Answer:
Yes, there is a loss of 8350 J of energy in the form of heat
Explanation:
The principle of energy conservation is described mathematically
as the energy conservation equation as follows:
ΔK+ΔU=Q+W
where:
ΔK: changes in kinetics energy [J]ΔU: changes in potential energy [J]Q: gain or loss of heat[J]W: work done by the system or in the system[J]Analyzing each term of the equation:
[tex]ΔK=\frac{1}{2}mv_{f} ^{2} - \x]frac{1}{2}mv_{i} ^{2}=\frac{1}{2}*100*15^{2} - \frac{1}{2}*100*0^{2}=11250 J[/tex]
ΔU=[tex]mgh_{f}-mgh_{i}=100*9,8*0-100*9,8*20=-19600[J][/tex]
Q=?
W=0 [J]
Replacing in the main equation:
11250-19600=Q+0
Q= -8350 [J]
So, the answer is YES, there is a loss of 8350 J of energy in the form of heat.
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 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 runner runs 4339 ft in 7.45 minutes. What is the runner's average speed in miles per hou 6.62 mi/hr O 0.110 mi/hr 6.618 mi/h e 582 mi/hr
Answer: The average speed of the runner is 6.618 miles/hr
Explanation:
Average speed is defined as the ratio of total distance traveled to the total time taken.
To calculate the average speed of the runner, we use the equation:
[tex]\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}[/tex]
We are given:
Distance traveled = 4339 ft
Time taken = 7.45 mins
Putting values in above equation, we get:
[tex]\text{Average speed of runner}=\frac{4330ft}{7.45min}=582.42ft/min[/tex]
To convert the speed into miles per hour, we use the conversion factors:
1 mile = 5280 ft
1 hr = 60 mins
Converting the speed into miles per hour, we get:
[tex]\Rightarrow \frac{528.42ft}{min}\times (\frac{1miles}{5280ft})\times (\frac{60min}{1hr})\\\\\Rightarrow 6.618mil/hr[/tex]
Hence, the average speed of the runner is 6.618 miles/hr
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]
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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267 moles of NADH are produced by the Citric Acid Cycle (CAC). How many moles of glucose needed to be broken down during glycolysis in order to produce enough pyruvate for the CAC to produce this many moles of NADH?
Answer:
45 moles
Explanation:
From glycolysis, 1 mole of glucose gives 2 moles of pyruvate which undergoes citric acid cycle.
1 mole of pyruvate undergoes citric acid cycle (After conversion to acetyl-CoA) gives 3 moles of NADH.
Also,
2 moles of pyruvate undergoes citric acid cycle (After conversion to acetyl-CoA) gives 6 moles of NADH.
Thus,
6 moles of NADH are produced from 2 moles of pyruvate or 1 mole of glucose.
1 mole of NADH is produced from 1/6 mole of glucose
267 moles of NADH are produced from [tex]\frac {1}{6}\times 267[/tex] moles of glucose.
Thus, moles of glucose needed to be broken ≅ 45 moles
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.
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
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%.
Calculate the answer to the following expression to the correct number of significant figures.
4 X 10^8/2.0 X 10^-5
Answer : The answer will be [tex]\Rightarrow 2\times 10^{13}[/tex]
Explanation :
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
The rule apply for the multiplication and division is :
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
As we are given the expression :
[tex]\frac{4\times 10^8}{2.0\times 10^{-5}}[/tex]
[tex]\Rightarrow 2\times 10^{13}[/tex]
In the given expression, [tex]4\times 10^8[/tex] has 1 significant figure and [tex]2.0\times 10^{-5}[/tex] has 2 significant figures. From this we conclude that 1 is the least significant figures in this problem. So, the answer should be in 1 significant figures.
Thus, the answer will be [tex]\Rightarrow 2\times 10^{13}[/tex]
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
The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi
The factor 0.01 corresponds to the prefix D. 'centi'.
In the metric system, prefixes are used to denote different multiples or fractions of a base unit. Here's a breakdown of the prefixes relevant to this question:
milli-: This prefix represents the factor $10⁻³ $ or 0.001.
deci-: This prefix represents the factor $10⁻¹ $ or 0.1.
deka-: This prefix represents the factor $10¹ or 10.
centi-: This prefix represents the factor $10⁻² $ or 0.01.
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.
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
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.
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
A 25 liter gas cylinder contains 10 mole of carbon dioxide. Calculate the pressure, in atm, of the gas at 325K.
Answer: The pressure of carbon dioxide gas is 11 atm
Explanation:
To calculate the pressure of gas, we use the equation given by ideal gas equation:
PV = nRT
where,
P = pressure of the gas = ?
V = Volume of gas = 25 L
n = number of moles of gas = 10 mole
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = 325 K
Putting values in above equation, we get:
[tex]P\times 25L=10mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 325K\\\\P=11atm[/tex]
Hence, the pressure of carbon dioxide gas is 11 atm
A house has an area of 234 m2. What is its area in each unit?
A)km2
B)dm2
C)cm2
Answer:
a) [tex]0.000324 km^2[/tex]
b) [tex]32400 dm^2[/tex]
c) [tex]3.24x10^6 cm^2[/tex]
Explanation:
To do the different conversions we need to know the follow:
[tex]1 km->1000 m so 1 km^2-> 1000^2 m^2= 1x10^6 m^2[/tex]
[tex]324 m^2*(1 km^2/1x10^6 m^2) = 0.000324 km^2[/tex]
[tex]1m-> 10 dm so 1m^2->10^2 dm^2=100 dm^2[/tex]
[tex]324 m^2*(100 dm^2/1 m^2)=32400 dm^2[/tex]
[tex]1m-> 100 cm so 1m^2->100^2 cm^2=10000 cm^2[/tex]
[tex]324 m^2*(10000 cm^2/1 m^2)=3.24x10^6 cm^2[/tex]
Answer :
(A) [tex]234m^2=2.34\times 10^8km^2[/tex]
(B) [tex]234m^2=2.34\times 10^4dm^2[/tex]
(C) [tex]234m^2=2.34\times 10^6cm^2[/tex]
Explanation :
The conversion used for area from [tex]m^2[/tex] to [tex]km^2[/tex] is:
[tex]1m^2=10^6km^2[/tex]
The conversion used for area from [tex]m^2[/tex] to [tex]dm^2[/tex] is:
[tex]1m^2=100dm^2[/tex]
The conversion used for area from [tex]m^2[/tex] to [tex]cm^2[/tex] is:
[tex]1m^2=10000cm^2[/tex]
Part (A):
As, [tex]1m^2=10^6km^2[/tex]
So, [tex]234m^2=\frac{234m^2}{1m^2}\times 10^6km^2=2.34\times 10^8km^2[/tex]
Part (B):
As, [tex]1m^2=100dm^2[/tex]
So, [tex]234m^2=\frac{234m^2}{1m^2}\times 100dm^2=2.34\times 10^4dm^2[/tex]
Part (C):
As, [tex]1m^2=10000cm^2[/tex]
So, [tex]234m^2=\frac{234m^2}{1m^2}\times 10000cm^2=2.34\times 10^6cm^2[/tex]
a 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.
Pyridine is a conjugate base which reacts with H to form pyridine hydrochloride. The hydrochloride dissociates to yield H' with a pKof 5.36. Describe the preparation of a pyridine buffer at pH 5.2 starting with 0.1M pyridine and 1.OM HCI. You may start with one liter of the 0.1 M pyridine.
Answer:
To one liter of the 0.1 M pyridine you need to add 41 mL of 1,0M HCl to obtain a buffer at 5,2
Explanation:
The reaction is:
pyridine-H⁺ ⇄ pyridine + H⁺ pka = 5,36; k = [tex]10^{-5,36}[/tex]
Using Henderson-Hasselbalch formula:
5,2 = 5,36 + log[tex]\frac{[Py-H^+]}{[Py]}[/tex]
0,692 = [tex]\frac{[Py-H^+]}{[Py]}[/tex] (1)
As total intial moles are 0,1:
0,1 = Py-H⁺ moles + Py moles (2)
Replacing (2) in (1) final moles of both Py-H⁺ and Py are:
Py: 0,059 moles
Thus:
Py-H⁺: 0,041 moles
Moles in reaction are:
Py: 0,1-x moles
H⁺: Y-x moles Y are initial moles of H⁺
Py-H⁺: x moles
Knowing x = 0,041 moles, pyridine volume is 1L and HCl molarity is 1 mol/L and [H⁺] = [tex]10^{-5,2}[/tex]
[tex]10^{-5,2}[/tex] = [tex]\frac{Y-0,041moles}{Y+1L}[/tex]
Y = 0,04100605 moles≡ 41 mL of 1,0M HCl
I hope it helps!
A mixture of methanol and methyl acetate contains 15.0 weight percent methanol. Determine the number of gmols of methanol in 110.0 kilograms of the mixture.
Answer:
There are 550.5 moles of methanol in 110.0 kilograms of the mixture.
Explanation:
A solution 15% weight of methanol means there is 15g of methanol per 100g of the mixture or 0.1kg of the mixture. Also, the molar mass of methanol (CH3OH) is:
[tex]m_{C} + 4xm_{H} + m_{O} = 12.0g/mol + 4x1.0g/mol + 16.0g/mol = 32.0g/mol[/tex]
Thus, dividing 15g by molar mass
[tex]15.0g / 32.0\frac{g}{mol} = 0.5moles[/tex] we find there is 0.5 moles of methanol per 0.1Kg of the mixture. Calculating the number of mols of methanol in 110.0 kilograms of the mixture:
[tex]\frac{110.1Kgx0.5moles}{0.1Kg} = 550.5 moles[/tex]
Therefore, there is 550.5 moles of methanol in 110.0 kilograms of the mixture.
How many gram-moles of formic acid, CH2O2, are in 225 g of the compound? Numbe g-mol CH,0, How many pound-moles of formic acid are in 225 g of the compound? Number Ib-moles CH,O,
Explanation:
(a) Mass given = 225 g
Molecular mass of [tex]CH_{2}O_{2}[/tex] = [tex]12 + 2 + 16 \times 2[/tex] = 12 + 2 + 32 = 46 g/mol
As, it is known that number of moles is equal to the mass divided by molar mass.
Mathematically, No. of moles = [tex]\frac{mass}{\text{molecular mass}}[/tex]
Hence, calculate the number of moles of formic acid as follows.
No. of moles = [tex]\frac{225 g}{46 g/mol}[/tex]
= 4.89 mol or g mol
Hence, in 225 g of the compound there are 4.89 mol or g mol.
(b) It is known that 1 g = 0.0022 pound
So, pounds present in 4.89 g will be calculated as follows.
4.89 \times 0.0022 pound
= 0.0107 pound mole
Hence, in 225 g of the compound there are 0.0107 pound mole.
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.
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.
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.
An astronaut uses a laboratory balance and weighs an object on earth and again on the moon Which statement below about the weight and mass of the object is true? A) The mass and weight will be identical on the earth and the moon. B) The mass will be the same on earth and moon but the weight will be less on the moon C) Both the mass and weight will be different on earth and moon. D) The weight will be the same on earth and moon but the mass will be less on the moon.
Answer:
B) The mass will be the same on earth and moon but the weight will be less on the moon
Explanation:
Mass of a substance is a intrinsic property. It is same throughout the universe. On the other hand,
Weight = Mass × Gravitational acceleration
Weight depends on the gravitational acceleration of the planet on which the substance is present.
Thus, the astronaut weight on Moon will be approximately six times less than the weight on Earth as gravitational acceleration on Moon is six times less than the gravitational acceleration on Earth the but number of the atoms in body of the astronaut has not changed and thus, mass is same same at two places.
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.
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
You have ice cubes in your freezer at home. The ice cubes melt into water. This type of change is called a... O a. nuclear change O b.physical change O c. chemical change
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.