Answer: The given mass in kilograms is 0.00399 kg
Explanation:
Gram and Kilograms are the units which are used to express the mass of a substance. These units are inter changeable.
We are given:
Mass of a substance = 3.99 g
To convert the given mass into kilograms, we use the conversion factor:
1 kg = 1000 g
Converting the given value, we get:
[tex]\Rightarrow 3.99g\times (\frac{1kg}{1000g}=0.00399kg[/tex]
Hence, the given mass in kilograms is 0.00399 kg
Convert 3.15 x 10 m to the equivalent length in nanometers. 3.15 x 10-ºm =
Answer:
∴ 3.15 * 10 m = 3.15 E10 nm
Explanation:
1 m ≡ 1 E9 nm⇒ 3.15 * 10 m = 31.5 m * ( 1 E9 nm /m ) = 3.15 E10 nm
If the half-life of a radioisotope is 10,000 years, the amount remaining after 20,000 years is
none
half
a fourth
double
answer is c
Answer:
Considering the half-life of 10,000 years, after 20,000 years we will have a fourth of the remaining amount.
Explanation:
The half-time is the time a radioisotope takes to decay and lose half of its mass. Therefore, we can make the following scheme to know the amount remaining after a period of time:
Time_________________ Amount
t=0_____________________x
t=10,000 years____________x/2
t=20,000 years___________x/4
During the first 10,000 years the radioisotope lost half of its mass. After 10,000 years more (which means 2 half-lives), the remaining amount also lost half of its mass. Therefore, after 20,000 years, the we will have a fourth of the initial amount.
Final answer:
After 20,000 years, or two half-lives, one-fourth of the original amount of a radioisotope with a half-life of 10,000 years would remain.
Explanation:
If the half-life of a radioisotope is 10,000 years, the amount of the isotope remaining after 20,000 years would be one-fourth of the original amount. This can be understood by realizing that with each half-life period, the quantity of the isotope is reduced by half. After the first 10,000 years, one half-life, half of the isotope will have decayed leaving 50%. After another 10,000 years, which makes two half-lives in total, half of the remaining 50% will have decayed, leaving 25% of the original amount. Hence, the correct answer to the question is a fourth, or 25%.
Perform unit conversions and determine Re for the case when a fluid with density of 92.8 lbm/ft3 and viscosity of 4.1 cP (centipoise) flows with a velocity of 237 ft/min in a pipe with 28 inch diameter.
Re = (density*velocity*diameter)/viscosity
Answer:
Re=309926.13
Explanation:
density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3
viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s
velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s
diameter=28inch*(0.0254m/1inch)=0.71m
Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s
Re=309926.13
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.
Spermine (structure shown) is a compound isolated from sperm. Which of the following statements correctly describes an aqueous solution of spermine?
NH2(CH2)3NH(CH2)4NH(CH2)3NH2
Question 2 options:
A)
The hydroxide ion concentration in the solution would be greater than that in pure water.
B)
The hydronium ion concentration in the solution would be greater than that in pure water.
C)
The solution would be acidic.
D)
The pH of the solution would be equal to that of pure water.
Answer:
A) The hydroxide ion concentration in the solution would be greater than that in pure water.
Explanation:
Spermine has four amino groups. In water, these amino groups will be protonate -In NH₂⁺ or NH₃⁺- decreasing the H⁺ concentration and, by water equilibrium, increasing OH⁻ concentration.
Thus, in an aqueous solution of spermine the hydroxide ion concentration in the solution would be greater than that in pure water.
I hope it helps!
Answer:
The correct answer is option A) The hydroxide ion concentration in the solution would be greater than that in pure water.
Explanation:
Spermine is basic in nature as it contains di-amine group (NH2), which is basic in nature that is, it easily gives off electrons.
Thus being basic spermine solution will have more concentration of Hydroxide ion [OH-].
There is inverse relation between pressure and diffusion in liquid phase?
•True
•False
Answer:
True
Explanation:
There is inverse relation between pressure and diffusion in liquid phase is true statement because ,with the increase in pressure of the liquid phase, the material in the pores of the find it difficult to penetrate and therefore it is difficult for them to diffuse. Hence, there is an inverse relation between pressure and diffusion.
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]
Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg C. The densities of hexane and octane are 0.655 g/cm3 and 0.703 g/cm3, respectively. Assume you have 1.0 L of octane. Calculate the required volume of hexane. Report your answer in liters.
Answer:
The required volume of hexane is 0.66245 Liters.
Explanation:
Volume of octane = v=1.0 L=[tex]1000 cm^3[/tex]
Density of octane= d = [tex]0.703 g/cm^3[/tex]
Mass of octane ,m= [tex]d\times v=0.703 g/cm^3\times 1000 cm^3=703 g[/tex]
Moles of octane =[tex]\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol[/tex]
Mole percentage of Hexane = 45%
Mole percentage of octane = 100% - 45% = 55%
[tex]55\%=\frac{6.166 mol}{\text{Total moles}}\times 100[/tex]
Total moles = 11.212 mol
Moles of hexane :
[tex]45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100[/tex]
Moles of hexane = 5.0454 mol
Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g
Density of the hexane,D = [tex]0.655 g/cm^3[/tex]
Volume of hexane = V
[tex]V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L[/tex]
(1 cm^3= 0.001 L)
The required volume of hexane is 0.66245 Liters.
Consider the gas reaction: H2 (g)2(g) 2HI (g) The equilibrium constant at 731 K is 50.3. Equal amounts of all three gases (0.100 M) are introduced in a container, calculate the concentration of each gas after the system reaches equilibrium. Express your results with the right number of significant figures.
Answer : The concentration of [tex]H_2,I_2[/tex] and [tex]HI[/tex] at equilibrium 0.033 M, 0.033 M and 0.234 M respectively.
Solution : Given,
[tex]K_c=50.3[/tex]
Concentration = 0.100 M
The given equilibrium reaction is,
[tex]H_2(g)+I_2(g)\rightleftharpoons 2HI(g)[/tex]
Initially conc. 0.100 0.100 0.100
At equilibrium (0.100-x) (0.100-x) (0.100+2x)
The expression of [tex]K_c[/tex] will be,
[tex]K=\frac{[HI]^2}{[H_2][I_2]}[/tex]
Now put all the given values in this expression, we get:
[tex]50.3=\frac{(0.100+2x)^2}{(0.100-x)\times (0.100-x)}[/tex]
By solving the term x, we get
[tex]x=0.067\text{ and }0.158[/tex]
From the values of 'x' we conclude that, x = 0.158 can not more than initial concentration. So, the value of 'x' which is equal to 0.158 is not consider.
Thus, the concentration of [tex]H_2[/tex] and [tex]I_2[/tex] at equilibrium = (0.100-x) = 0.100 - 0.067 = 0.033 M
The concentration of [tex]HI[/tex] at equilibrium = (0.100+2x) = 0.100 + 2(0.067) = 0.234 M
A liquid mixture of density 0.85 g/mL flowing at a rate of 700 lbm/h is formed from mixing a stream of benzene S = 0.879 and another of n-hexane S = 0.659. Determine the volumetric flow rates of all three streams as well as the mass flow rate of benzene and n-hexane.
Answer:
Volumetric flow:
Mixture = 373.55 l/h; Benzene = 310.05 l/h; n-hexane = 63.5 l/h
Mass flow
Mixture = 317.52 kg/h; Benzene = 275.66 kg/h; n-hexane = 41.85 kg/h
Explanation:
First, we can express the mixture mass flow as
[tex]M=700\frac{lbm}{h}*\frac{453.6g}{1lbm} = 317,520g/h=317,52kg/h[/tex]
The volumetric flow of the mixture is
[tex]V=M/\rho = \frac{317520g/h}{0.85g/ml} =373,553ml/h=373.55l/h[/tex]
The mass balance can be written as
[tex]M_1+M_2=M\\[/tex]
And the volumetric flow as
[tex]V_1+V_2=V\\\\\rho_1*M_1+\rho_2*M_2=\rho*M\\\\\rho_1(M-M_2)+\rho_2*M_2=\rho*M\\\\(\rho_2-\rho_1)*M_2=(\rho-\rho_1)*M\\\\M_2=\frac{(\rho-\rho_1)}{(\rho_2-\rho_1)} *M[/tex]
If fluid 2 is n-hexane, we have
[tex]M_2=\frac{(\rho-\rho_1)}{(\rho_2-\rho_1)} *M\\\\M_2=\frac{(0.85-0.879)}{(0.659-0.879)} *317.52kg/h\\\\M_2=0.1318*317.52kg/h=41.85kg/h[/tex]
The mass flow of n-hexane is 41.85 kg/h.
Its volumetric flow is 63.5 litre/h
[tex]V=M/\rho=41.85\frac{kg}{h}*\frac{1}{0.659g/ml} *\frac{1000g}{1kg}\\\\V = 63505 ml/h=63.5 l/h[/tex]
The mass flow of benzene can be deducted by difference:
[tex]M_1= M-M_2= 317.52-41.85=275.66 kg/h\\\\V_1=V-V_2= 373.55 - 63.5=310.05 l/h[/tex]
A permeation membrane separates an inlet air stream, F, (79 mol% N2, 21 mol% O2), into a permeate stream, M, and a reject stream, J. The inlet stream conditions are 293 K, 0.5 MPa, and 2 mol/min; the conditions for both outlet streams are 293 K and 0.1 MPa. If the permeate stream is 50 mol% O2, and the reject stream is 13 mol% O2, what are the volu- metric flowrates (L/min) of the two outlet streams?
Answer:
Permeate stream: M=10,47 L/minReject stream: J=38,24 L/minExplanation:
Using the law of mass conservation, we have that the moles of any component in the inlet air stream equals the moles of this compenent in the two outlets stream. Being A the mole flowrate for the permeate stream and B the mole flowrate for the reject stream (both in mol/min), we have equation (1) and (2):
Oxygen:2 mol/min * 0,21 = A mol/min * 0,50 + B mol/min * 0,13 ......................(1)
0,42 = 0,5A + 0,13B...........................(1)
Nitrogen:2 mol/min * 0,79 = A mol/min * 0,50 + B mol/min * 0,87 ......................(2)
1,58 = 0,5A + 0,87B...........................(2)
Solving this system:
1,58 = 0,5A + 0,87B...........................(2)
0,42 = 0,5A + 0,13B...........................(1)
1,16 = 0A + 0,74B...............................(2) - (1)
1.16=0,74B
1,16/0,74=B
B=1,57 mol/min
And now replacing B in equation (1):
1,58 = 0,5A + 0,87*1,57
1,58-1,37=0,5A
0,21/0,5=A
A=0,43 mol/min
To calculate the volumetric flowrate of the permeate stream (M) and the reject stream (J), we just replace the given and calculate data into the ideal gas equation:
PV=nRT,.................... V = nRT/P
Permeate stream:[tex]M= \frac{0,43 \frac{mol}{min}*8,314\frac{m3Pa}{molK}*293K }{0,1 MPa*\frac{10^{6}Pa }{1 MPa} }=0,01047\frac{m^{3} }{min} * \frac{1000 L}{1 m^{3} }=10,47 L/min[/tex]
Rejct stream:[tex]J= \frac{1,57 \frac{mol}{min}*8,314\frac{m3Pa}{molK}*293K }{0,1 MPa*\frac{10^{6}Pa }{1 MPa} }=0,0382\frac{m^{3} }{min} * \frac{1000 L}{1 m^{3} }=38,24 L/min[/tex]
a Draw a Lewis structure for CH_4 Explicitly draw all H atoms. Include all valence lone pairs in your answer. Include all nonzero formal charges. C opy P aste H ** H н. H H
Answer : The Lewis-dot structure of [tex]CH_4[/tex] is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, [tex]CH_4[/tex]
As we know that carbon has '4' valence electrons and hydrogen has '1' valence electron.
Therefore, the total number of valence electrons in [tex]CH_4[/tex] = 4 + 4(1) = 8
According to Lewis-dot structure, there are 8 number of bonding electrons and 0 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :
[tex]\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}[/tex]
[tex]\text{Formal charge on C}=4-0-\frac{8}{2}=0[/tex]
[tex]\text{Formal charge on }H_1=1-0-\frac{2}{2}=0[/tex]
[tex]\text{Formal charge on }H_2=1-0-\frac{2}{2}=0[/tex]
[tex]\text{Formal charge on }H_3=1-0-\frac{2}{2}=0[/tex]
[tex]\text{Formal charge on }H_4=1-0-\frac{2}{2}=0[/tex]
Hence, the Lewis-dot structure of [tex]CH_4[/tex] is shown below.
Question 6 1.75 pts The following reaction 2H2S(g)=2H2(g)+S2(g), Kc=1.625x10-7 at 800°C is carried out at the same temperature with the following initial concentrations: [H,S]=0.162M, [H2]=0.184 M, and [S2]=0.00 M. Find the equilibrium concentration of S2. Express the molarity to three significant figures. Answer in units of nM.
Answer:
[S₂] = 1.27×10⁻⁷ M
Explanation:
2 H₂S(g) ⇄ 2 H₂(g) + S₂(g), Kc=1,625x10⁻⁷
The equation of this reaction is:
1,625x10⁻⁷ = [tex]\frac{[H_2]^2[S_2]}{[H_{2}S]^2}[/tex]
The equilibrium concentrations are:
[H₂S] = 0,162 - 2x
[H₂] = 0,184 + 2x
[S₂] = x
Replacing:
1,625x10⁻⁷ = [tex]\frac{[0,184+2x]^2[x]}{[0,162-2x]^2}[/tex]
Solving:
4x³ + 0,736x² + 0,033856x - 4,3x10⁻⁹
x = 1.27×10⁻⁷
Thus, concentration of S₂ is:
[S₂] = 1.27×10⁻⁷ M
Which of the following is not a postulate of the kinetic molecular theory? Select one: a. Gas particles have most of their mass concentrated in the nucleus of the atom. b. The moving particles undergo perfectly elastic collisions with the walls of the container. c. The forces of attraction and repulsion between the particles are insignificant. d. The average kinetic energy of the particles is directly proportional to the absolute temperature. e. All of the above are postulates of the kinetic molecular theory.
Answer:
a. Gas particles have most of their mass concentrated in the nucleus of the atom
Explanation:
The main postulates of the molecular kinetic theory are:
A gas consists of a set of small particles that move with rectilinear motion and obey Newton's laws (this means that this theory doesn't divide the atom in a nucleus and electrons, this is why a. is not correct)
The molecules of a gas do not occupy volume (this is assumed because between the molecules of a gas is a lot of space).
Collisions between the molecules are perfectly elastic (this means that energy is not gained or lost during the crash).
There're no forces of attraction or repulsion between the molecules (because each molecule is "far" of the others).
The average kinetic energy of a molecule is [tex]E_k=\frac{3}{2} \times k\times T[/tex] (where T is the absolute temperature and k the Boltzmann constant. This means that kinetic energy is directly proportional to temperature).
The kinetic molecular theory of gases has several postulates that describe the behavior of gas particles.
Explanation:The kinetic molecular theory of gases has several postulates that describe the behavior of gas particles. The postulates are as follows:
Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size. The particles of a gas may be either atoms or molecules.Gas particles are in constant rapid motion in random directions. The fast motion of gas particles gives them a relatively large amount of kinetic energy.Collisions between gas particles and between particles and the container walls are elastic collisions. An elastic collision is one in which there is no overall loss of kinetic energy.There are no forces of attraction or repulsion between gas particles. It is assumed that the particles of an ideal gas have no such attractive forces.The average kinetic energy of gas particles is dependent upon the temperature of the gas. As the temperature of a sample of gas is increased, the speeds of the particles are increased, resulting in an increase in the kinetic energy of the particles.Based on these postulates, the answer to the question is:
a. Gas particles have most of their mass concentrated in the nucleus of the atom
A major component of gasoline is octane . When liquid octane is burned in air it reacts with oxygen gas to produce carbon dioxide gas and water vapor. Calculate the moles of carbon dioxide produced by the reaction of of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Final answer:
The balanced equation for the combustion of octane is 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O. From this equation, we can determine the moles of carbon dioxide produced from a given amount of octane.
Explanation:
The balanced equation for the combustion of octane is:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
From the equation, we can see that for every 2 moles of octane burned, 16 moles of carbon dioxide are produced. Therefore, if we know the number of moles of octane burned, we can calculate the moles of carbon dioxide produced.
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
Cyclopropane, C3H6, is used as a general anesthetic. If a sample of cyclopropane stored in a 2.36-L container at 10.0 atm and 25.0°C is transferred to a 7.79-L container at 5.56 atm, what is the resulting temperature? Enter your answer in the provided box
Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 2.36 L, [tex]T_{1}[/tex] = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K,
[tex]P_{1}[/tex] = 10.0 atm, [tex]V_{2}[/tex] = 7.79 L,
[tex]T_{2}[/tex] = ?, [tex]P_{2}[/tex] = 5.56 atm
And, according to ideal gas equation,
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Hence, putting the given values into the above formula to calculate the value of final temperature as follows.
[tex]\frac{10.0 atm \times 2.36 L}{298 K} = \frac{5.56 atm \times 7.79 L}{T_{2}}[/tex]
[tex]T_{2}[/tex] = [tex]\frac{43.3124}{0.0792}[/tex] K
= 546.87 K
Thus, we can conclude that the final temperature is 546.87 K.
Final answer:
The resulting temperature when transferring cyclopropane from one container to another is approximately 12.4 °C.
Explanation:
To find the resulting temperature when transferring cyclopropane from one container to another, we can use the ideal gas law: PV = nRT. We are given the initial and final pressures and volumes, and we need to find the final temperature.
First, we can calculate the initial number of moles using the ideal gas law: n1 = (P1 * V1) / (R * T1). Then, we can use this number of moles to calculate the final temperature using the ideal gas law: T2 = (P2 * V2) / (n1 * R).
Substituting the given values, we have: T2 = (5.56 atm * 7.79 L) / ((10.0 atm * 2.36 L) / (0.0821 atm L/mol K)). Solving this equation, we find that the resulting temperature is approximately 12.4 °C.
Calculate the frequency of the light emitted by ahydrogen atom
during a transition of its electron from the n = 4 tothe n = 1
principal energy level.
Final answer:
The frequency of light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level is approximately -2.06 x 10^14 Hz.
Explanation:
The frequency of light emitted by a hydrogen atom during a transition of its electron can be calculated using the formula:
f = R(1/n1^2 - 1/n2^2)
where f is the frequency, R is a constant (R = 3.29 x 10^15 Hz), and n1 and n2 are the initial and final principal energy levels, respectively.
In this case, the electron transitions from n = 4 to n = 1. Plugging the values into the formula, we get:
f = 3.29 x 10^15 (1/4^2 - 1/1^2)
f = 3.29 x 10^15 (1/16 - 1/1)
f = 3.29 x 10^15 (0.9375 - 1)
f = 3.29 x 10^15 (-0.0625)
f = -2.06 x 10^14 Hz
The frequency of the light emitted by the hydrogen atom during this transition is approximately -2.06 x 10^14 Hz.
An organic chemist measures the temperature T of a solution in a reaction flask. Here is the result. T = 149.206 °C Convert T to Sl units. Round your answer to 3 decimal places.
If an organic chemist measures the temperature T of a solution in a reaction flask. The temperature in SI units is approximately 422.356 K.
What is the temperature in SI units?To convert temperature from Celsius (°C) to SI units (Kelvin, K), lets make use of the following formula:
K = °C + 273.15
Given that T = 149.206 °C, we can calculate the temperature in Kelvin:
K = 149.206 + 273.15
= 422.356 K
Therefore the temperature in SI units is approximately 422.356 K.
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The temperature T in SI units is 422.356 Kelvin (K).
Temperature is a measure of the average kinetic energy of the particles in a substance. It represents how hot or cold an object or environment is.
To convert the temperature T from degrees Celsius (°C) to SI units, we need to use the Kelvin scale. The Kelvin scale is an absolute temperature scale where 0 Kelvin (0 K) represents absolute zero, the lowest possible temperature.
Calculation:
T = 149.206 °C
T (in Kelvin) = T (in °C) + 273.15
T (in Kelvin) = 149.206 °C + 273.15
T (in Kelvin) = 422.356 K.
Therefore, the temperature T in SI units is 422.356 Kelvin (K).
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What is the name of the engineer who investigated the efficiency of steam engines? Michael Faraday O Frederic Tudor OSadi Carnot O Robert Boyle
Answer:
The correct option is: Sadi Carnot
Explanation:
Steam engine is the machine that converts the energy of the steam to mechanical energy in order to perform mechanical work. It is a type of of heat engine.
The efficiency of a steam engine was first given by the French military scientist and physicist, Nicolas Léonard Sadi Carnot. The efficiency of steam engine is the ratio of the output energy of mechanical work and the input energy put provided to engine by burning fuel.
Energy absorbed or released as heat in a chemical or physical change is measured by a(n) a. thermometer. b. scale. 26. c. incubator. d. calorimeter
Answer:
d. calorimeter
Explanation:
Calorimeter -
It is the instrument use for the measurement of the heat that is absorbed or released during a physical change or in chemical change .
The instrument or devise is used for the study thermodynamics , biochemistry and chemistry .
Hence , From the question information , the instrument which can be used is a Calorimeter .
Energy absorbed or released as heat during chemical or physical changes is measured by a calorimeter, which calculates this based on temperature changes and the specific heat and mass of the substances involved.
The correct answer is d. Calorimeter. Energy absorbed or released as heat in a chemical or physical change is measured by a calorimeter. A calorimeter is a device designed specifically for measuring the amount of heat involved in chemical or physical processes. For instance, during an exothermic reaction, where heat is produced, the temperature of the solution within the calorimeter increases.
Conversely, for an endothermic reaction, where heat is required, the temperature of the solution decreases as it absorbs heat from the thermal energy of the solution. These temperature changes, in conjunction with specific heat and mass calculations, allow the calorimeter to accurately measure the heat absorbed or released. Calorimetry relies on capturing the temperature changes that occur as a result of heat transfer during these processes, using them to calculate the total amount of heat involved.
400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.0861 atm at 25.0 °C. Calculate the molar mass of the protein. Round your answer to 3 significant digits. mol x 6 ?
To determine the molar mass of an unknown protein, first calculate the molar concentration using the osmotic pressure formula. Next, calculate the moles of solute from this molarity. Finally, determine the molar mass by dividing the mass of the protein by the moles of solute.
Explanation:First, we need to calculate the molarity (M) of the solution using the osmotic pressure formula II = MRT. Here, 'II' represents the osmotic pressure in atm, which is 0.0861 atm, 'R' is the gas constant (0.08206 L atm/mol K), and 'T' is the temperature in Kelvin. Convert the given temperature 25.0 °C to Kelvin by adding 273.15, which results in 298.15 K.
Substituting the known values into the formula, we get M = II / RT. Now, compute the molar concentration (M) from the osmotic pressure.
Next, the molar mass of the protein can be determined. Divide the mass of the solute by the number of moles in that mass. In this case, the mass of the solute is 400mg, which equals 0.4 g (as 1 g = 1000 mg). By using the calculated molarity, you can get the moles of solute, from which the molar mass can be determined by dividing the mass of solute by moles of solute.
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To determine the molar mass of the protein, use the osmotic pressure equation. After solving for molarity and then moles, the molar mass is found to be approximately 22,700 g/mol. This process involves converting units and applying the ideal gas constant.
To find the molar mass of the protein, we use the formula for osmotic pressure (π):
π = MRT
π (osmotic pressure) = 0.0861 atm
M (molarity) = n/V (number of moles/volume)
R (gas constant) = 0.0821 L·atm/(K·mol)
T (temperature) = 25 + 273 = 298 K
First, convert the mass of the protein to grams:
400 mg = 0.400 g
The volume of the solution is 5.00 mL, which is 0.00500 L.
We can rearrange the formula to find the molarity (M):
M = π / (RT)
M = 0.0861 atm / (0.0821 L·atm/(K·mol) × 298 K)
M = 0.00352 mol/L
Now, the number of moles of the protein (n) can be related to its molarity and the volume of the solution:
n = M × V
n = 0.00352 mol/L × 0.00500 L = 1.76 x 10⁻⁵ mol
Finally, the molar mass (M) is given by the mass of the protein divided by the number of moles:
Molar mass = mass / n
Molar mass = 0.400 g / 1.76 x 10⁻⁵ mol ≈ 22727 g/mol
Rounded to three significant digits, the molar mass of the protein is 22,700 g/mol.
Question 5 Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.) Note: you are writing the molecular, and not the net ionic equation. → + Ca Cl O 4 2 aq H 2 O l Clears your work. Undoes your last action. Provides information about entering answers.
Explanation:
A chemical reaction equation that contains equal number of atoms on both reactant and product side is known as a balanced chemical equation.
For example, when the products are given as [tex]Ca(ClO_{4})_{2}(aq) + H_{2}O(l)[/tex]
So, reactants for this equation will be as follows.
[tex]HClO_{4}(aq) + Ca(OH)_{2}(aq) \rightarrow Ca(ClO_{4})_{2}(aq) + H_{2}O(l)[/tex]
Number of atoms present on reactant side are as follows.
H = 3
Cl = 1
O = 6
Ca = 1
Number of atoms present on product side are as follows.
H = 2
Cl = 2
O = 9
Ca = 1
Therefore, to balance this equation we multiply [tex]HClO_{4}[/tex] by 2 on reactant side and multiply [tex]H_{2}O[/tex] by 2 on product side.
Hence, the balanced chemical reaction equation will be as follows.
[tex]2HClO_{4}(aq) + Ca(OH)_{2}(aq) \rightarrow Ca(ClO_{4})_{2}(aq) + 2H_{2}O(l)[/tex]
The reactants for the given chemical reaction are likely to be calcium carbonate (CaCO₃) and hydrochloric acid (2HCl). These react to form the given products: calcium chloride (CaCl₂) and water (2H₂O), with the balanced equation being: CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂.
Explanation:Predicting the reactants of the chemical reaction requires understanding of chemistry concepts. First, let's rearrange the given product side which is CaCl₂(aq) + 2H₂O(l). From this, we can find that the reactants are likely CaCO₃ + 2HCl, as calcium carbonate generally reacts with hydrochloric acid to form calcium chloride and water.
In terms of steps, the HCl will dissolve the limestone (CaCO₃), and the resulting calcium ions (Ca₂+) will combine with the chloride ions (Cl-) to form calcium chloride (CaCl₂). Meanwhile, the carbonate ions (CO₃₂-) will combine with the hydrogen ions (H+) from the HCl to form water (H₂O) and carbon dioxide (CO₂).
The balanced molecular equation would therefore be: CaCO₃ + 2HCl --> CaCl₂+ H₂O + CO₂.
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How many grams of F are in 12.56 g of SF6? h.
Answer:
9.80 g
Explanation:
The molecular mass of the atoms mentioned in the question is as follows -
S = 32 g / mol
F = 19 g / mol
The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol
The mass of 6 F = 6 * 19 = 114 g /mol .
The percentage of F in the compound =
mass of 6 F / total mass of the compound * 100
Hence ,
The percentage of F in the compound = 114 g /mol / 146 g / mol * 100
78.08 %
Hence , from the question ,
In 12.56 g of the compound ,
The grams of F = 0.7808 * 12.56 = 9.80 g
Write 0.00057824 in Engineering Notation with 3 significant figures.
Answer:
[tex]578.2\times 10^{-6}[/tex]
Explanation:
Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.
For example, [tex]1000^2[/tex] is to be written as [tex]10^6[/tex] in engineering notation.
The given number:
0.00057824 can be written as [tex]578.24\times 10^{-6}[/tex]
Answer upto 4 significant digits = [tex]578.2\times 10^{-6}[/tex]
When will a precipitate form? a. When the concentration of the cation is greater than the concentration of the anion. b. When the concentration of the anion is greater than the concentration of the cation. c. When the reaction quotient is greater than the solubility-product constant. d. When the solubility-product constant is greater than the reaction quotient. e. None of these.
Answer:
c. when the raction quotient is greater than the Ksp
Explanation:
AxBy ↔ xA+ + yB-∴ Ksp = [ A+ ]∧x * [ B- ]∧y.....solubility product constant
∴ I.P = xA+ * yB-............ionic product
⇒ the relationship between Ksp and I.P, will give us the precipitation conditions.
∴ I.P > Ksp ⇒ precipitation
∴ I.P < Ksp ⇒ no precipitation (disolution)
∴ I.P = Ksp ⇒ equilibrium
⇒ the ionic product is directly related to the reaction quotient Q, where Q = [A+] * [B-] / [AxBy] = PI / [AB], then the higher this product the higher the reaction quotient will be and therefore will be greater than the value of the constant and will have precipitated, the correct answer is the c
using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical reaction:
P4O10 + 6H2O → 4H3PO4
Round your answer to zero decimal places.
Answer: The value of [tex]\Delta S^o[/tex] for the reaction is -206 J/K
Explanation:
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change of a reaction is:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]P_4H_{10}(s)+6H_2O(l)\rightarrow 4H_3PO_4(s)[/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(4\times \Delta S^o_{(H_3PO_4)})]-[(1\times \Delta S^o_{(P_4O_{10})})+(6\times \Delta S^o_{(H_2O)})][/tex]
We are given:
[tex]\Delta S^o_{(H_3PO_4)}=110.5J/K.mol\\\Delta S^o_{(H_2O(l))}=69.91J/K.mol\\\Delta S^o_{(P_4H_{10})}=228.86J/K.mol[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(4\times (110.5))]-[(1\times (228.86))+(6\times (69.91))]\\\\\Delta S^o_{rxn}=-206.32J/K=-206J/K[/tex]
Hence, the value of [tex]\Delta S^o[/tex] for the reaction is -206 J/K
To calculate the standard reaction entropy, ΔS°, find the Standard Molar Entropies of the products and reactants from the ALEKS Data tab, multiply them by their respective coefficients, and then subtract the sum of products from the sum of reactants. The result should be rounded to zero decimal places.
Sulfur hexafluoride gas is collected at 14.0°C in an evacuated flask with a measured volume of 5.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.090atm . Calculate the mass and number of moles of sulfur hexafluoride gas that were collected. Be sure your answer has the correct number of significant digits.
Answer: The mass and number of moles of sulfur hexafluoride gas is 2.8 g and 0.019 moles respectively.
Explanation:
To calculate the moles of sulfur hexafluoride gas, we use the equation given by ideal gas:
PV = nRT
where,
P = Pressure of sulfur hexafluoride = 0.090 atm
V = Volume of the sulfur hexafluoride gas = 5.0 L
n = number of moles of gas = ?
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
T = Temperature of sulfur hexafluoride gas = [tex]14^oC=[273+14]K=287K[/tex]
Putting values in above equation, we get:
[tex]0.090atm\times 5.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 287K\\\\n=0.019mol[/tex]
To calculate the mass of sulfur hexafluoride, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of sulfur hexafluoride = 146.0 g/mol
Moles of sulfur hexafluoride = 0.0191 moles
Putting values in equation 1, we get:
[tex]0.019mol=\frac{\text{Mass of sulfur hexafluoride}}{146.0g/mol}\\\\\text{Mass of sulfur hexafluoride}=2.8g[/tex]
In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits. Here, the least precise number of significant digits are 2.
Hence, the mass and number of moles of sulfur hexafluoride gas is 2.8 g and 0.019 moles respectively.
To calculate the mass and number of moles of sulfur hexafluoride gas, we use the ideal gas law equation PV = nRT. By substituting the given values and solving for n, we find that 0.1903 moles of sulfur hexafluoride were collected. The mass of the gas is determined by multiplying the moles by the molar mass of sulfur hexafluoride, resulting in approximately 27.8 grams.
Explanation:To calculate the mass and number of moles of sulfur hexafluoride gas, we can use the ideal gas law equation: PV = nRT. Rearranging the equation to solve for moles (n), we have n = PV/RT. We are given the pressure (0.090 atm), volume (5.0 L), temperature (14.0°C which needs to be converted to Kelvin), and the ideal gas constant (R = 0.0821 L·atm/(mol·K)). After converting the temperature to Kelvin (287 K), we substitute the values and solve for n. n = (0.090 atm * 5.0 L) / (0.0821 L·atm/(mol·K) * 287 K). This gives us n = 0.1903 moles of sulfur hexafluoride gas.
To calculate the mass, we use the formula mass = moles * molar mass. The molar mass of sulfur hexafluoride (SF6) is SF6 = (32.06 g/mol * 6) + (19.00 g/mol * 1). This gives us a total molar mass of 146.06 g/mol. We can now calculate the mass by multiplying the moles by the molar mass: mass = 0.1903 moles * 146.06 g/mol. The mass of sulfur hexafluoride gas that were collected is approximately 27.8 grams.
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Limiting Reactant For the general reaction stoichiometry of 2A + 3B → D+E, which reactant is the limiting reactant (A or B), if the initial number of moles is 3 moles of A and 4 moles of B?
Answer:
B
Explanation:
Limiting reagent is the one which is present in small molar amount. It got exhausted at the end of the reaction and the formation of the products is governed by it.
The given reaction:
2A + 3B → D+E
2 moles of A react with 3 moles of B
1 mole of A react with 3 / 2 moles of B
Given moles of A = 3 moles
Moles of B = 4 moles
So,
3 moles of A will react with 1.5 * 3 moles of B
Moles of B = 4.5
Since, only we have 4 moles of B. So, B is the limiting reagent.
The metal Cadmium tends to form Cd2+ ions. The following observations are made: i) When a strip of zinc metal is placed in Cd2+(aq), cadmium metal is deposited onto the strip, ii) When a strip cadmium metal is placed into Ni2+(aq), nickel metal is deposited onto the strip; A) Write net-ionic equations to explain each of these observations. B) What can you conclude about the position of cadmium in the activity series relative to zinc and nickel?
A)
The first reaction can be represented by the following equation:
i) Zn(s) + Cd²⁺(aq) → Zn²⁺(aq) + Cd(s)
The second reaction can be represented by the following equation:
ii) Cd(s) + Ni²⁺(aq) → Cd²⁺(aq) + Ni(s)
B) The activity series is a tool used to predict displacement reactions. More reactive metals (above in the series) can displace less reactive ones (below in the series) from the salts they are part of in solutions.
In i) we can see zinc displacing cadmium. Therefore, cadmium is below zinc in the activity series.
In ii) we can see cadmium displacing nickel. Thus, cadmium is above nickel in the activity series.
The student's question involves writing net-ionic equations for redox reactions with cadmium and determining its position in the activity series. Zinc can replace cadmium ions, and cadmium can replace nickel ions, indicating that cadmium is less reactive than zinc but more reactive than nickel.
Explanation:The student is asking about redox reactions involving the metal cadmium, zinc, and nickel, which can be explained using net-ionic equations and the activity series of metals.
A) Net-ionic equations:
B) Conclusion about cadmium's position:
Based on the observations, cadmium is less reactive than zinc but more reactive than nickel within the activity series. This is deduced because zinc can replace cadmium ions and cadmium can replace nickel ions in solution, indicating a middle position for cadmium in the series between zinc and nickel.