Answer:
[tex]t=5.84\times 10^{-5}s[/tex]
Explanation:
First rearrange the whole equation by keeping t in one side and rest parameters in another side of equationThen plug-in all the given values for each parameters and get the solution[tex]ln(\frac{A}{B})=-kt[/tex]
or, [tex]t=\frac{1}{-k}\times ln(\frac{A}{B})[/tex]
Now plug-in all given values:
[tex]t=\frac{1}{-4.80\times 10^{4}s^{-1}}\times ln(\frac{1.00\times 10M}{1.65\times 10^{2}M})[/tex]
So, [tex]t=5.84\times 10^{-5}s[/tex]
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]
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]
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!
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.
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
1. Make a 100 mL volume of 100 mM acetic acid (HAC) HAC (glacial, 100%, 17.4 M, 1.05 g/ml, MW 60.05) ddh,0 ML add to 100 mL Measure 250 ml beaker. Add mL of ddH20 using a 100-ml graduated cylinder and transfer this volume to a ul of 100 mM HAC to the water. Mix with a stir bar.
Final answer:
To prepare a 100 mM acetic acid solution, calculate the moles needed, measure the glacial acetic acid based on its density and volume, and dilute with distilled water to the final volume. Mix well for homogeneity.
Explanation:
The question involves preparing a solution of acetic acid and relates to the subjects of molarity and solution preparation in chemistry. The solution is to be made to a specified concentration using a direct dilution from a more concentrated stock.
Steps to Prepare 100 mM Acetic Acid Solution
First, we calculate the number of moles of acetic acid required for 100 mL of a 100 mM solution using the molar mass of acetic acid (MW 60.05). Second, we use the density of glacial acetic acid (1.05 g/mL) and the volume needed (from the calculated moles and molarity) to determine how much of the glacial acetic acid to measure out. Finally, we add this measured amount of acid to a beaker and dilute to the mark with distilled water (ddH2O) to achieve the desired final concentration. Mixing with a stir bar ensures a homogeneous solution.
To calculate the mole fraction of acetic acid in a solution, we need the number of moles of acetic acid and the number of moles of water in the solution. This can be accomplished by converting the mass of each component to moles using their respective molar masses.
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.
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.
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
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
Two moles of a monatomic ideal gas are contained at a pressure of 1 atm and a temperature of 300 K; 34,166 J of heat are transferred to the gas, as a result of which the gas expands and does 1216 J of work against its surroundings. The process is reversible. Calculate the final temperature of the gas.
Answer:
Final temperature is 302 K
Explanation:
You can now initial volume with ideal gas law, thus:
V = [tex]\frac{n.R.T}{P}[/tex]
Where:
n are moles: 2 moles
R is gas constant: 0,082 [tex]\frac{atm.L}{mol.K}[/tex]
T is temperature: 300 K
P is pressure: 1 atm
V is volume, with these values: 49,2 L
The work in the expansion of the gas, W, is: 1216 J - 34166 J = -32950 J
This work is:
W = P (Vf- Vi)
Where P is constant pressure, 1 atm
And Vf and Vi are final and initial volume in the expansion
-32950 J = -1 atm (Vf-49,2L) × [tex]\frac{101325 J}{1 atm.L}[/tex]
Solving: Vf = 49,52 L
Thus, final temperature could be obtained from ideal gas law, again:
T = [tex]\frac{P.V}{n.R}[/tex]
Where:
n are moles: 2 moles
R is gas constant: 0,082 [tex]\frac{atm.L}{mol.K}[/tex]
P is pressure: 1 atm
V is volume: 49,52 L
T is final temperature: 302 K
I hope it helps!
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
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]
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.
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.
Calculate the number of mol of helium in a 2.01-L balloon at 29°C and 2.71 atm of pressure. Be sure to answer all parts. Imol
Answer:
0.220 mol He
Explanation:
Hello,
Based on the ideal gas equation of state:
[tex]PV=nRT\\n=\frac{PV}{RT} \\n=\frac{2.71 atm * 2.01 L}{0.082\frac{atm*L}{mol*K}*302K } \\n=0.220 mol[/tex]
Best regards.
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
Question: Consider the molecule cyclohexene, which
containsonly hydrogen and carbon atoms.
When a sample is burned in oxygen, 4.822g of CO(2) and1.650g
of H(2)O are obtained.
What is its simplest formula?
Answer:
The simplest formula is C₃H₅.
Explanation:
The moles of CO₂ are the moles of Carbon because oxygen is in excess. Thus, the double of moles of H₂O are the moles oh hydrogen (Double because H₂O has two hydrogens for each mole).
4,822g CO₂ × (1mol / 44,01 g) = 0,1096 moles of C
1,650g H₂O × (1mol / 18,01 g) × 2= 0,1832 moles of H
The next step is divide these numbers between the smaller one, thus:
0,1096 moles / 0,1096 moles = 1
0,1832 moles / 0,1096 moles = 1,672
The last step is to find a number to multiple that gives integers:
1 × 3 = 3 Carbon
1,672 × 3 = 5 hydrogen
Thus, the simplest formula is C₃H₅.
I hope it helps!
When preparing 1-bromobutane from 1-butanol and HBr, why
mustit be washed with H2O, cold NaOH and NaCl?
Answer:
Answer has been given below
Explanation:
During reaction of 1-butanol and HBr, some unreacted amount of HBr might present in reaction mixture. Hence cold NaOH is used to neutralize unreacted HBr.Washing with [tex]H_{2}O[/tex] is done to remove the salt produced during neutralization i.e. NaBr from organic layer of 1-bromobutane.NaCl is added to remove water from organic layer. Because NaCl is more soluble in water than organic layer.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.
A water main made out of galvanized iron has an inner pipe diameter of 0.1 meters, an inlet pressure of 1000 kPa, outlet pressure of 500 kPa, water flow rate of 5 m3/min, and 10 90 degree elbows. What length galvanized iron line should be used to reach the desired outlet pressure?
Explanation:
The given data is as follows.
Diameter = 0.1 m, [tex]P_{1}[/tex] = 1000 kPa
[tex]P_{2}[/tex] = 500 kPa
Change in pressure [tex]\Delta P[/tex] = 1000 kPa - 500 kPa = 500 kPa
Since, 1000 Pa = 1 kPa. So, 500 kPa will be equal to [tex]500 \times 10^{3}[/tex].
Q = 5 [tex]m^{3}/min[/tex] = [tex]\frac{5}{60} m^{3}/sec[/tex] = 0.0833 [tex]m^{3}/sec[/tex]
It is known that Q = [tex]A \times V[/tex]
where, A = cross sectional area
V = speed of the fluid in that section
Hence, calculate V as follows.
V = [tex]\frac{Q}{A}[/tex]
= [tex]\frac{Q \times 4}{\pi \times d^{2}}[/tex]
= [tex]\frac{0.0833 \times 4}{3.14 \times (0.1)^{2}}[/tex]
= 10.61 m/sec
Also it is known that Reynold's number is as follows.
Re = [tex]\frac{\rho \times V \times d}{\mu}[/tex]
= [tex]\frac{1000 \times 10.61 \times 0.1}{10^{-3}}[/tex]
= 1061032.954
As, it is given that the flow is turbulent so we cannot use the Hagen-Poiseuille equation as follows. Therefore, by using Blasius equation for turbulent flow as follows.
[tex]\Delta P = \frac{0.241 \times \rho^{0.75} \times \mu^{0.25} \times L}{D^{4.75}} \times Q^{1.75}[/tex]
[tex]500 \times 10^{3} = \frac{0.241 \times (1000)^{0.75} \times (10^{-3})^{0.25} \times L}{(0.1)^{4.75}} \times (0.0833)^{1.75}[/tex]
[tex]500 \times 10^{3} = 5535.36 \times L[/tex]
L = [tex]\frac{500 \times 10^{3}}{5535.36}[/tex]
= 90.328 m
Thus, we can conclude that 90.328 m length galvanized iron line should be used to reach the desired outlet pressure.
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 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 woman informs her engineer husband that "hot water will freeze faster than cold water." He calls this statement nonsense but he does so in a polite sensitive manner. She answers by saying that she has actually timed the freezing process for ice trays in the home refrigerator and found that hot water does indeed freeze faster. As a friend, you are asked to settle the argument and preserve family harmony. Is there any logical explanation for the woman's observation
Answer:
Explanation:
It could be as a result of evaporation from the hot water causing it to have less mass than usual. Evaporation will allow the volume of the hot water to decrease and by so doing will cool faster than cold water which is at a higher mass. This is how I will explain my observations to them.
Final answer:
The concept of hot water freezing faster than cold water is known as the Mpemba effect and is influenced by various factors including evaporation and thermal energy transfer. Specific heat capacity plays a role in cooling processes, affecting the rate at which materials like metal are cooled when in contact with ice or water.
Explanation:
The observation that hot water freezes faster than cold water under certain conditions has been noted historically and is often referred to as the Mpemba effect. While it may seem counterintuitive, various experiments have shown that hot water can indeed freeze more quickly than cold water due to a number of factors, including evaporation (which reduces the mass of hot water), thermodynamic properties of water (such as specific heat capacity and the behavior of water molecules at different temperatures), and the environment in which the water is being frozen (like the surrounding air temperature and properties of the freezing surface).
Regarding the specific scenarios described, mixing hot and cold water or the implications of adding cold water to a hot liquid like tea or coffee, the result is affected by the thermal energy transfer between the substances. The outcome is influenced by the specific heat of the materials, the current temperatures, and the mass of the substances involved. For instance, pouring hot water into cold water would result in heat transfer from the hot water to both the cold water and its surroundings.
As for the question of cooling a hot piece of metal, the method of cooling (using ice versus cold water) will affect the rate at which the metal cools down, with the specific heat capacities playing a crucial role. Water generally has a higher specific heat capacity than ice, making it more efficient at absorbing heat.
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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The density of a certain type of plastic is 0.81 g/cm3. If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick, what is its mass?
mass = _____ x 10_____g
g
Enter your answer in scientific notation.
Answer: The mass of plastic sheet is [tex]8.1\times 10^4g[/tex]
Explanation:
The plastic sheet is in the form of cuboid. To calculate the volume of cuboid, we use the equation:
[tex]V=lbh[/tex]
where,
V = volume of cuboid
l = length of cuboid = 10.0 m = 1000 cm (Conversion factor: 1 m = 100 cm)
b = breadth of cuboid = 1.0 m = 100 cm
h = height of cuboid = 1 cm
Putting values in above equation, we get:
[tex]V=1000\times 100\times 1=10^5cm^3[/tex]
To calculate mass of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
We are given:
Density of plastic sheet = [tex]0.81g/cm^3[/tex]
Volume of plastic sheet = [tex]10^5cm^3[/tex]
Putting values in equation 1, we get:
[tex]0.81g/cm^3=\frac{\text{Mass of plastic sheet}}{10^5cm^3}\\\\\text{Mass of plastic sheet}=8.1\times 10^4g[/tex]
Hence, the mass of plastic sheet is [tex]8.1\times 10^4g[/tex]
The mass of the plastic sheet is found by multiplying the density (0.81 g/cm³) with its volume (100,000 cm³), resulting in a mass of 81,000 g, or 8.1 × 10⁴ g in scientific notation.
To calculate the mass of a plastic sheet with a given density, you need to multiply its density by its volume. The density of the plastic is given as 0.81 g/cm³. To find the volume, convert all dimensions into the same unit, in this case, centimeters, because the density is given in g/cm³. The sheet is 10.0 m long (1000 cm), 1.0 m wide (100 cm), and 1 cm thick. Therefore, the volume is 1000 cm × 100 cm × 1 cm = 100,000 cm³.
Now, calculate the mass using the formula: mass = density × volume, so mass = 0.81 g/cm³ × 100,000 cm³. The mass of the plastic sheet is thus 81,000 g. To express this in scientific notation, we write it as 8.1 × 10⁴ g.
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
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.
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l) In an industrial synthesis of urea, a chemist combines 132.0 kg of ammonia with 211.4 kg of carbon dioxide and obtains 172.7 kg of urea. determine the limiting reactant
Answer:
NH₃ is the limiting reagent
Explanation:
Mass of ammonia = 132.0kg
Mass of carbon dioxide = 211.4kg
Mass of urea = 172.7kg
Unknown:
Limiting reagent = ?
Solution
First, we write the balanced stoichiometeric equation:
2NH₃ + CO₂ → CH₄N₂O + H₂O
The reactant that is present in short supply determines the amount of product that is formed in a reaction. This reactant is called the limiting reagent.
To establish the limiting reagent, we need to go find out what is happening at the start of the reaction:
Convert the masses of the reactants to moles.
Number of moles of NH₃ = [tex]\frac{mass}{molar mass}[/tex]
Molar mass of NH₃ = 14 + (3x1) = 17g/mol
Number of moles of NH₃ = [tex]\frac{132}{17}[/tex] = 7.765mole
Number of moles of CO₂ = [tex]\frac{mass}{molar mass}[/tex]
Molar mass of CO₂ = 12 + (2 x 16) = 44g/mol
Number of moles of CO₂ = [tex]\frac{211.4}{44}[/tex] = 4.805mole
From the reaction equation:
2 moles of NH₃ reacted with 1 mole of CO₂
so 7.765 mole of NH₃ will require [tex]\frac{7.765}{2}[/tex]mole, 3.883 of CO₂
But we are given 4.805mole of CO₂.
Therefore, CO₂ gas is in excess and NH₃ is the limiting reagent.
The limiting reactant for the synthesis of CH₄N₂O given the data is ammonia, NH₃
Balanced equation2NH₃ + CO₂ —> CH₄N₂O + H₂O
Molar mass of NH₃ = 17 g/mole
Mass of NH₃ from the balanced equation = 2 × 17 = 34 g = 0.034 Kg
Molar mass of CO₂ = 44 g/mole
Mass of CO₂ from the balanced equation = 1 × 44 = 44 g = 0.044 Kg
SUMMARY
From the balanced equation above,
0.034 Kg of NH₃ reacted with 0.044 Kg of CO₂
How to determine the limiting reactantFrom the balanced equation above,
0.034 Kg of NH₃ reacted with 0.044 Kg of CO₂
Therefore,
132 Kg of NH₃ will react with = (132 × 0.044) / 0.034 = 170.8 of Kg of CO₂
From the calculation made above, we can see that only 170.8 of Kg of CO₂ out of 172.7 Kg given is needed to react completely with 132 Kg of NH₃.
Thus, NH₃ is the limiting reactant
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