Ranjit titrates a sample 10.00 mL of Ba(OH)2 solution to the endpoint using 12.58 mL of 0.1023 M H2SO4.
Based on this data, calculate the concentration of the barium hydroxide solution.
[Ba(OH)2] = ___ M
Answer: The concentration of [tex]Ba(OH)_2[/tex] comes out to be 0.129 M.
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]
We are given:
[tex]n_1=2\\M_1=0.1023M\\V_1=12.58mL\\n_2=2\\M_2=?M\\V_2=10.00mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.1023\times 12.58=2\times M_2\times 10.00\\\\M_2=0.129M[/tex]
Hence, the concentration of [tex]Ba(OH)_2[/tex] comes out to be 0.129 M.
You are given a protein solution with a concentration of 0.15 mg/ml.
iii. If the molecular weight of the protein is 22,000 Da, express its initial concentration in moles/liter, micromoles/mL, and micromoles/microliters. If we want 20 micromoles of proteins for a reaction, what volume do we need to prepare?
Answer:
We need 2.933 L of 0.15 mg /mL of protein solution.
Explanation:
Concentration of given solution[tex]C_1 = 0.15 mg/mL[/tex]
1 mg = 0.001 g , 1 mL = 0.001 L
[tex]C_1=\frac{0.15\times 0.001 mg}{1\times 0.001 L}=0.15 g/L[/tex]
Molecular weight of protein = 22,000 Da =22,000 g/mol
Initial concentration in moles/liter:
[tex]C_1=\frac{0.15 g/L}{22,000 g/mol}=6.8182\times 10^{-6} mol/L[/tex]
Initial concentration in micromoles/mL :
1 L = 1000 mL
[tex]C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000 mL}=6.8182\times 10^{-3} \mu mole/ mL[/tex]
Initial concentration in micromoles/microLiter :
1 L = 1000,000 μL
[tex]C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000000 \mu L}=6.8182\times 10^{-6}\mu mol/\mu L[/tex]
Moles of protein required = 20 μmoles
n(Moles)=C(concentration) × V(Volume of solution)
[tex]20 \mu mol=6.8182\times 10^{-6}\mu mol/\mu L\times V[/tex]
[tex]V =\frac{20 \mu mol}{6.8182\times 10^{-6}\mu mol/\mu L}[/tex]
[tex]V=2.933\times 10^6 \mu L = 2.933 L[/tex]
We need 2.933 L of 0.15 mg /mL of protein solution.
What advantage does the "coolant" (mixture of water and ethylene glycol) have over water as a liquid to cool the car radiator?
Answer:
We use coolant instead of water because when heated, water can start to accumulate the minerals that it has, and this can corrode the metal on your car. Other solution could be to use distilled water, but is will also corrode the metal of your radiator. One advantage of ethylene glycol is that prevents the corrosion in your radiator providing a protective coating to the metals.
Determine the transition length at the entrance to a 10 mm tube through which 100 percent glycerol at 60 C is flowing at a velocity of 0.5 m/s. The density of a glycerol is 1240 kg/m Data: The dynamic viscosity is 0.0813 Pa-s. a) 18.13 mm b) 50 mm c) 45.43 mm d) 38.13 mm
Answer: Option (d) is the correct answer.
Explanation:
The given data is as follows.
Tube diameter d = 10 mm = 0.01 m
Velocity of glycerol, v = 0.5 m/s
Density of glycerol ([tex]\rho[/tex]) = 1240 kg/m3
Dynamic viscosity of glycerol ([tex]\mu[/tex]) = 0.0813 pa.s
Reynolds number (Re) = [tex]\rho \times velocity \times \frac{density}{\mu}[/tex]
= [tex]1240 \times 0.5 \times \frac{0.01}{0.0813}[/tex]
= 76.26
Therefore, according to Reynolds number we can say that flow is laminar.
Lt = [tex]0.05 \times Re \times d[/tex]
= [tex]0.05 \times 76.26 \times 0.01[/tex]
= [tex]0.03813 m[/tex]
As it is known that 1 m = 1000 mm. Hence, in 0.03813 m will be equal to [tex]0.03813 m \times \frac{1000 mm}{1 m}[/tex]
= 38.13 mm
Thus, we can conclude that the transition length of glycerol is 38.13 mm.
Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this process, the gas absorbed 800J of heat from the surroundings. Please calculate the internal energy change of the gas, AE.
Explanation:
The given data is as follows.
P = 3 atm
= [tex]3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}[/tex]
= [tex]3.03975 \times 10^{5} Pa[/tex]
[tex]V_{1}[/tex] = 9 L = [tex]9 \times 10^{-3} m^{3}[/tex] (as 1 L = 0.001 [tex]m^{3}[/tex]),
[tex]V_{2}[/tex] = 15 L = [tex]15 \times 10^{-3} m^{3}[/tex]
Heat energy = 800 J
As relation between work, pressure and change in volume is as follows.
W = [tex]P \times \Delta V[/tex]
or, W = [tex]P \times (V_{2} - V_{1})[/tex]
Therefore, putting the given values into the above formula as follows.
W = [tex]P \times (V_{2} - V_{1})[/tex]
= [tex]3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})[/tex]
= 1823.85 Nm
or, = 1823.85 J
As internal energy of the gas [tex]\Delta E[/tex] is as follows.
[tex]\Delta E[/tex] = Q - W
= 800 J - 1823.85 J
= -1023.85 J
Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.
500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. CaCl2 = Ca++ + 2Cl-
a) Express the concentration of the solution in % w/v
b) Express the concentration in ratio strength
c) Express the concentration in molarity (M)
d) Express the concentration in molality (m)
e) How many equivalents of calcium chloride would be in 1.5 L of the solution?
The attention measures for a CaCl2 result with 1000 mg in 500 mL are0.2 w/ v, a rate strength of 2 g/ L,0.01818 M, and molality isn't directly reliable from the given data. For1.5 L of this result, there are0.05454 coequals of CaCl2.
Computations for attention of CaCl2 result In order to calculate the colorful attention measures of the given result containing calcium chloride( CaCl2), we should first understand the ensuing delineations w/ v- Percent weight per volume expresses attention as grams of solute per 100 mL of result. rate strength- The quantum of solute in grams per 1000 mL( 1 L) of result. Molarity( M)- The number of intelligence of solute per liter of result. Molality( m)- The number of intelligence of solute per kilogram of detergent. Given that the molecular weight of CaCl2 is 110 g/ spook, and we've 1000 mg( 1 g) in 500 mL, we can calculate the following w/ v( 1 g/0.5 L) * 100 = 0.2 w/ v rate strength 1 g in 500 mL equals 2 g in 1000 mL or 2 g/L. Molarity( M)( 1 g/ 110 g/ spook)/(0.5 L) = 0.01818 M Molality( m) can not be calculated without the mass of the detergent; still, assuming the specific graveness is 1, molality would be equal to molarity for veritably dilute results like this bone
Why solar energy is called an inexhaustible resource?
Explanation:
Inexhaustible resource -
It is the resource which never get reduced or deplete on constant usage , they are mainly the type of naturally occurring resources , hence they reappear in nature again .
The example of inexhaustible resources are - the wind energy , tides , solar energy and the geothermal energy .
Hence , the solar energy is also a form of inexhaustible resource .
Define "Monosaccharide", "Disaccharide", "Oligosaccharide", "Polysaccharide", and "Glucoconjugate"
Explanation:
Monosaccharides are the elementary form of the sugar and most basic units of the carbohydrates. These sugars cannot be further hydrolyzed to form the simpler chemical compounds. The general formula is [tex]C_nH_{2n}O_n[/tex]. Example: Glucose and fructose.
Disaccharide is sugar which is formed when the two monosaccharides are joined by the glycosidic linkage. Disaccharides are soluble in the water. Examples: sucrose and lactose.
Oligosaccharide is the saccharide polymer which contains small number of the monosaccharides. They can have many functions like the cell recognition and the cell binding. Example: glycolipids which have role in immune response.
Polysaccharides are the polymeric carbohydrate molecules which are composed of the long chains of the monosaccharide units that are bound together by the glycosidic linkages which on the hydrolysis give constituent monosaccharides or th eoligosaccharides. Example: Starch.
Glycoconjugates is general classification for the carbohydrates which are covalently linked with the other chemical species such as peptides, proteins, saccharides and lipids. Example: Blood proteins
What is the value for the kinetic energyfor a n = 2 Bohr orbit electron in Joules?
Answer:
K.E. = 5.4362 × 10⁻¹⁹ J
Explanation:
The expression for Bohr velocity is:
[tex]v=\frac{Ze^2}{2 \epsilon_0\times n\times h}[/tex]
Applying values for hydrogen atom,
Z = 1
Mass of the electron ([tex]m_e[/tex]) is 9.1093×10⁻³¹ kg
Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C
[tex]\epsilon_0[/tex] = 8.854×10⁻¹² C² N⁻¹ m⁻²
h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s
We get that:
[tex]v=\frac {2.185\times 10^6}{n}\ m/s[/tex]
Given, n = 2
So,
[tex]v=\frac {2.185\times 10^6}{2}\ m/s[/tex]
[tex]v=1.0925\times 10^6\ m/s[/tex]
Kinetic energy is:
[tex]K.E.=\frac {1}{2}\times mv^2[/tex]
So,
[tex]K.E.=\frac {1}{2}\times 9.1093\times 10^{-31}\times ({1.0925\times 10^6})^2[/tex]
K.E. = 5.4362 × 10⁻¹⁹ J
Question 8 If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is Selected Answer: Correct Answer. D. Mn304 D. Mn304
Answer:
The compound is [tex]Mn_{3}O_{4}[/tex]
Explanation:
The mass percentage of Mn is 72.1% and the mass percentage of O is 27.9%.
The mass percentage of a compound is given by:
[tex]percentage_{A}=\frac{n*MM_{A}}{MM_{C}} *100[/tex]
where:
n is its coefitient in the compund formula
MMa=Molar mass of the element A
MMc=Molar mass of the compound
So, we can figure out which compound is by dividing the percentage by its molar mass
Mn=72.1÷54.938045=1.31239
O=27.9÷15.9994=1.74382
Then, we divide each result by the smaller one (Mn)
Mn=1.31239÷1.31239=1
O=1.74382÷1.31239=1.3287
Each the realation of Mn:O is 1:1.3287
Then we multiply each result by 3:
Mn=1×3=3
O=1.3287×3=3.986≈4
Finally we figure out that the compound has 3 atoms of Mn and 4 atoms O. Result= [tex]Mn_{3}O_{4}[/tex]
Mn3O4 is sometimes used as a starting material in the production of soft ferrites e.g. manganese zinc ferrite, and lithium manganese oxide, used in lithium batteries.
The empirical formula of a compound with 72.1% Mn and 27.9% O is MnO.
Explanation:The empirical formula of a compound can be determined based on its mass percentage composition of different elements. In this case, the compound is composed of 72.1% Mn (manganese) and 27.9% O (oxygen).
To find the empirical formula, we need to consider the relative number of atoms in the compound. Assuming we have 100g of the compound, we have 72.1g of Mn and 27.9g of O.
Using the molar masses of Mn (54.938 g/mol) and O (15.999 g/mol), we can calculate the number of moles of each element:
Moles of Mn = 72.1g / 54.938 g/mol = 1.314 molMoles of O = 27.9g / 15.999 g/mol = 1.744 molNow, we divide the number of moles by the smallest value to get the mole ratio:
Mn:O = 1.314 mol / 1.314 mol = 1:1O:Mn = 1.744 mol / 1.314 mol = 1.326:1Since we want whole-number ratios, we round the ratios to the nearest whole number:
Mn:O = 1:1Therefore, the empirical formula of the compound with 72.1% Mn and 27.9% O is MnO.