Answer:
C. Electron cloud
the electron is around 1/2000 times the size of the proton.
If you imagine the proton a a marble in the middle of a football field, the electrons will revolve around the last row
Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
Answer:
The correct answer is 1332 KJ.
Explanation:
Based on the given information,
ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.
Now the balanced equation is:
C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)
ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2
ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)
ΔH°rxn = -1411.9 KJ
Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.
Now based on the balanced chemical reaction,
ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2
ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)
ΔS°rxn = -267.68 J/K or -0.26768 KJ/K
T = 25 °C or 298 K
Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get
ΔG = -1411.9 - 298.0 × (-0.2677)
ΔG = -1332 KJ.
Thus, the maximum work, which can obtained is 1332 kJ.
Using only sodium carbonate, Na2CO3, sodium bicarbonate, NaHCO3, and distilled water determine how you could prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3. The total molarity of the ions should be 0.20 M. The Ka of the hydrogen carbonate ion, HCO3 - , is 4.7 x 10-11 .
Answer:
Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.
Explanation:
The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:
[tex] pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]
We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:
[tex] 10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]
[tex] \frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex] (1)
Also, we know that:
[tex] [Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M [/tex] (2)
From equation (2) we have:
[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] [/tex] (3)
By entering (3) into (1):
[tex] \frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]
[tex] 0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20 [/tex]
[tex] [NaHCO_{3}] = 0.103 M [/tex]
Hence, the [Na_{2}CO_{3}] is:
[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M [/tex]
Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:
[tex]m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g[/tex]
[tex]m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g[/tex]
Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.
I hope it helps you!
2) What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 230 mL of solution?
When scientists are ready to publish the result of their experiments why is it important for them to include a description of the procedure they used
Answer: So other scientist can replicate the experiment and see if they get the same results in other words, test reliability.
Explanation:
The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 8.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M
Answer:
0.9718M
Explanation:
Rate constant, k = 0.255 M-1s-1
time, t = 8.00 s
Initial concentration, [A]o = 1.33 M
Final concentration, [A] = ?
These quantities are represented by the equation;
1 / [A] = 1 / [A]o + kt
1 / [A] = 1 /1.33 + (0.255 * 8)
1 / [A] = 0.7519 + 2.04
[A] = 1 / 2.7919 = 0.3582 M
How much of NO2 decomposed is obtained from the change in concentration;
Change in concentration = Initial - Final
Change = 1.33 - 0.3582 = 0.9718M
Acetonitrile (CH3CN) is an important industrial chemical. Among other things, it is used to make plastic moldings, which have multiple uses, from car parts to Lego bricks. Which one of the following statements about acetonitrile is not correct?a. Acetonitrile has 16 valence electrons in its Lewis structure. b. Acetonitrile has one triple bond. c. Acetonitrile has one pair of nonbonding electrons. d. All atoms satisfy the octet rule in acetonitrile. e. One carbon atom and the nitrogen atom have nonzero formal charges.
Answer:
One carbon atom and the nitrogen atom have nonzero formal charges.
Explanation:
The compound Acetonitrile has sixteen valence electrons as is easily San from its structure. It contains a carbon nitrogen triple bond with a lone pair of electrons on nitrogen. All atoms satisfy the octet rule and there is no hyper valent atom in the molecule.
The formal charge an carbon and nitrogen is calculated as follows;
No. of valence electron on atom - [non bonded electrons + no. of bonds]
Therefore, for carbon and nitrogen, we have;
formal charge on carbon = 4 - (0 + 4) = 0
formal charge on nitrogen = 5 - (2 + 3) = 0
Hence carbon and nitrogen both possess zero formal charges.
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
Br(g)
Cl2(g)
I2(g)
F2(g)
B. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2S(g)
H2O(g)
H2O2(g)
C. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous)
C(s, diamond)
C(s, graphite)
Answer:
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Explanation:
Hello,
In this case, we can apply the following principles to explain the order:
- The greater the molar mass, the larger the standard molar entropy.
- The greater the molar mass and the structural complexity, the larger the standard molar entropy.
- The greater the structural complexity, the larger the standard molar entropy.
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
This is due to the fact that the greater the molar mass, the larger the standard molar entropy.
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).
Regards.
The literature value for the Ksp of Ca(OH)2 at 25 °C is 4.68E−6. Imagine you ran the experiment and got a calculated value for Ksp which was too high. Select all of the possible circumstances which would cause this result.
A. The HCl was more concentrated than the labeled molarity (0.0500 M).
B. The Ca[OH]2 solution may have been supersaturated.
C. The HCl was less concentrated than the labeled molarity (0.0500 M).
D. The Ca[OH]2 solution may have been unsaturated.
E. The titration flask may have not been clean and had a residue of a basic solution.
F. The titration flask may have not been clean and had a residue of an acidic solution.
Answer:
D. The Ca[OH]2 solution may have been unsaturated
Explanation:
The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.
If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.
A 25.00 mL sample of unknown concentration of HNO3 solution requires 22.62 mL of 0.02000 M NaOH to reach the equivalence point. What is the concentration of the unknown HNO3 solution
Answer:The concentration of the unknown HNO3 solution = 0.01809 M
Explanation:
For the acid-base reaction, HNO3 + NaOH-----> NaN03 + H20
we have that
C1 V1 = C2 V2
Where ,
C1 = concentration of HNO3=?
V1 = volume of HNO3 = 25.00 mL,
V2 = volume of NaOH = 22.62 mL,
C2 = concentration of NaOH = 0.02000 M
Therefore ,
25.00 mL x C1 = 22.62 mL x 0.02000 M
= (22.62 mL / 25.00 mL) x 0.02000 M = 0.01809 M
The concentration of the unknown HNO3 solution = 0.01809 M
Human blood typically contains 1.04 kg/L of platelets. A 1.89 pints of blood would contain what mass (in grams) of platelets
A 1.89 pints of blood would contain 873 grams of platelets.
To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:
1 pint = 473.2 mL[tex]1.89 \times 473.2 = 894.3 mL[/tex]
1000 L = 1mL
[tex]\frac{894.3}{1000}= 0.84L[/tex]
Now, just calculate the amount of platelets present in 0.84L:
[tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]
x = 873 grams
So, a 1.89 pints of blood would contain 873 grams of platelets.
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Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na 2CrO 4 are added to 250 mL of 0.12 M AgNO 3? [K sp(Ag 2CrO 4) = 1.1 × 10 –12] What is the concentration of the silver ion remaining in solution?
Answer:
A precipitate will form.
[Ag⁺] = 2.8x10⁻⁵M
Explanation:
When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:
Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)
Ksp is defined as:
Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]
Where the concentrations [] are in equilibrium
Reaction quotient, Q, is defined as:
Q = [Ag⁺]² [CrO₄²⁻]
Where the concentrations [] are the actual concentrations
If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,
The actual concentrations are -Where 500mL is the total volume of the solution-:
[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M
[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M
And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴
As Q > Ksp; a precipitate will form
In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:
[Ag⁺] = 0.06M - 2X
[CrO₄²⁻] = 0.165M - X
Where X is defined as the reaction coordinate
Replacing in Ksp expression:
1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]
Solving for X:
X = 0.165M → False solution. Produce negative concentrations.
X = 0.0299986M
Replacing, equilibrium concentrations are:
[Ag⁺] = 0.06M - 2(0.0299986M)
[CrO₄²⁻] = 0.165M - 0.0299986M
[Ag⁺] = 2.8x10⁻⁵M[CrO₄²⁻] = 0.135M
A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (see sketch at right).
Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits.
You'll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999/gcm3. You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.
Answer:
Amount of salt in 1 L seawater = 34 g
Explanation:
According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater
mass of freshwater = density * volume
1 cm³ = 1 mL
mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g
mass of freshwater + cup = 734.265 + 25 = 759.265 g
Therefore, mass of equal volume of seawater = 759.265 g
Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)
1 liter = 1000 cm³ = 1000 mL;
Density of seawater = mass / volume
Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L
Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L
mass of 1 Liter seawater = 1033.01 g
mass of 1 Liter freshwater = 999 g
mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g
Therefore, amount of salt in 1 L seawater = 34 g
If 2.9g of water is heated from 23.9C to 98.9C, how much heat (in calories) was added to the water?
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
By heating a 93% pure kclo3 sample, what percentage of its mass is reduced?
2KCLO3---->2KCL+3O2
Explanation:
free your mind drink water and go outside take fresh air you will get answers
A sample is found to contain 1.29×10-11 g of salt. Express this quantity in picograms
Answer:12.9e-12g or in short 12.9pg
Explanation:as p=1e-12
A balloon has an initial volume of 2.954 L containing 5.50 moles of helium. More helium is added so that the balloon expands to 4.325 L. How much helium (moles) has been added if the temperature and pressure stay constant during this process.
Answer:
8.05 moles
Explanation:
5.50 / 2.954 = x / 4.325
x = 8.05
According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added so that the balloon expands to 4.325 L.
What is ideal gas equation?The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.
In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314 so n=V/R=4.325/8.314=0.520 moles.
Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.
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An aqueous solution of potassium bromide, KBr, contains 4.34 grams of potassium bromide and 17.4 grams of water. The percentage by mass of potassium bromide in the solution is 20 %.
Answer:
True
Explanation:
The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:
Percentage by mass = mass of substance in solution/mass of solution x 100
In this case;
mass of KBr = 4.34 grams
mass of water = 17.4 grams
mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74
Percentage by mass of KBr = 4.34/21.74 x 100
= 19.96 %
19.96 is approximately 20%.
Hence, the statement is true.
Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.
Answer:
There are no forces holding the gas particles together.
Explanation:
Compound A is an alkene that was treated with ozone to yield only (CH3CH2CH2)2C=O. Draw the major product that is expected when compound A is treated with a peroxy acid (RCO3H) followed by aqueous acid (H3O+).
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
What do we know about the unknown alkene?
We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
What is the product with the peroxyacid?
This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)
What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the ions in solution were [Cd2+] = 0.5 M and [Zr4+] = 0.5 M at 298K?
Answer:
1.05 V
Explanation:
Since;
E°cell= E°cathode- E°anode
E°cathode= -0.40 V
E°anode= -1.45 V
E°cell= -0.40-(-1.45) = 1.05 V
Equation of the process;
2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)
n= 8 electrons transferred
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]
Since log 1=0
Ecell= E°cell= 1.05 V
When equation for neutralization of HBr by Ca(OH)2 is correctly balanced, how many molecules of water will be formed
Answer:
When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed
Explanation:
A neutralization reaction is one in which an acid (or acidic oxide) reacts with a base (or basic oxide). In the reaction a salt is formed and in most cases water is formed. A Salt is an ionic compound formed by the union of ions and cations through ionic bonds.
In the reactions of a strong acid (those substances that completely dissociate) with a strong base (they dissociate completely, giving up all their OH-), the complete neutralization of the species is carried out:
2 HBr (aq) + Ca(OH)₂ (s) → CaBr₂ (aq) + 2 H₂O (l)
The reaction is already balanced, complying with the law of conservation of matter. This law states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reactants must be equal to the number of atoms present in the products.
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), 2 moles of water H₂O are formed.
On the other hand, Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.
Then you can apply the following rule of three: if 1 mole of H₂O contains 6.023*10²³ molecules, 2 moles of H₂O, how many molecules does it contain?
[tex]amount of molecules=\frac{2moles*6.023*10^{23}molecules }{1 mole}[/tex]
amount of molecules= 1.2046*10²⁴ molecules
When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed
A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)+H2O(l)↽−−⇀H3O+(aq)+A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.260 M , [H3O+]=4.00×10−4 M , and [A−]=4.00×10−4 M . Calculate the Ka value for the acid HA.
Answer:
Ka = 6.15x10⁻⁷
Explanation:
Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:
HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)
And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:
Ka = [H₃O⁺] [A⁻] / [HA]
You don't take water in the equilibrium beacuse is a pure liquid
Replacing with the concentrations of the problem:
Ka = [H₃O⁺] [A⁻] / [HA]
Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]
Ka = 6.15x10⁻⁷
Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite under these conditions is ·0.710J·g−1K−1 . Be sure your answer has the correct number of significant digits.
Answer:
7.23 J
Explanation:
Step 1: Given data
Mass of graphite (m): 566.0 mgInitial temperature: 5.2 °CFinal temperature: 23.2 °CSpecific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)
Q = 7.23 J
Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value?
Answer:
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
Rate constant:
t(1/2) = ln 2 / K
As half-life of Cesium-137 is 30.2 years:
30.2 years = ln 2 / K
K = 0.02295 years⁻¹
Replacing this result and with the given data of the problem:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.02295 years⁻¹* t + ln[A]₀
Ln ([A] / [A₀]) = -0.02295 years⁻¹* t
As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:
Ln (0.2) = -0.02295 years⁻¹* t
70.1 years = t
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value1. Explain what the police siren sounds like to Jane:
2. Explain what the police siren sounds like to John:
3. Explain why the police siren sounds different between Jane and John:
Answer:
1. the siren has a lower pitch to Jane
2. the siren has a higher pitch to John
3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter
Explanation:
The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.
This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.
What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 4.0 g of urea in 32.0 g of methanol, CH3OH
Answer:
0.0630
Explanation:
The molar mass of urea = 60 g/mol
we all know that:
[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
Then; the number of moles of urea
= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]
= 0.0667 mol
Similarly; the number of moles of methanol
= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]
= 0.9988 mol
The total number of moles = (0.0667 + 0.9988) mol
= 1.0655 mol
Finally,the mole fraction of urea [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]
[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]
= 0.0630
How much work (in Joules) is required to expand the volume of a pump from 0.00 L to 2.50 L against an external pressure of 1.10 atm
Answer:
W = 278.64375 Joules
Explanation:
The information given in this problem are;
Initial volume = 0L
Final volume = 2.50L
ΔV = 2.50 - 0 = 2.50 L
External pressure, P = 1.10 atm
Work = ?
These parameters are related by the equation;
w = - P ΔV
W = - (1.10 )(2.50)
W = 2.75 L atm
Upon conversion to joules;
1 liter atmosphere is equal to 101.325 joule
W = 278.64375 Joules
Which of the following chemical equations corresponds to the standard molar enthalpy of formation of Na_2CO_3(s)?
a. 2 NA(s) + C(s) + 3 O(g) ------------> Na_2CO_3(s)
b. Na_2O(s) + CO_2(g) --------------->Na_2CO_3 (s)
c. Na_2(s) + C(s) + 3 O(g) -------------> Na_2CO_3 (s)
d. Na_2O(s) + CO(g) ---------------> Na_2CO_3(s)
e. 2 Na(s)+ C(s) + 3/2 O_2(g) ------------> Na_2CO_3(s)
Answer:
2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)
Explanation:
The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)
For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:
2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)Write the equation for the reaction described: A solid metal oxide, , and hydrogen are the products of the reaction between metal and steam. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
Pb + 2H2O --> PbO2 + 2H2
Explanation:
Products:
Solid metal; PbO2
Hydrogen; H
Reactants:
Metal; Pb
Steam; H2O
Reactants --> Products
Pb + H2O --> PbO2 + H2
Upon balancing we have;
Pb + 2H2O --> PbO2 + 2H2
3. What is the mass of an object with a volume of 4 L and a density of 1.25 g/mL?
Answer:
5000g
Explanation:
mass= density × volume
Since the unit of density here is g/mL, we need to convert the volume to mL.
1L= 1000mL
4L= 4 ×1000 = 4000 mL
Mass of object
= 1.25 ×4000
= 5000g
Answer:
5,000 grams
Explanation:
The mass of an object can be found by multiplying the volume by the density.
mass= volume * density
The density is 1.25 g/mL and the volume is 4 L.
First, we must convert the volume to mL. The density is given in grams per milliliter, but the volume is given in liters.
There are 1,000 mL per L. The volume is 4 L. Therefore, we can multiply 4 and 1,000.
4 * 1,000 = 4,000
The volume is 4,000 mL.
Now, find the mass of the object.
mass= volume * density
volume = 4,000
density= 1.25
mass= 4,000 * 1.25 = 5,000
Add the appropriate units for mass, in this case, grams, or g.
mass= 5,000 g
The mass of the object is 5,000 grams.
