Answer: (E) 300 bq
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.
Half life is represented by [tex]t_{\frac{1}{2}[/tex]
Half life of Thallium-208 = 3.053 min
Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. [tex]\frac{2400}{2}=1200[/tex], after second half life, the activity would be reduced to half of 1200 i.e. [tex]\frac{1200}{2}=600[/tex], and after third half life, the activity would be reduced to half of 600 i.e. [tex]\frac{600}{2}=300[/tex],
Thus the activity 9 minutes later is 300 bq.
You place a glass of water in the freezer. Hours later the glass of water has frozen solid. What happened to the particles in the glass?
A. They started to move freely.
B. They slowed down and packed together.
C. They moved faster and packed together.
D. They slowed down and moved freely.
Answer:
The answer is B
Explanation:
The answer to this question is based on the molecular kinetic theory. The more you diminish the temperature, the slower the kinetic energy of molecules (i.e their movement) will be. Therefore, if the particles aren't allowed to move freely, they'll package themselves and the mean distance between two particles will be fixed and the shortest you'd be able to find.
Titanium has an HCP unit cell for which the ratio of the lattice parameters cais 1.58. If the radius of the Be atom is 0.1445 nm, (a) determine the unit cell volume, and (b) calculate the theoretical density of Ti, given that its atomic weight is 47.87 g/mol
Answer :
(a) The volume of unit cell is, [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]
(b) The theoretical density of Ti is, [tex]4.81g/cm^3[/tex]
Explanation :
(a) First we have to calculate the volume of the unit cell.
Formula used :
[tex]V=6r^2c\sqrt {3}[/tex]
where,
V = volume of unit cell = ?
r = atomic radius = [tex]0.1445nm=1.445\times 10^{-8}cm[/tex]
conversion used : [tex](1nm=10^{-7}cm)[/tex]
Ratio of lattice parameter = c : a = 1.58 : 1
So, c = 1.58 a
And, a = 2r
c = 1.58 × 2r
Now put all the given values in this formula, we get:
[tex]V=6\times r^2\times (1.58\times 2r)\sqrt {3}[/tex]
[tex]V=6\times r^3\times (1.58\times 2)\sqrt {3}[/tex]
[tex]V=6\times (1.445\times 10^{-8}cm)^3\times (1.58\times 2)\sqrt {3}[/tex]
[tex]V=9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]
The volume of unit cell is, [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]
(b) Now we have to calculate the density of Ti.
Formula used for density :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
[tex]\rho=\frac{Z\times M}{N_{A}\times V}[/tex] ..........(1)
where,
[tex]\rho[/tex] = density of Ti = ?
Z = number of atom in unit cell = 6 atoms/unit cell (for HCP)
M = atomic mass = 47.87 g/mol
[tex](N_{A})[/tex] = Avogadro's number = [tex]6.022\times 10^{23}atoms/mole[/tex]
[tex]a^3=V[/tex] = volume of unit cell = [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]
Now put all the values in above formula (1), we get:
[tex]\rho=\frac{6\times 47.87}{(6.022\times 10^{23})\times (9.91\times 10^{-23})}[/tex]
[tex]\rho=4.81g/cm^3[/tex]
The theoretical density of Ti is, [tex]4.81g/cm^3[/tex]
Platinum, Pt, is one of the rarest of the metals. Worldwide annual production is only about 130 tons. Platinum has a density of 21.4g/cm3. If thieves were to steal platinum from a bank using a small truck with a maximum payload capacity of 900 lb, how many 1 L bars of the metal could they take?(A) 19 bars(B) 2 bars(C) 42 bars(D) 1 bars(E) 47 bars
Answer:
(A) 19 bars
Explanation:
First off, we calculate the mass of platinum contained in one 1 L bar. To do that we convert 1 L into cm³ -1 L equals to 1000 cm³-.
21.4 g/cm³ * 1000 cm³ = 21,400 g
Each bar of platinum weighs 21,400 grams.
Now we convert the maximum payload capacity of the truck, into grams (1 lb equals to 453,592 g):
[tex]900lb*\frac{453,592g}{1lb}=408232.8g[/tex]
Then we divide the weight of one bar by the maximum payload capacity:
408232.8 / 21400 =19.09
Thus the thieves could carry 19 1 L bars
A system that had work done on it but which receives or loses no heat from or to the surroundings has
w < 0, ΔE > 0
w = - ΔE
w > 0, ΔE < 0
w = ΔE
Answer:
w >0, ΔE < 0
Explanation:
As per the first law of thermodynamics,
ΔE = Q - W
Where,
ΔE = Change in internal energy
Q = Heat receive or heat loss
W = work done
Work done by the system is always -ve.
Work done on the system is always +ve.
It is given that work done on the system.
W = +ve or W > 0
As, there is no heat receive or heat loss
So, Q = 0
Now, as per the first law of thermodynamics.
ΔE = Q - W
Q = 0
ΔE = - W
or ΔE < 0
So, answer would be w > 0, ΔE < 0
A macromolecule is added at a concentration of 18 g L−1 to water at a temperature of 10°C. If the resulting osmotic pressure of this solution is found to be equal to 12 mmHg, estimate the molecular weight of the macromolecule in grams per mole
Answer: 26138g/mol
Explanation:
[tex]\pi =CRT[/tex]
[tex]\pi[/tex] = osmotic pressure = 12 mmHg =[tex]\frac{1}{760}\times 12=0.016[/tex] atm (760 mmHg= 1atm)
C= concentration in Molarity
R= solution constant = 0.0821 Latm/Kmol
T= temperature = [tex]10^0C=(10+273)K=283K[/tex]
For the given solution: 18 g of macromolecule is dissolved to make 1 L of solution.
[tex]Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}[/tex]
[tex]C=\frac{18\times 1000}{M\times 1000ml}=\frac{18}{M}[/tex]
[tex]0.016=\frac{18}{M}\times 0.0821\times 283}[/tex]
[tex]M=26138g/mol[/tex]
The molecular weight of the macromolecule in grams per mole is 26138.
As a technician in a large pharmaceutical research firm, you need to produce 350. mL of 1.00 M potassium phosphate buffer solution of pH = 7.07. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution?
Answer:
You need to add 203 mL of 1,00M KH₂PO₄ and 147 mL of 1,00M K₂HPO₄.
Explanation:
It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:
pH = pka + log₁₀
Where A⁻ is conjugate base and HA is conjugate acid
The equilibrium of phosphate buffer is:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,21
Thus, Henderson–Hasselbalch equation for 7,07 phosphate buffer is:
7,07 = 7,21 + log₁₀ [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]
0,7244 = [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex] (1)
As the buffer concentration must be 1,00 M:
1,00 = [H₂PO₄⁻] + [HPO4²] (2)
Replacing (2) in (1):
[H₂PO₄⁻] = 0,5799 M
Thus:
[HPO4²] = 0,4201 M
To obtain these concentrations you need to add:
0,5799 M × 0,350 L × [tex]\frac{1L}{1mol}[/tex] = 0,203 L ≡ 203 mL of 1,00M KH₂PO₄
And:
0,4201 M × 0,350 L × [tex]\frac{1L}{1mol}[/tex] = 0,203 L ≡ 147 mL of 1,00M K₂HPO₄
I hope it helps!
Predict the direction and the relative value of the dipole moment of the following bond:
(A) C←Cl, electronegativity difference > 0.5
(B) C←Cl, electronegativity difference < 0.5
(C) C→Cl, electronegativity difference > 0.5
(D) C→Cl, electronegativity difference < 0.5
Final answer:
The dipole moment points toward the more electronegative atom, Cl, in a C-Cl bond. The greater the electronegativity difference, the larger the dipole moment. Thus, (A) C←Cl with an electronegativity difference > 0.5 correctly predicts the dipole direction and suggests a significant dipole moment.
Explanation:
The direction and relative value of the dipole moment in a bond between two atoms depends on the difference in their electronegativities. Electronegativity is a measure of an atom's ability to attract electrons in a chemical bond. When two atoms form a bond, the more electronegative atom will attract the bonding electrons more strongly and will acquire a partial negative charge, while the less electronegative atom will have a partial positive charge.
Regarding the options given:
(A) C←Cl, with an electronegativity difference > 0.5 would have a dipole moment pointing towards the Cl, because Cl is more electronegative than C.
(B) C←Cl, with an electronegativity difference < 0.5 might suggest a relatively nonpolar bond, which is uncommon for C-Cl and might not be a realistic scenario.
(C) C→Cl, with an electronegativity difference > 0.5 would incorrectly suggest that C is more electronegative than Cl, which is not the case; thus this depiction of the dipole direction is incorrect.
(D) C→Cl, with an electronegativity difference < 0.5 would also be incorrect as it suggests the incorrect direction of the dipole moment.
The relative value of the dipole moment will be greater when the electronegativity difference is greater, leading to a stronger separation of charges, and hence a larger dipole moment.
Methane is burned to complete combustion with 30% excess air.
The air enters at 25oC and 1 atm (absolute) at 50%
relative humidity.
a) Calculate the flow of O2 per mole of methane
entering the process.
b) Calculate the flow of N2 per mole of methane
entering the process.
c) Calculate the mole fraction of H2O in the humid
air stream entering the process.
d) Calculate the flow of H2O per mole of methane
entering the process.
Answer:
a) 2.6 mole of O2 per mole of methane
b) 9.78 mole of N2 per mole of methane
c) Mole fraction of H2O = 0.33
d) 6.12 mole of H2O pero mole of methane
Explanation:
The first thing you have to do is the balanced reaction of the methane combustion:
CH4 +2 O2 -->2 H2O + CO2
You calculate the moles of O2 needed to react with 1 mole of methane and add the excess factor
2 x 1.3 = 2.6 moles of O2
For the N2 moles you calculate it with the air composition (21 % O2, 79% N2)
2.6 moles of O2 * 0.79/0.21 = 9.78 moles of N2
With the total stream of air = 12.38 moles you add the humidity factor
12.38 * 1.5 = 18.57 moles So 6.12 are moles of H20 entering per mole of CH4.
To calculate the mole fraction yo divide the moles of water among the moles of the stream:
6.12/18.57 --> 0.33
The solubility of acetanilide is 0.53 g in 100 mL of ice‑cold water, and 5.50 g in 100 mL of boiling water. What is the maximum percent recovery that can be achieved for the recrystallization of acetanilide from water?
Answer:
MPR=90,36%
Explanation:
The recrystallization is a purification process where the solid to purify is dissolved in an appropriated dissolvent and then, changing the conditions the solubility changes and that solid (that was in solution before) precipitates and form crystals.
In this case, for boiling water 5.50 g of acetanilide could be dissolved and then cold the water, so the mass of crystals formed will be
[tex]m_{crystal}=\left(5.5-0.53\right)g=4.97g[/tex]
The maximum percent recovery is then
[tex]MPR=\frac{mass\ of\ solid\ recovered}{mass\ of\ solid\ disolved}\times 100\% = \frac{4.97g}{5.5g}\times 100\%=90,36\%[/tex]
A chemistry student needs 50.0 mL of chloroform for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of chloroform is 1.48 g.cm . Calculate the mass of chloroform the student should weigh out. Round your answer to 3 significant digits. x 6 ?
The mass of chloroform the student should weigh out will be 74.0g of chloroform.
What is chloroform?Chloroform is a name of the gas whose chemical name is nitrous oxide. It is a gas that is used to freeze the area or sense of a body part when there is any operation or treatment.
"The mass per unit volume is known as density. A scalar quantity, density. It is represented by the letter D, and the Greek letter rho is used as the sign for density". "Mass divided by volume is how density is computed."
"Mass is a physical body's total amount of matter. Mass is defined as the sum of the moles of the material and the compound's molar mass".
Density relates to mass and volume, and 1 cm⁻³ = 1 mL:
1.48 g of chloroform 1 cm⁻³ chloroform
m = 50.0 cm ⁻³ chloroform
m = 74.0 g of chloroform.
Therefore, the chemistry student will need to weigh 74.0g of chloroform.
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The molar mass of chloroform is calculated using the ideal gas law and the conditions provided (mass, volume, temperature, pressure). By converting the conditions to appropriate units and applying the formula, the calculated molar mass should closely approximate the given value of 119.37 amu.
Explanation:To calculate the molar mass of chloroform (CHCl3), we use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. First, we need to convert the given temperature of 99.6 °C to Kelvin by adding 273.15, resulting in 372.75 K. Next, we convert the pressure from mm Hg to atmospheres by dividing by 760. So, the pressure is 0.9765 atm. Using the ideal gas law and rearranging for n (n = PV/RT), we can find the number of moles of chloroform. Finally, we calculate the molar mass by dividing the mass of the chloroform sample (0.494 g) by the number of moles calculated.
The density of chloroform mentioned is not directly needed for calculating molar mass in this context. The provided molecular mass of chloroform, 119.37 amu, serves as a reference and validation of our calculation.
A lead atom has a mass of 3.14 x 10 to the negative 22nd g.How
many lead atoms are in a cube of lead that has a volume of 2.00cm
cubed, if the density of lead is 11.3 g/cm cubed?
Explanation:
The given data is as follows.
Mass of a lead atom = [tex]3.14 \times 10^{-22}[/tex]
Volume = 2.00 [tex]cm^{3}[/tex]
Density = 11.3 [tex]g/cm^{3}[/tex]
As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.
So, in 2 cubic centimeter there will be [tex]2 \times 11.3 g = 22.6 g[/tex] of lead atoms.
One lead atom has a mass of [tex]3.14 \times 10^{-22}[/tex]. Therefore, number of atoms present in 22.6 g of lead will be as follows.
[tex]\frac{22.6 g}{3.14 \times 10^{-22}}[/tex]
= [tex]7.197 \times 10^{22} atoms[/tex]
Thus, we can conclude that there are [tex]7.197 \times 10^{22} atoms[/tex] of lead are present.
Give the ΔH value for the formation of binary compounds as shown in the reaction H2(s)+Br2(g)→2HBr(s)+36.3kJ.
Express your answer using three significant figures. If the value is positive, do not include the + sign in your answer.
The ΔH value for the formation of Hydrogen Bromide (HBr) from Hydrogen and Bromine is -36.3 kJ, indicative of an exothermic reaction.
Explanation:In the given chemical reaction, the formation of the binary compound Hydrogen Bromide (HBr) is exothermic, meaning it releases energy. This is denoted by the negative ΔH value, which refers to the change in enthalpy or total energy of the system. Given that the reaction releases 36.3 kJ, the ΔH of the reaction is -36.3 kJ. Expressing this with three significant figures, the ΔH value becomes -36.3 kJ. This value is negative which indicates that the reaction is exothermic—energy is released in the formation of the compounds.
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Consider the reaction: 2A(g)+B(g)→3C(g).
Part A
Determine the expression for the rate of the reaction with respect to each of the reactants and products.
a) Rate=−13Δ[A]Δt=−Δ[B]Δt=12Δ[C]Δt
b) Rate=−12Δ[A]Δt=−Δ[B]Δt=13Δ[C]Δt
c) Rate=−Δ[A]Δt=−12Δ[B]Δt=13Δ[C]Δt
d) Rate=12Δ[A]Δt=12Δ[B]Δt=13Δ[C]Δt
Part B
When A is decreasing at a rate of 0.100 M⋅s−1 , how fast is B decreasing?
Part C
How fast is C increasing?
Answer:
Part A
[tex]Rate = -\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t} = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]
Part B
[tex]-\frac{\Delta B}{\Delta t}= 0.0500 M s^{-1} [/tex]
Part C
[tex]\frac{\Delta C}{\Delta t} = 0.15 M s^{-1}[/tex]
Explanation:
For a general reaction,
[tex]aA(g) + bB(g) \rightarrow cC(g)[/tex]
Rate is given by:
Rate: [tex]Rate = -\frac{1}{a}\frac{\Delta A}{\Delta t} =-\frac{1}{b}\frac{\Delta B}{\Delta t} = \frac{1}{c}\frac{\Delta C}{\Delta t}[/tex]
So, for the given reaction:
[tex]2A(g) + B(g) \rightarrow 2C(g)[/tex]
[tex]Rate = -\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t} = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]
Part B
[tex]-\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t}[/tex]
Given: [tex]-\frac{\Delta A}{\Delta t} = 0.100\ Ms^{-1}[/tex]
[tex] \frac{1}{2}\frac{0.100}{\Delta t} =-\frac{\Delta B}{\Delta t}[/tex]
[tex]-\frac{\Delta B}{\Delta t}[/tex] = 0.0500 M s^-1
Part C
[tex]-\frac{\Delta B}{\Delta t} =\frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]
[tex]-\frac{\Delta B}{\Delta t} =\frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]
[tex]0.0500 = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]
[tex]\frac{\Delta C}{\Delta t} = 3 \times 0.0500 = 0.15 M s^{-1}[/tex]
The rate equation is Rate=−1/2Δ[A]/Δt=−Δ[B]/Δt=1/3Δ[C]/Δt. The rate of decrease of B is 0.100 M⋅s−1, and the rate of increase of C is 0.150 M⋅s−1.
Explanation:The correct expression for the rate of the given reaction 2A(g)+B(g)→3C(g) is: Rate=−1/2Δ[A]/Δt=−Δ[B]/Δt=1/3Δ[C]/Δt. This is because the rate of disappearance of A and B and the rate of appearance of C are all proportional to the stoichiometric coefficients in the balanced chemical equation.
For Part B, since A is decreasing at a rate of 0.100 M⋅s−1, B will be decreasing at the same rate because the rate is proportional to their coefficients in the balanced equation. So, B is decreasing at a rate of 0.100 M⋅s−1.
For Part C, if A is decreasing at a rate of 0.100 M⋅s−1, since the rate of increase of product C is 1.5 times the rate of decrease of reactant A (according to their coefficients). Hence, C is increasing at a rate of 0.150 M⋅s−1.
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The speed of light in a vacuum is 2.998 x 108 m/s. What is its speed in kilometers per hour (km/h)? העתK speed = What is its speed in miles per minute (mi/min)? speed = mi/min
Final answer:
The speed of light in a vacuum converted to kilometers per hour (km/h) is 1.079 x 10^12 km/hr, and in miles per minute (mi/min) is 11184.71 mi/min.
Explanation:
The speed of light in a vacuum is 2.998 x 108 m/s. To convert this speed into kilometers per hour (km/h), you multiply by the number of meters in a kilometer (1000) and the number of seconds in an hour (3600). This calculation gives us:
2.998 x 108 m/s x 1000 m/km x 3600 s/hr = 1.079 x 1012 km/hr.
Similarly, to find the speed in miles per minute (mi/min), you must use the conversion factor that 1 meter is approximately equal to 0.000621371 miles, and there are 60 seconds in a minute:
2.998 x 108 m/s x 0.000621371 mi/m x 60 s/min = 11184.71 mi/min.
Pick the correct pair of species that can form hydrogen bond with water. (A) CH, HCOOH (B) F, HCOOH (C) F .CH,OCH, (D) Both (B) and (C)
Answer:
(B) F⁻, HCOOH
Explanation:
(A) CH₄, HCOOH
(B) F⁻, HCOOH
(C) F⁻, CH₃-O-CH₃
The hydrogen bonds are formed when the hydrogen is found between two electronegative atoms such as oxygen (O), nitrogen (N) or florine (F).
O····H-O, F····H-O, O····H-N
(A) CH₄, HCOOH
- here methane CH₄ is not capable to form hydrogen bond with water
- formic acid HCOOH can form hydrogen bonds with water
H-C(=O)-O-H····OH₂
(B) F⁻, HCOOH
-both floride (F⁻) and formic acid can form hydrogen bonds with water
F····OH₂
H-C(=O)-O-H····OH₂
(C) F⁻, CH₃-O-CH₃
- dimethyl-ether CH₃-O-CH₃ is not capable to form hydrogen bond with water
- floride (F⁻) can form hydrogen bonds with water
F····OH₂
The correct pairs of species that can form hydrogen bonds with water are option (B) F, HCOOH and option (C) F.CH, OCH. This is because hydrogen bonds are formed between hydrogen and a highly electronegative atom such as Oxygen, Nitrogen, or Fluorine.
Explanation:The correct pairs of species that can form hydrogen bonds with water are option (B) F, HCOOH and option (C) F .CH, OCH. Hydrogen bonds are primarily formed between hydrogen and a highly electronegative atom such as Oxygen, Nitrogen, or Fluorine, which are present in both HCOOH (formic acid) and OCH (a group from a larger molecule such as methanol).
Hydrogen bonds form due to the attraction between the slightly positive Hydrogen of one molecule and the slightly negative Oxygen, Nitrogen, or Fluorine of another. For example, in a water molecule, the oxygen atom carries a slight negative charge due to its higher electronegativity while hydrogen atoms carry a slight positive charge.
Option (A) CH cannot form a hydrogen bond with water as it is a nonpolar molecule and lacks an electronegative atom. Also, individual fluorine atoms as given in option (B) do not form hydrogen bonds as they lack the H-F bond necessary for hydrogen bonding.
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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, and 49.00 amu. The lightest-weight isotope has a natural abundance of 10.0%. What is the percent abundance of the heaviest isotope?
Answer : The percent abundance of the heaviest isotope is, 78 %
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
As we are given that,
Average atomic mass = 48.68 amu
Mass of heaviest-weight isotope = 49.00 amu
Let the percentage abundance of heaviest-weight isotope = x %
Fractional abundance of heaviest-weight isotope = [tex]\frac{x}{100}[/tex]
Mass of lightest-weight isotope = 47.00 amu
Percentage abundance of lightest-weight isotope = 10 %
Fractional abundance of lightest-weight isotope = [tex]\frac{10}{100}[/tex]
Mass of middle-weight isotope = 48.00 amu
Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %
Fractional abundance of middle-weight isotope = [tex]\frac{(90-x)}{100}[/tex]
Now put all the given values in above formula, we get:
[tex]48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})][/tex]
[tex]x=78\%[/tex]
Therefore, the percent abundance of the heaviest isotope is, 78 %
Answer:
78 %
Explanation:
The atomic mass is the weighted average of the atomic masses of each isotope.
In a weighted average, we multiply each value by a number representing its relative importance.
In this problem, the percent abundance represents the relative importance of each isotope.
Data:
X-47: mass = 47.00 u; abundance = 10.0 % = 0.100
X-48: mass = 48.00 u
X-49: mass = 49.00 u
Calculations:
Let x = abundance of X-49
Then 0.900 - x = abundance of X-48
[tex]\begin{array}{cccc}\\\textbf{Isotope} & \textbf{Mass/u} & \textbf{Abundance} & \textbf{Contribution/u}\\\text{X-47} & 47.00 & 0.100 & 4.700\\\text{X-48} & 48.00 & 0.900 - x & 48.00(0.900 - x)\\\text{X-49} & 49.00 & x & 49.00x\\& \text{TOTAL} & = & \mathbf{48.68}\\\end{array}[/tex]
[tex]\begin{array}{rcr}4.700 + 48.00(0.900 - x) + 49.00x & = & 48.68\\4.700 + 43.20 - 48.00x + 49.00x & = & 48.68\\47.90 +x & = & 48.68\\x & = & \mathbf{0.78}\\\end{array}[/tex]
The heaviest isotope has an abundance of 78 %.
Aluminium does not corrode in a normal atmosphere because: a) Aluminum is unreactive with oxygen b) The reaction with oxygen has too high and activation energy. c) The formation of an oxide layer inhibits further diffusion and corrosion d) None of the above
Answer:
The correct option is: c) The formation of an oxide layer inhibits further diffusion and corrosion
Explanation:
Aluminium is a chemical element which belongs to the group 13 of the periodic table and has atomic number 13. It is a soft metal and a member of the p-block.
When aluminium is exposed to the normal atmosphere, the top layer of the metal gets oxidized to form a thin protective layer of aluminium oxide. The thin and hard protective aluminium oxide layer then protects the metal from getting further corroded. This process is called passivation.
Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.
Repeat the calculation for the time until overflow for the case of water entering a 4.00-m3 tank at a rate of 6.83 kg/s and withdrawn at a rate of 3.50 kg/s. The tank is initially two thirds full.
Answer:
(a) The time until overflow is 649 s
(b) The time until overflow is 355 s
Explanation:
The volume as a function of time can be expressed as
[tex]V(t) = V_0+(q_i-q_o)*t[/tex]
If the tank is initially half full, V(0) = V0 = 4/2 = 2 m3.
With ρ=1000 kg/m3, the volume flows are
Flow in = 6.33 kg/s * 0.001 m3/kg = 0.00633 m3/s
Flow out = 3.25 kg/s * 0.001 m3/kg = 0.00325 m3/s
The time until overflow (V(t)=4 m3) is
[tex]V(t) = 2+(0.00633 - 0.00325)*t=2+0.00308*t=4\\\\t=(4-2)/0.00308 = 649.4 s=11 min[/tex]
If the flows are
Flow in = 6.83 kg/s * 0.001 m3/kg = 0.00683 m3/s
Flow out = 3.50 kg/s * 0.001 m3/kg = 0.0035 m3/s
And the tank is initially 2/3 full (V(0)=2.67 m3)
The time until overflow is
[tex]V(t) = 2.67+(0.00683 - 0.00350)*t=2.67+0.00375*t=4\\\\t=(4-2,67)/0.00375 = 354.67 s =6 min[/tex]
In a particular experiment at 300 ∘C, [NO2] drops from 0.0100 to 0.00800 M in 100 s. The rate of appearance of O2 for this period is ________ M/s. In a particular experiment at 300 , drops from 0.0100 to 0.00800 in 100 . The rate of appearance of for this period is ________ . 4.0×10−3 2.0×10−3 2.0×10−5 4.0×10−5 1.0×10−5
Answer:
[tex]1\times{10}^{-5}\frac{M}{s}[/tex]
Explanation:
The stoichiometry for this reaction is
[tex]2NO_2\rightarrow2NO+O_2[/tex]
The rate for this reaction can be written as
[tex]-r_{NO_2}=-\frac{d\left[NO_2\right]}{dt}=\frac{(0.01-0.008)M}{100s}=2\times{10}^{-5}\frac{M}{s}[/tex]
This rate of disappearence of [tex]NO_2[/tex] can be realated to the rate of appearence of [tex]O_2[/tex] as follows (the coefficients of each compound are defined by the stoichiometry of the reaction)
[tex]-r_{O_2}=-r_{NO_2}\times\frac{coefficient\ O_2\ }{coefficient\ NO_2}=2\times{10}^{-5}\frac{M}{s}\times\frac{1\ mole\ O_2\ }{2\ mole\ NO_2}=1\times{10}^{-5}\frac{M}{s}[/tex]
A mixture is 20.00 mole% methyl alcohol, 60.0 mole% methyl acetate, and 20.0 mole% acetic acid.
What is the mass of a sample containing 45.0 kmol of methyl acetate?
Answer:
4714.950 kilograms is the mass of a sample containing 45.0 kmol of methyl acetate.
Explanation:
Moles of methyl acetate =[tex]n_1[/tex]=45.0 kmol= 45000 mol
Mole percentage of methyl acetate = 60.0%
Total moles in the sample = n
[tex]60.0\%=\frac{45000 mol}{n}\times 100[/tex]
[tex]n=\frac{45000 mol}{60.0}\times 100=75000 mol[/tex]
Mole percentage of methyl alcohol = 20.0%
Moles of methyl alcohol = n_2
[tex]20.0\%=\frac{n_2}{75000 mol}\times 100[/tex]
[tex]n_2=15,000 mol[/tex]
Mass of methyl alcohol = [tex]n_2\times 32.04 g/mol[/tex]
=[tex]15000 mol\times 32.04 g/mol=480,600 g[/tex]
Mole percentage of acetic acid = 20.0%
Moles of acetic acid = n_3
[tex]20.0\%=\frac{n_3}{75000 mol}\times 100[/tex]
[tex]n_3=15,000 mol[/tex]
Mass of acetic acid= [tex]n_3\times 60.05 g/mol[/tex]
[tex]15000 mol\times 60.05 g/mol=900,750 g[/tex]
Mass of methyl methyl acetate= [tex]n_1\times 74.08 g/mol[/tex]
[tex]45000 mol\times 74.08 g/mol =3,333,600 g[/tex]
Mass of sample: 480,600 g + 3,333,600 g + 900,750 g = 4714950 g
4714950 g = 4714.950 kg
(1 kg = 1000 g)
To find the mass of a sample with 45.0 kmol of methyl acetate, multiply the kmol amount by the molar mass of methyl acetate (74.08 g/mol), resulting in 3333.6 kg.
Explanation:To calculate the mass of a sample containing 45.0 kmol (kilo moles) of methyl acetate, we first need to know the molar mass of methyl acetate. Methyl acetate (C3H6O2) has a molar mass of approximately 74.08 g/mol. Knowing this, we can calculate the mass of the methyl acetate in the sample.
The calculation is as follows:
Determine the molar mass of methyl acetate: 74.08 grams per mole (g/mol).Multiply the amount of substance (in moles) by the molar mass: 45.0 kmol x 74.08 g/mol = 3333600 grams or 3333.6 kilograms.This calculation reveals that a sample containing 45.0 kmol of methyl acetate has a mass of 3333.6 kilograms.
Consider the combustion of carbon monoxide (CO) in
oxygengas
2CO + O2 ----> 2CO2
Starting with 3.60 moles of CO, calculate the
numberof moles of CO2 procduced if there is enoughoxygen
gas to react with all of the CO.
Answer:
3.60 mol CO₂
Explanation:
Balanced chemical reaction:
2CO + O₂ ⇒ 2CO₂
The molar ratio between CO₂ and CO is 1:1
2CO₂/2CO = CO₂/CO
Thus, the moles of CO₂ produced from 3.60 moles of CO is 3.60 moles:
(3.60 mol CO)(CO₂/CO) = 3.60 mol CO₂
You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?
a.12.3 g
b.44.0 g
c.24.5 g
d.88.0 g
Please show all steps, so I can understand how to do this, thanks!
Answer:
24.525 g of sulfuric acid.
Explanation:
Hello,
Normality (units of eq/L) is defined as:
[tex]N=\frac{eq_{solute}}{V_{solution}}[/tex]
Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:
[tex]eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq[/tex]
Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:
[tex]m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4[/tex]
Best regards.
A compound consists of 47.5% S and 52.5% Cl by mass. Draw the Lewis structure based on the empirical formula and comment on its deficiencies. Draw a more plausible structure with the 7. same ratio of S and Cl
Answer:
Look picture
Explanation:
The empirical formula is obtained with average of each atom by mass over Atomic weight, thus:
47,5% S ÷ 32,045 g/mol = 1,48 mol S
52,5% Cl ÷ 35,45 g/mol = 1,48 mol Cl
Thus, empirical formula is:
[tex]S_{1,48} Cl_{1,48}[/tex] ≡ SCl.
The Lewis structure for this SCl molecule is in the picture. You can see one unpaired electron in S. These unpaired electrons are very unstable doing SCl an improbable molecule.
The more pausible structure with the same S:Cl ratio is S₂Cl₂ (Look picture). In this molecule you don't have unpaired electrons doing this compound more stable (In fact, exist, and its name is Disulfur dichloride)
I hope it helps!
Answer:
See explanation.
Explanation:
Hello,
In this case, one could identify the subscripts in the empirical formula, based on the given percentages as shown below:
[tex]n_S=\frac{47.5g}{32g/mol}=1.48 \\n_{Cl}=\frac{52.5g}{35.45g/mol}=1.48\\ S=\frac{1.48}{1.48} =1;Cl=\frac{1.48}{1.48} =1[/tex]
Thus, the empirical formula is:
[tex]SCl[/tex]
Nevertheless, such empirical formula does not respect the sulfur's octet, based on the first drawing on the attached picture (that is a deficiency), that is why a more plausible structure is based on the following formula:
[tex]S_2Cl_2[/tex]
Which actually respect the octet based on the second drawing on the attached picture.
Best regards.
Enter your answer in the provided box. Calculate the number of moles of CrCl, that could be produced from 49.4 g Cr202 according to the equation Cr2O3(s) + 3CC14(7) ► 2CrC13(s) + 3COCl(aq) D mol CrCiz
Answer:
0.4694 moles of CrCl₃
Explanation:
The balanced equation is:
Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)
The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.
The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:
MCr = 52 g/mol
MCl = 35.5 g/mol
MO = 16 g/mol
So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.
The number of moles is the mass divided by the molar mass, so:
n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.
For the stoichiometry:
1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃
0.2347 mol of Cr₂O₃----------- x
By a simple direct three rule:
x = 0.4694 moles of CrCl₃
Final answer:
The number of moles of CrCl3 that can be produced from 49.4 g of Cr2O3 is calculated by dividing the mass of Cr2O3 by its molar mass and then using the stoichiometry of the balanced chemical equation, resulting in approximately 0.65 moles of CrCl3.
Explanation:
To calculate the number of moles of CrCl3 that could be produced from 49.4 g of Cr2O3, we need to first find the molar mass of Cr2O3. Chromium has an atomic mass of approximately 52.00 g/mol and oxygen has an atomic mass of approximately 16.00 g/mol, which makes the molar mass of Cr2O3 to be about (2*52.00 g/mol) + (3*16.00 g/mol) = 152.00 g/mol.
Next, we divide the mass of Cr2O3 by its molar mass to find the number of moles:
Number of moles of Cr2O3 = mass of Cr2O3 / molar mass of Cr2O3Number of moles of Cr2O3 = 49.4 g / 152.00 g/mol ≈ 0.325 moles of Cr2O3Using the balanced chemical equation, we see that 1 mole of Cr2O3 produces 2 moles of CrCl3. Therefore, 0.325 moles of Cr2O3 will produce 0.325 x 2 moles of CrCl3, which is approximately 0.65 moles of CrCl3.
Final answer:
The number of moles of CrCl3 that can be produced from 49.4 g of Cr2O3 is calculated by dividing the mass of Cr2O3 by its molar mass and then using the stoichiometry of the balanced chemical equation, resulting in approximately 0.65 moles of CrCl3.
Explanation:
To calculate the number of moles of CrCl3 that could be produced from 49.4 g of Cr2O3, we need to first find the molar mass of Cr2O3. Chromium has an atomic mass of approximately 52.00 g/mol and oxygen has an atomic mass of approximately 16.00 g/mol, which makes the molar mass of Cr2O3 to be about (2*52.00 g/mol) + (3*16.00 g/mol) = 152.00 g/mol.
Next, we divide the mass of Cr2O3 by its molar mass to find the number of moles:
Number of moles of Cr2O3 = mass of Cr2O3 / molar mass of Cr2O3Number of moles of Cr2O3 = 49.4 g / 152.00 g/mol ≈ 0.325 moles of Cr2O3Using the balanced chemical equation, we see that 1 mole of Cr2O3 produces 2 moles of CrCl3. Therefore, 0.325 moles of Cr2O3 will produce 0.325 x 2 moles of CrCl3, which is approximately 0.65 moles of CrCl3.
A 32.50-g sample of a solid is placed in a flask. Toluene, in which the solid is insoluble, is added to the flask so that the total volume of solid and liquid together is 55.00 mL . The solid and toluene together weigh 58.68 g . The density of toluene at the temperature of the experiment is 0.864 g/mL. What is the density of the solid?
Answer:
The density of the solid is 1,316 g/mL
Explanation:
The weight of both Toluene and the solid insoluble is 58,68 g. And the weight of the solid is 32,50 g. Thus, weight of toluene is:
58,68 g - 32,50 g = 26,18 g of Toluene
To know how much volume that toluene occupy you must use density thus:
26,18 g of toluene × ( 1 mL / 0,864 g) = 30,30 mL of toluene
The volume of both Toluene and the solid is 55,00 mL and the volume of toluene is 30,30 mL. Thus, the volume of the solid is:
55,00 mL - 30,30 mL = 24,70 mL
Knowing both volume and weight it is possible to know the density thus:
32,50 g / 24,70 mL = 1,316 g/mL
I hope it helps!
How many liters of 0.1107 M KCI contain 15.00 g of KCI (FW 74.6 g/mol)? 0.02227 L O 0.5502 L 1.661 L O 1816 L 18.16 L ent Navigator PrtScr Delete FB F9 F10 F11 F12 Insert Backspace 6 U P
Answer: The volume of solution required is 1.816 L
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
We are given:
Mass of solute (KCl) = 15.00 g
Molar mass of potassium chloride = 74.6 g/mol
Molarity of solution = 0.1107 M
Putting values in above equation, we get:
[tex]0.1107M=\frac{15.00g}{74.6g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=1.82L[/tex]
Hence, the volume of solution required is 1.816 L
To determine the volume of 0.1107 M KCl solution containing 15.00 g of KCl, we can use the formula Molarity (M) = moles of solute / volume of solution (in liters). By calculating the moles of KCl and using the molarity formula, the volume of the solution is found to be 1.814 L.
Explanation:To calculate the number of liters of 0.1107 M KCl that contain 15.00 g of KCl, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, we need to calculate the moles of KCl using its molecular weight:
Moles of KCl = mass of KCl / molecular weight of KCl
Moles of KCl = 15.00 g / 74.6 g/mol = 0.201 moles of KCl
Now, we can use the formula to calculate the volume of solution:
Volume of solution = moles of solute / molarity
Volume of solution = 0.201 moles / 0.1107 M = 1.814 L
bre: 1073/2900 Resources A sample of metal has a mass of 24.02 g, and a volume of 5.02 mL. What is the density of this metal? g/cm privacy policy terms of use contact us hele MacBook Pro
Answer:
4.78 g/cm³
Explanation:
Density is expressed as mass per unit volume:
D = m/V
D = (24.02 g) / (5.02 mL) = 4.78 g/mL
The density should be expressed in g/cm³, so mL must be converted to cm³. The conversion ratio is 1 mL = 1 cm³.
(4.78 g/mL)(1 mL/1 cm³) = 4.78 g/cm³
A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is: i. 64 grains/in ii. 500 grains/in? iii. 20 grains/in-
Answer:
A) M = 100X
B) M = 36X
C) M = 178.88X
Explanation:
Given data:
ASTM grain size number 7
a) total grain per inch^2 - 64 grain/inch^2
we know that number of grain per square inch is given as
[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]
where M is magnification, n is grain size
therefore we have
[tex]64 = 2^{7-1}(\frac{100}{M})^2[/tex]
solving for M we get
M = 100 X
B) total grain per inch^2 = 500 grain/inch^2
we know that number of grain per square inch is given as
[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]
where M is magnification, n is grain size
therefore we have[tex]500 = 2^{7-1}(\frac{100}{M})^2[/tex]
solving for M we get
M = 36 X
C) Total grain per inch^2 = 20 grain/inch^2
we know that number of grain per square inch is given as
[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]
where M is magnification, n is grain size
therefore we have[tex]20 = 2^{7-1}(\frac{100}{M})^2[/tex]
solving for M we get
M = 178.88 X
Define stereochemistry
Answer:
Stereochemistry is a branch of chemistry that studies the spatial arrangement of atoms or groups in a molecule.
The molecules with the same molecular formula, bond connectivity, and reactivity but a different arrangement of atoms in the space are known as stereoisomers. These molecules interact differently in a chiral environment or optical light.
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a new solution in which [K+] is 0.846 M, what is the concentration of the potassium dichromate used to make the new solution?
Answer:
The concentration of the Potassium Dichromate solution is 0.611 M
Explanation:
First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:
KBr (aq) → K+ (aq) + Br- (aq)
K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)
According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.
Having said that, we calculate the moles of potassium ions coming from the KBr solution:
0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:
1000 mL solution ----- 0.19 moles of KBr
253 mL solution ----- x = 0.04807 moles of KBr
As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.
Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):
1000 mL solution ----- 0.846 moles K+
694 mL solution ----- x = 0.587124 moles K+
This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:
0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7
If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:
2 K+ moles ----- 1 K2Cr2O7 mole
0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles
Now this quantity of potassium dichromate moles came from the respective solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:
441 mL ----- 0.269527 K2Cr2O7 moles
1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M
The concentration of the original potassium dichromate solution used in the mixture is 1.222 M. This was calculated by analyzing the total potassium concentration in the final solution, subtracting the contribution of potassium bromide, and then considering the volume of the potassium dichromate solution.
Explanation:This is a problem related to the principle of conservation of matter focusing on concentration and volume in a chemical solution. As per the question, the potassium ion concentration ([K+]) in the final solution is 0.846 M. We know that the total amount of potassium (in moles) comes from both the potassium bromide and the potassium dichromate.
First, we calculate the moles of potassium from the potassium bromide: volume (L) x concentration (M) = 0.253 L x 0.19 M = 0.04807 moles. Now, consider that the total volume of the solution is 253 mL + 441 mL = 694 mL or 0.694 L. Since the given final concentration of the mixed solution is 0.846 M, the total moles of potassium in the solution would be: 0.694 L x 0.846 M = 0.587124 moles. We subtract the moles of potassium from the potassium bromide to find the moles contributed by potassium dichromate: 0.587124 moles - 0.04807 moles = 0.539054 moles.
This is the amount of potassium in the potassium dichromate solution. To find concentration, we divide this by the volume of the potassium dichromate solution: 0.539054 moles / 0.441 L = 1.222 M. So, the concentration of the original potassium dichromate solution is 1.222 M.
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