Answer : The change in molar entropy of the sample is 10.651 J/K.mol
Explanation :
To calculate the change in molar entropy we use the formula:
[tex]\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p,m}dT}{T}[/tex]
where,
[tex]\Delta S[/tex] = change in molar entropy
n = number of moles = 1.0 mol
[tex]T_f[/tex] = final temperature = 300 K
[tex]T_i[/tex] = initial temperature = 273 K
[tex]C_{p,m}[/tex] = heat capacity of chloroform = [tex]91.47+7.5\times 10^{-2}(T/K)[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=1.0\int\limits^{300}_{273}{\frac{(91.47+7.5\times 10^{-2}(T/K))dT}{T}[/tex]
[tex]\Delta S=1.0\times [91.47\ln T+7.5\times 10^{-2}T]^{300}_{273}[/tex]
[tex]\Delta S=1.0\times 91.47\ln (\frac{T_f}{T_i})+7.5\times 10^{-2}(T_f-T_i)[/tex]
[tex]\Delta S=1.0\times 91.47\ln (\frac{300}{273})+7.5\times 10^{-2}(300-273)[/tex]
[tex]\Delta S=8.626+2.025[/tex]
[tex]\Delta S=10.651J/K.mol[/tex]
Therefore, the change in molar entropy of the sample is 10.651 J/K.mol
Does Biuret test work for most proteins, if not which ones and why?
Answer:
Biuret test does not work for all kinds of proteins. It only works with proteins and peptides that have 2 or more peptide bonds.
Explanation:
The Biuret test is used to determine proteins and polypeptides that contain 2 or more peptide bonds.
It is made of potassium hydroxide and cupric sulfate, along with sodium and potassium tartrate.
It consists of treating a protein with Cu ++ in alkaline medium, producing a violet coloration by forming a coordination complex between the Cu ++ and the free electron pairs of the nitrogens of the amino groups of the peptide junction. At least two peptide bonds are necessary for the reaction to take place.
Therefore, this test does not serve to determine the amount of amino acids and peptides with a single peptide bond such as: tyrosine (amino acid), tryptophan (amino acid), alanine (amino acid), aspartame (peptide)
The energy transferred between samples of matter because of a difference in their temperatures is called a. heat. b. thermochemistry c. chemical kinetics. d. temperature.
Answer:
The energy transferred between samples of matter because of a difference in their temperatures is called a. heat.
Explanation:
The first law of thermodynamics establishes that when two bodies with different temperatures are put in contact they will find thermic equilibrium to a final temperature by transferring heat. Thus the correct answer is (a).
Thermochemistry is the study of the transformations of heat energy on the chemical reactions. Chemical kinetics is the study of the rate of chemical reactions. And temperature is the measure of the heat.
Write 274,541,005,000 in Engineering Notation with 3 significant figures.
Answer : The correct answer is, [tex]0.274\times 10^{12}[/tex]
Explanation :
Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power. It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.
The engineering notation written in the form:
[tex]a\times 10^b[/tex]
where,
a = the number which is greater than 0 and less than 999
b = an integer multiple of 3
For example : [tex]45.89\times 10^3[/tex] or [tex]56.45\times 10^{-6}[/tex]
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
As we are given the 274,541,005,000 in standard notation.
Now converting this into engineering notation, we get:
[tex]\Rightarrow 274,541,005,000=0.274\times 10^{12}[/tex]
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the correct answer is, [tex]0.274\times 10^{12}[/tex]
Final answer:
To write 274,541,005,000 in Engineering Notation with 3 significant figures, we need to move the decimal point so that we have a number between 1 and 10 multiplied by a power of 10. The number becomes 2.75 × 10¹¹.
Explanation:
Engineering Notation is a way of expressing numbers in scientific notation but with the power of 10 always being a multiple of 3. To write 274,541,005,000 in Engineering Notation with 3 significant figures, we need to move the decimal point so that we have a number between 1 and 10 multiplied by a power of 10. The number becomes 2.75 × 10¹¹.
At constant temperature 10.0L of N, at 0.983 atm is compressed to 2.88L.What is the final pressure of N,? 0.283 atm b) 1.51 atm c) 3.41 atm d 29.3 atm a)
Answer:
c) 3.41 atm
Explanation:
We can calculate the final pressure using Boyles Law, P₁V₁ = P₂V₂.
P₁V₁ = P₂V₂
(0.983 atm)(10.0 L) = P₂(2.88 L)
9.83 ÷ 2.88 = P₂
P₂ = 3.41 atm
The final pressure can be determined by using Boyles law and will be 3.41 atm.
What is pressure?The force delivered perpendicularly to a surface of the structure per unit area throughout whom that force would be dispersed is known as pressure.
What is Boyles law?
According to Boyle's law, the relationship between a gas's pressure as well as volume seems to be inverse.
Given data:
[tex]P_{1} = 0.983 atm\\V_{1} = 10 L\\V_{2} = 2.88 L[/tex]
Calculation of pressure.
The formula of Boyles law:
[tex]P_{1} V_{1} = P_{2} V_{2}[/tex]
Put the value of given data:
[tex](0.983) (10 ) = P_{2} (2.88)\\P_{2} = 3.41 atm[/tex]
Therefore, the final pressure will be 3.41 atm.
To know more about Boyle's law and pressure
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calcualte pressure at STP in 10.0 L vessel after reaction of 1.0 L hydrochloride acid (concentration 35% and density 1.28 g/cub.cm) with 1.0 kg of limestone. yield of reaction 97%
Answer:
The pressure in the vessel is 13,3 atm.
Explanation:
The reaction that occurs in vessel (where limestone is 96% of CaCO₃) is:
2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)
The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.
We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:
HCl:
1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol
(Concentration) (L to cm³) (cm³ to g) (g to mol) (moles of reaction)
moles of HCl= 6,14 mol
CaCo₃:
1,0 kg × (96/100) × (1000 g/1kg) × (1 mol/100,09g)
(Limestone) (CaCo₃ in limestone) (kg to g) (g to mol)
moles of CaCo₃= 9,59 mol
So, reactant limit is HCl
This reaction have a yield of 97%. So, the CO₂ moles are:
6,14 mol × 97÷ = 5,96 mol CO₂
The ideal gas formula to obtain pressure is:
P = nRT/V
Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L
Replacing this values in the equation the pressure is
P = 13,3 atm
I hope it helps!
A 1 m ID spherical cooler made of 2 cm thick foam insulation (k = 0.15 Wm-1K-1) is used to store iced water at 0°C. The tank is located outdoors at 25 °C and is subjected to horizontal cross winds at 1 m/s. Determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0 °C that melts during a 24 h period. Assume the inner surface of the cooler is at 0 °C and take the thermal resistance of the cooler and heat transfer by radiation into consideration. Assume the average surrounding surface temperature for radiation exchange to be 20 °C and the outer surface of the cooler to have an emissivity of 0.85. The heat of fusion of water at atmospheric pressure is hif = 333.7 kJ/kg.
The answer is unknown
A sample of 02 gas occupies 346 mL at 45°C and 1.50 atm. What is the volume of this O2 gas sample at STP? Enter your answer in the provided box. L L
Answer: The volume of oxygen gas at STP is 446 mL
Explanation:
STP conditions are:
Pressure of the gas = 1 atm
Temperature of the gas = 273 K
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas
[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas
We are given:
[tex]P_1=1.50atm\\V_1=346mL\\T_1=45^oC=(45+273)K=318K\\P_2=1atm\\V_2=?\\T_2=273K[/tex]
Putting values in above equation, we get:
[tex]\frac{1.50atm\times 346mL}{318K}=\frac{1atm\times V_2}{273K}\\\\V_2=446mL[/tex]
Hence, the volume of oxygen gas at STP is 446 mL
Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 346 mL, [tex]T_{1}[/tex] = [tex]45.0^{o}C[/tex] = (45 + 273) K = 318 K,
[tex]P_{1}[/tex] = 1.50 atm, [tex]V_{2}[/tex] = ?, [tex]T_{2}[/tex] = 273 K,
[tex]P_{2}[/tex] = 1 atm
And, according to ideal gas equation,
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Now, putting the given values into the above formula and we will calculate the final volume as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]\frac{1.50 atm \times 346 mL}{318 K} = \frac{1 atm \times V_{2}}{273 K}[/tex]
[tex]V_{2}[/tex] = 445.56 mL
Thus, we can conclude that the volume of this [tex]O_{2}[/tex] gas sample at STP is 445.56 mL.
When is Raoult's law not a good approximation for a particular compound? When the compound is very dilute in the liquid phase. When the compound has a high mole fraction in the liquid phase. When the compound has a high mole fraction in the vapor phase.
Answer:
Raoult's law is not a good approximation when the compound is very dilute in the liquid phase.
Explanation:
Thermodynamics defines ideal mixture as a mixture in which all the molecules of the different species involved are so similar between each other that intermolecular forces between different molecules are the same to those of similar molecules. Thus, Raoult's law takes this definition and stablish that when a compound is highly concentrated in the solution behaves as if it is in a pure solution.
Also, the Raoult's law stablish that the partial pressure ([tex]P_{i}[/tex]) of every component in an ideal mixture of liquids is equal to the pressure of the pure component ([tex]P_{i}^{*}[/tex]) multiplicated by the molar fraction in the solution of that component ([tex]x_{i}[/tex])
[tex]P_{i} = P_{i}^{*}. x_{i}[/tex]
The answer "When the compound has a high mole fraction in the liquid phase" is wrong because this is the case that follows the Raoult's law. And the answer "When the compound has a high mole fraction in the vapor phase." talks about the fraction in the vapor, and Raoult's law use the molar fraction in the solution.
What is the energy in joules of one photon of
microwaveradiation with a wavelength 0.122m?
Final answer:
One photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules, calculated using Planck's equation (E=hc/λ).
Explanation:
The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy of the photon in joules, h is Planck's constant (6.626 x 10⁻³⁴ J·s), c is the speed of light in a vacuum (3.00 x 10⁸ m/s), and λ is the wavelength of the photon in meters. Using the given wavelength of 0.122 meters for the microwave radiation, we can calculate the energy of one photon by substituting the values into the equation:
E = (6.626 x 10⁻³⁴ J·s)(3.00 x 10⁸ m/s) / 0.122 m
E = 1.63 x 10⁻²⁴ Joules per photon. Therefore, one photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules.
Grunge is a rock style from Detroit. True False
Answer:
FALSE
Explanation:
Grunge refers to the genre of rock music and the fashion inspired by it. It originated in the mid-1980s in Seattle, Washington State.
Grunge was described as the fusion of punk rock and heavy metal.
This genre of music became popular in the early mid-1990s and included lyrics based on the theme of emotional and social alienation, betrayal, abuse, trauma etc.
The statement that grunge is a rock style from Detroit is false; grunge originated in Seattle, Washington. This music genre is characterized by a stripped-down aesthetic, distorted guitars, and socially conscious lyrics, gaining popularity in the late 1980s and early 1990s.
Grunge is indeed a style of rock music, but it is not from Detroit; that statement is false. Grunge, often referred to as the Seattle sound, originated in Seattle, Washington and became widely popular in the late 1980s and early 1990s. It represented a departure from the extravagant stage productions and fashion of contemporary rock bands, favoring a more stripped-down sound with an emphasis on distorted electric guitars and direct, often socially conscious lyrics.
The grunge movement produced several iconic bands, including Nirvana, Pearl Jam, and Soundgarden, which significantly shaped the alternative rock landscape. Seattle's geographic isolation played a pivotal role in the development of grunge music. Local bands evolved their unique sound and aesthetic away from the influence of mainstream music industry expectations, which at the time were centered around Los Angeles. This isolation contributed to grunge's authenticity and appeal when it eventually burst onto the national and international stages.
Contrary to the glamorous rock scene in Los Angeles, grunge musicians typically wore simple, unremarkable clothing, often sourced from second hand stores, and their lyrics addressed themes like political issues, mental health, and substance abuse, resonating with the disaffected Generation X.
Which of the following statements is correct about the use of alcohols as chemical control agents? Isopropyl alcohol is used in a 10% concentration. Ethyl alcohol and isopropyl alcohol are the most commonly used agents. DIpping small instruments in ethyl alcohol for a few seconds will result in disinfection. Alcohols will eliminate bacterial spores.
Answer:
Ethyl alcohol and isopropyl alcohol are the most commonly used agents.
Explanation:
Isopropyl alcohol is used in a 10% concentration.
- not true, to be useful as a chemical control agent its is the most potent with a (water) solution containing 70% isopropyl alcohol.
Ethyl alcohol and isopropyl alcohol are the most commonly used agents.
-true
Dipping small instruments in ethyl alcohol for a few seconds will result in disinfection.
- ethyl alcohol is denaturing proteins, inhibiting metabolic processes so it has bactericidal and fungicidal properties. However the proces is not so fast so the instruments are not disinfected in few seconds.
Alcohols will eliminate bacterial spores.
-alcohols are not sporicidal, however they do inhibit the processes of sporulation and germination but they do not eliminate the spores.
Calculate the molecular weight of a substance. In which the solution of this substance in the water has a concentration of 7 percent by weight, has a freezing point
Equal to -0.89 ° C, set Kf value of water = 1.86 ° C / m
Answer : The molecular weight of a substance is 157.3 g/mol
Explanation :
As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.
Mass of solute = 7 g
Mass of solution = 100 g
Mass of solvent = 100 - 7 = 93 g
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=k_f\times\frac{\text{Mass of substance(solute)}\times 1000}{\text{Molar mass of substance(solute)}\times \text{Mass of water(solvent)}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]T_f^o[/tex] = temperature of pure water = [tex]0^oC[/tex]
[tex]T_f[/tex] = temperature of solution = [tex]-0.89^oC[/tex]
[tex]K_f[/tex] = freezing point constant of water = [tex]1.86^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex](0-(-0.89))^oC=1.86^oC/m\times \frac{7g\times 1000}{\text{Molar mass of substance(solute)}\times 93g}[/tex]
[tex]\text{Molar mass of substance(solute)}=157.3g/mol[/tex]
Therefore, the molecular weight of a substance is 157.3 g/mol
Balance the equation and show the calculation of the number of moles and grams of Ca3(PO42 formed from 10.8 grams of Ca(OH)2. Show your answers to 3 significant figures Ca(OH)2 H3PO4-- Ca3(PO4)2 + H2O]
Answer:
The number of moles of calcium phosphate is 0.0485 mol. The mass is 15.1 g.
Explanation:
The balanced equation is:
[tex]3 Ca(OH)_{2} + 2 H_{3}PO_{4} => Ca_{3}(PO_{4})_{2} + 6H_{2}O[/tex]
As we can see, 3 moles of calcium hidroxide produced 1 mol of calcium phosphate. So the quantity of moles of Calcium posphate is:
[tex]10.8g Ca(OH)_{2}*\frac{1 molCa(OH)_{2}}{74.093gCa(OH)_{2}} *\frac{1mol Ca_{3}(PO_{4})_{2}}{3molCa(OH)_{2}} =0.0485 molCa_{3}(PO_{4})_{2} [/tex]
The mass of calcium phosphate in grams is:
[tex]0.0485 molCa_{3}(PO_{4})_{2}*\frac{310.176g}{mol} =15.1 g[/tex]
You have a 0.5 L of a 1.0 M buffer solution with pka = 5.14 and current pH = 5.23. Calculate the new pH when 1.50 mL of 5.00 M HCl is added. Be mindful of units. (How can you tell if your answer makes sense?) pH = pka + 109 U base 5.23 - 5.14+ log ( )
Answer:
pH = 5,22
Explanation:
A buffer consist in a solution with both conjugate acid (HA) and conjugate base (A⁻). To know the concentration of both A⁻ and HA you use Henderson-Hasselbalch equation:
pH = pka + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]
In the problem:
5,23 = 5,14 + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]
1,23 = [tex]\frac{A^{-} }{HA}[/tex] (1)
Also, you know buffer concentration is 1,0 M:
1,0 M = A⁻ + HA (2)
Replacing (2) in (1):
HA = 0,45 M, thus:
A⁻ = 0,55 M.
The addition of 1.50 mL of 5.00 M HCl represents a concentration of:
1,50x10⁻³ L×[tex]\frac{5,00 mol}{L}[/tex] ÷ 0,5 L = 0,015 M of HCl
The buffer equilibrium is:
HA ⇄ A⁻ + H⁺ ka = [tex]10^{-5,14}[/tex]
The concentrations in equilibrium are:
[HA] = 0,45 M + x
[A⁻] = 0,55 M - x
[H⁺] = 0,015 M -x
Because the equilibrium is displaced to the left.
The equation of equilibrium is:
[tex]10^{5,14}[/tex] = [tex]\frac{[0,015M -x][0,55M-x]}{[0,45M-x]}[/tex]
You will obtain:
x² - 0,565x + 8,24674x10⁻³ = 0
Solving:
x = 0,5500061 ⇒ No physical sense
x = 0,01499391
Thus, [H⁺] = 0,015 - 0,01499391 = 6,09x10⁻⁶
As pH = -log₁₀ [H⁺]
pH = 5,22
I hope it helps!
Calculate the fraction of atom sites that are vacant for silver at 552°C. Assume an energy for vacancy formation of 0.63 eV/atom.
Answer:
[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]
Explanation:
given data:
temperature = 552 degree celcius = 825 Kelvin
energy for vacancy formation is given as 0.63 eV/atom
fraction of atom can be obtained from following formula
[tex]N_C = N e^{\frac{-Q_V}{kT}}[/tex]
Where, K IS BOLTZMAN CONSTANT [tex]= 8.62\times 10^{-5}eV/K[/tex]
[tex]\frac{N_C}{N} =e^{\frac{-0.63}{8.62\times 10^{-5} \times 825}[/tex]
[tex]\frac{N_C}{N} = e^{-8.8858}[/tex]
[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]
The fraction of atom sites that are vacant for silver at 552°C is approximately [tex]\( 7.347 \times 10^{-5} \).[/tex]
The fraction of atom sites that are vacant for silver at a given temperature can be calculated using the formula derived from the statistics of a canonical ensemble:
[tex]\[ f = \exp\left(-\frac{\Delta G_{\text{form}}}{k_B T}\right) \][/tex]
Now, we can plug in the values:
[tex]\[ f = \exp\left(-\frac{0.63 \text{ eV/atom}}{(8.617 \times 10^{-5} \text{ eV/K}) \times (825.15 \text{ K})}\right) \] \[ f = \exp\left(-\frac{0.63}{8.617 \times 10^{-5} \times 825.15}\right) \] \[ f = \exp\left(-\frac{0.63}{0.0658}\right) \] \[ f = \exp\left(-9.574\right) \] \[ f \approx \exp\left(-9.574\right) \] \[ f \approx 7.347 \times 10^{-5} \][/tex]
The head loss in a turbulent flow in a pipe varies Approximant as square of velocity • Direct as the velocity • Invers as square of velocity • Invers as square of dimeter
Answer:
Head loss in turbulent flow is varying as square of velocity.
Explanation:
As we know that head loss in turbulent flow given as
[tex]h_F=\dfrac{FLV^2}{2gD}[/tex]
Where
F is the friction factor.
L is the length of pipe
V is the flow velocity
D is the diameter of pipe.
So from above equation we can say that
[tex]h_F\alpha V^2[/tex]
It means that head loss in turbulent flow is varying as square of velocity.
We know that loss in flow are of two types
1.Major loss :Due to surface property of pipe
2.Minor loss :Due to change in momentum of fluid.
Answer:
The correct answer is head varies directly with square of velocity of flow
Explanation:
The head loss in a pipe as given by Darcy Weisebach equation is
[tex]h_L=\frac{flv^2}{2gD}[/tex]
where
'f' is friction factor whose value depends on the nature of flow (Laminar/turbulent)
'L' is the length of the section in which the head loss is calculated
'v' is the velocity of the flow
'D' is the diameter of the duct
Thus we can see that the head loss varies with square of velocity of the fluid.
How does a buffer resist change in pH upon addition of a strong acid? The strong acid reacts with the strong base in the buffer to form a salt, which produces few H+ ions in solution and therefore only a little change in pH. The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH. The strong acid reacts with the weak acid in the buffer to form a weak base, which produces few H+ ions in solution and therefore only a little change in pH.
Answer:
The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH.
Explanation:
When a strong acid is added to the buffer, the acid dissociates and furnish hydrogen ions which combine with the conjugate of the weak acid, forming weak acid. The weak acid dissociates to only some extent and can furnish only some protons and there is no significant change in the pH.
Hence, option B is correct.
In a buffer, a strong acid reacts with the weak base to form a weak acid, yielding fewer H+ ions, so only a slight change in pH is noted. Buffer solutions, including weak conjugate acid-base pairs, resist pH changes. However, the buffering ability may be disrupted if large amounts of acid or base are added, a concept known as buffer capacity.
Explanation:A buffer essentially resists changes in pH when an acid or base is added to the solution. This is due to the presence of a weak conjugate acid-base pair. In the buffer, a strong acid will react with the weak base to form a weak acid, thus producing fewer H+ ions. This, in essence, leads to only a slight change in pH.
A typical example of a buffer might include a solution of acetic acid and sodium acetate. This buffer consists of a weak acid and its salt. Similarly, a solution of ammonia and ammonium chloride could serve as a buffer consisting of a weak base and its salt.
It's crucial to note, however, that buffers do not have infinite capacity to resist pH changes. This is known as the buffer capacity. If a large amount of acid or base is added to the buffer solution, potentially lowering the concentration of the conjugate acid or base pair, the ability of the buffer to regulate pH may be disrupted.
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You have 3.27g of O2. How many grams of CO2 can be made?
Answer:
Amount of Carbon dioxide equals 4.49625 grams.
Explanation:
From the basic stichometric reaction between carbon and oxygen we know that 1 mole of carbon combines with 1 mole of oxygen to form 1 mole of carbon dioxide.
Thus we can say that 12 grams of carbon combines with 32 grams of oxygen to form 44 grams of carbon dioxide.
In the given question assuming that there is no limited supply of carbon we can find the find the amount of carbon dioxide formed from 3.27 grams of Oxygen using ratio and proportion method.
As we can see that 32 grams of oxygen form 44 grams of carbon dioxide thus we can say 1 gram of oxygen yields [tex]\frac{44}{32}grams[/tex] Carbon Dioxide
Thus the carbon dioxide formed by 3.27 grams of Oxygen equals
[tex]3.27\times \frac{44}{32}=4.49625grams[/tex]
A chemist determines by measurements that 0.0200 moles of iodine solid participate in a chemical reaction. Calculate the mass of iodine solid that participates. Round your answer to 3 significant digits. $ x 6 ?
To calculate the mass of iodine solid that participates in a chemical reaction, multiply the number of moles of iodine by its molar mass.
Explanation:To calculate the mass of iodine solid that participates in a chemical reaction, we need to use the concept of moles and the molar mass of iodine. Given that 0.0200 moles of iodine solid participate in the reaction, we can convert moles to grams by multiplying the number of moles by the molar mass of iodine. The molar mass of iodine (I2) is 253.8 grams per mole. Multiplying 0.0200 moles by 253.8 grams/mole gives us a mass of 5.08 grams of iodine solid that participates in the chemical reaction.
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Calculate the volume of 8,410 g of a substance whose density is 0.71 g/mL. Express your answer in milliliters using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
12 mL
Explanation:
Density is expressed as mass divided by volume, so the expression can be rearranged to solve for volume:
D = m/V ⇒ V = m/D
V = m/D = (8.410 g)/(0.71g/mL) = 12 mL
Final answer:
The volume of a substance with a mass of 8,410 g and a density of 0.71 g/mL is 12,000 mL, rounded to two significant figures to match the precision of the given density.
Explanation:
To calculate the volume of a substance, you can use the density formula: Density = Mass / Volume. Given the density of the substance is 0.71 g/mL and the mass is 8,410 g, the volume can be found by rearranging the formula to Volume = Mass / Density.
Volume = 8,410 g / 0.71 g/mL = 11,845.07042 mL. This raw calculation has more significant figures than is justified by the precision of the given data. Since the density value has two significant figures, the volume should also be reported with two significant figures, resulting in 12,000 mL.
Consider the following reaction: A+B C+D According to Le Chatelier's Principle what would happen if you added more D to the reaction? From the point of view of the molecules and atoms why would the reaction respond this way? Your answer may use a maximum of 15 words. Using more than 15 words will result in a 0 score
Explanation:
According to the Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the direction of equilibrium where there is less stress or disturbance.
For example, [tex]A + B \rightleftharpoons C + D[/tex]
So, when we increase the concentration of D in this reaction then disturbance will increase on the product side.
And, frequency of successful collisions between the reactant atoms will decrease. Hence, there will be less stress on reactant side.
As a result, equilibrium will shift in the backward direction that is, on the reactant side it will shift.
A worker is told her chances of being killed by a particular process are 1 in every 300 years. Should the worker be satisfied or alarmed? What is the FAR (assuming normal working hours) and the deaths per person per year? What should her chances be, assuming an average chemical plant?
Answer:
(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.
(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.
(c) The expected number of fatalities on a average chemical plant are one in 12500 years.
Explanation:
Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.
In 300 years we have 600,000 hours.
With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.
If we have 2000 work hours a year, it is expected 0.0033 deaths/year.
[tex]1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year[/tex]
The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).
In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:
[tex]FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8} hours=1.67*10^{2} =167[/tex]
A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.
If we assume FAR=4, as in an average chemical plant, we expect
[tex]4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}[/tex]
This is equivalent to say
[tex]\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}[/tex]
The expected number of fatalities on a average chemical plant are one in 12500 years.
In a combustion furnace, 2094 standard ft3 per hour
of natural gas (Methane) is burned with 6% excess air.
How many standard ft3 of air are drawn from outside
per hour by the fan that supplies the air?
Explanation:
The chemical reaction is as follows.
[tex]CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O[/tex]
It is given that 2094 [tex]ft^{3}/hr[/tex]. And, it is known that 1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]
Hence, convert 2094 [tex]ft^{3}/hr[/tex] into [tex]m^{3}/s[/tex] as follows.
[tex]\frac{2094 ft^{3}/hr}{127133 ft^{3}/hr} \times 1 m^{3}/s[/tex]
= [tex]0.0165 m^{3}/s[/tex]
As ideal gas equation is PV = nRT. So, calculate the number of moles as follows.
n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{1 atm \times 0.0165 m^{3}}{0.0821 \times 298 K}[/tex]
= 0.673 mol/sec
According to the stoichiometry of the given reaction, 1 mol of methane reacts with 2 mol of oxygen.
So, 1 mol [tex]CH_{4}[/tex] = 2 mol [tex]O_{2}[\tex]
Hence, [tex]O_{2}[\tex] required theoretically = [tex]2 \times 0.673 mol/s[/tex] = 1.346 mol/s
Hence, air required theoretically = [tex]\frac{1.346}{0.21}[/tex] = 6.4095 mol/s.
Since, 6% of excess air is being supplied. Therefore, total air supplied will be calculated as follows.
Total air supplied = [tex]6.4095 mol/s [1 + \frac{6}{100}][/tex]
= 6.794 mol/s
Now, calculate the volume using ideal gas law equation as follows.
PV = nRT
[tex]1 atm \times V = 6.794 mol/s \times 8.21 \times 10^{-5} Latm/K mol \6 times 298 K[/tex]
V = 0.166229 [tex]m^{3}/s[/tex]
Converting calculated volume into [tex]ft^{3}/hr[/tex] as follows.
1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]
So, 0.166229 [tex]m^{3}/s[/tex] = [tex]0.166229 m^{3}/s \times 127133 \frac{ft^{3}/hr}{1 m^{3}/s}[/tex]
= 21133.191 [tex]ft^{3}/hr[/tex]
Thus, we can conclude that 21133.191 [tex]ft^{3}/hr[/tex] of air are drawn from outside per hour by the fan that supplies the air.
The wavelength of the bright red line in the spectrum of atomic
hydrogen is 652 nm. What is the energy of the photon generated in
the transition?
Answer:
The energy of the photon generated in the transition is 3.14*10⁻¹⁹ J
Explanation:
There are two equation that we need to use in order to solve this problem:
The first one is Planck's equation, which describes the relationship between energy and frequency:
E = h*v eq. 1)
Where E is energy, h is Planck's constant (6.626 * 10⁻³⁴ J*s) and v is the radiation frequency.
In order to know the frequency, we use the second equation, which is the wave equation:
c = λ*v eq. 2)
Where c is the speed of light in vacuum (aprx 3 * 10⁸ m/s), and λ is the wavelength. If we solve that equation for v we're left with
v=c/λ eq. 3)
We replace v in eq. 1):
E= c*h/λ
Lastly we put the data we know and solve the equation, keeping in mind the correct use of units (converting 652 nm into m):
[tex]E=\frac{3*10^{8}ms^{-1} *6.626*10^{-34}Js}{633*10^{-9}m }=3.14*10^{-19}J[/tex]
Final answer:
The energy of the photon generated in the transition using the given formula. Calculating gives approximately 3.04 x 10^-19 joules.
Explanation:
The energy of the photon generated in the transition can be calculated using the formula:
E = hc / λ
Where E is the energy of the photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the light. Substituting the given values:
E = (6.63 x 10^-34 J s x 3 x 10^8 m/s) / (652 x 10^-9 m)
Calculating this gives the energy of the photon as approximately 3.04 x 10^-19 joules.
The change in enthalpy for the reaction A - P is x kJ/mol. What does the enthalpy change for the reaction P -A? (A) -x kJ/mol (B) +x kJ/mol (C) 0 kJ/mol (D) -2x kJ/mol
Answer : The correct option is, (A) -x kJ/mol
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction is,
[tex]A\rightarrow P[/tex] [tex]\Delta H=x\text{ kJ/mole}[/tex]
Now we have to determine the value of [tex]\Delta H[/tex] for the following reaction i.e,
[tex]P\rightarrow A[/tex] [tex]\Delta H'=?[/tex]
According to the Hess’s law, if we reverse the reaction then the sign of [tex]\Delta H[/tex] change.
So, the value [tex]\Delta H'[/tex] for the reaction will be:
[tex]\Delta H'=-(x\text{ kJ/mole})[/tex]
[tex]\Delta H'=-x\text{ kJ/mole}[/tex]
Hence, the value of [tex]\Delta H[/tex] for the reaction is -x kJ/mole.
A 1.92 mol sample of hydrogen gas occupies 22.2 L. How many mol of hydrogen are needed to fill a 102.1 L container at the same pressure and temperature? Enter your answer in the provided box. I mol H2
Answer:
8.83 moles of Hydrogen
Explanation:
We are told that 1.92 moles of H2 occupies 22.2 L at a certain pressure and temperature, and we are asked to calculate the quantity of moles of H2 are necessary to fill a container of 102.1 L. So, if the pressure and temperature are constant, then the equality expressed before 1.92 H2 moles = 22.2 L maintains. Therefore, we can calculate what is being asked as follows:
22.2 L ----- 1.92 H2 moles
102.1 L ---- x = (102.1 L × 1.92 H2 moles)/22.2 L = 8.83 H2 moles
This means that 8.83 H2 moles are necessary to fill a 102.1 L container at a certain pressure and temperature.
If 5.00 g helium gas is added to a 1.00 L balloon containing 1.00 g of helium gas, what is the new volume of the balloon? Assume no change in temperature or pressure. Enter your answer in the box provided. L
Explanation:
The given data is as follows.
[tex]mass_{1}[/tex] = 5 g, [tex]Volume_{1}[/tex] = 1 L
[tex]mass_{2}[/tex] = 1 g, [tex]Volume_{1}[/tex] = ?
No. of moles of helium present in 5 g helium gas are as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{5 g}{4 g/mol}[/tex]
= 1.25 mol
No. of moles of helium present in 1 g helium gas are as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1 g}{4 g/mol}[/tex]
= 0.25 mol
According to the ideal gas equation, PV = nRT. And, since temperature and pressure are held constant. Therefore,
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]
[tex]\frac{1 L}{1.25 mol} = \frac{V_{2}}{0.25 mol}[/tex]
[tex]V_{2}[/tex] = 0.2 L
Thus, we can conclude that the new volume of the balloon is 0.2 L.
The final volume of the balloon after adding 5.00 g of helium to the already present 1.00 g of helium, under constant temperature and pressure, is 6.00 L.
Explanation:The subject of your question pertains to the ideal gas law which is represented by the formula PV = nRT, where P is pressure, V is volume, n is the number of moles, R refers to the gas constant, and T is temperature. Since the question states that there is no change in temperature or pressure, you can use Avogadro's law which states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules.
In your problem, you are adding more helium to the balloon. Therefore, the volume of the balloon will increase proportionally. Since the initial volume of 1.00 L corresponds to 1.00 g of Helium, 5.00 g of Helium will correspond to 5.00 L. This is because the amount of Helium has increased by a factor of 5, and thus, the volume will also increase by a factor of 5.
Therefore, the final volume of the balloon will be the sum of the initial volume and the increase, which equals 1.00 L + 5.00 L = 6.00 L.
Learn more about Ideal Gas Law here:https://brainly.com/question/30458409
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how could you easily and accurately determine the glass
transition and melting temperature of a polymer?
Answer:
You can use a thermal analysis called Differential Scanning Calorimetry (DSC)
Explanation:
The glass transition temperature (Tg) and melting temperature (Tm) are thermal properties of substances, such as polymers. You can determine them with a very low sample quantity by differential scanning calorimetry (DSC).
In DSC, there are two aluminium pans. In one pan, you put the sample, and the other pan is empty because is the reference pan. The pans are placed in a chamber of a DSC calorimeter and they are heated at a temperature range you select at a given speed (usually in ºC/min). The system register the difference of heat between the pans (sample pan and reference pan) and it shows you a plot of heat flow (in J/s) vs temperature.
Tg is commonly detected as a step in the plot
Tm is usually an endothermic peak (upwards in the plot), because is a process in which heat is absorbed.
Differential scanning calorimetry (DSC) under quasi-isothermal conditions is utilized to accurately determine the glass transition temperature and melting temperature of a polymer, reflecting its chain interactions and molecular weight.
Determining Glass Transition and Melting Temperature
To easily and accurately determine the glass transition temperature (Tg) and melting temperature (Tm) of a polymer, differential scanning calorimetry (DSC) is the most commonly used technique. The glass transition is a measure of the change in the polymer's heat capacity (Cp) as it crosses certain heat energy thresholds, and it can be more accurately determined under quasi-isothermal conditions as these conditions produce precise Cp measurements as a function of time. When a polymer is heated to its Tg, it enters a molten state, and upon cooling, becomes brittle. If the heating continues beyond Tg, the polymer will soften, allowing it to be molded or deformed.
The glass transition temperature can also provide insight into a polymer's chain length and structure, as it reflects the number of van der Waals or entangling chain interactions. Additionally, the molecular weight of a polymer influences Tg, typically following the Flory-Fox equation. The melting temperature (Tm) is the temperature at which a polymer transitions from a crystalline to a molten state and is detectable as a peak on a DSC trace above Tg.
Overall, by using DSC and applying quasi-isothermal steps across the apparent glass transition range, we can observe changing Cp rates that indicate the exact Tg range. Similarly, for Tm, we can detect the transition between crystalline and molten states as the temperature rises above Tg.
Liquid dimethyl disulfide (CH3SSCH3) flows through a pipe with a mass flow rate of 86.0 g's. Given that the density of dimethyl disulfide is 1.0625 g/cm, find: The molar flow rate in molimin: Number mol/min The volumetric flow rate in L/hr.
Explanation:
Molar mass of [tex]CH_{3}SSCH_{3}[/tex] is 94 g/mol. As it is known that number of moles is equal to mass of a substance divided by its molar mass.
Then, calculate the number of moles as follows.
No. of moles = [tex]\frac{86.0 g}{94 g/mol}[/tex] in 1 s
= 0.914 mol
So, in 60 sec number of moles will be equal to 0.914 x 60 = 54.89 mol/min.
Hence, the molar flow rate = 54.89 mol/min
Also, density is equal to mass of a substance divided by its volume.
Density = [tex]\frac{mass}{volume}[/tex]
Volume = [tex]\frac{mass}{Density}[/tex]
= [tex]\frac{86.0 g}{1.0625 g/cm^{3}}[/tex]
= 80.941 [tex]cm^{3}[/tex]
As, 80.941 [tex]cm^{3}[/tex] of volume flows in 1 s . Therefore, flow of volume in 1 hour will be calculated as follows.
In 1 hr = 80.941 [tex]cm^{3} \times 3600[/tex]
= 291388.24 [tex]cm^{3}/hr[/tex]
Since, 1 [tex]cm^{3}[/tex] = 0.001 L.
So, 291388.24 [tex]cm^{3}/hr \times 0.001 L/cm^{3}[/tex]
= 291.38824 L/hr
Thus, we can conclude that molar flow rate in mol/min is 54.89 mol/min and the volumetric flow rate in L/hr is 291.38824 L/hr.
Find the molar concentration of sulfuric acid in a 20.0 wt% solution of sulfuric acid in water (SG=1.13).
Answer: The molarity of sulfuric acid solution is 2.3 M
Explanation:
The relationship between specific gravity and density of a substance is given as:
[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]
Specific gravity of sulfuric acid solution = 1.13
Density of water = 1.00 g/mL
Putting values in above equation we get:
[tex]1.13=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.13\times 1.00g/mL)=1.13g/mL[/tex]
We are given:
20% (m/m) sulfuric acid solution. This means that 20 g of sulfuric acid is present in 100 g of solution
To calculate volume of a substance, we use the equation:[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.13 g/mL
Mass of Solution = 100 g
Putting values in above equation, we get:
[tex]1.13g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.13g/mL}=88.5mL[/tex]
To calculate the molarity of solution, we use the equation:[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Mass of solute (sulfuric acid) = 20 g
Molar mass of sulfuric acid = 98 g/mol
Volume of solution = 88.5 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{20g\times 1000}{98g/mol\times 88.5mL}\\\\\text{Molarity of solution}=2.3M[/tex]
Hence, the molarity of sulfuric acid solution is 2.3 M