Answer:
0.00642mole
Explanation:
Molar Mass of Cu = 63.5g/mol
Mass of Cu from the question = 0.4076g
Number of mole =?
Number of mole = Mass /Molar Mass
Number of mole of Cu = 0.4076/63.5 = 0.00642mole
The following Lewis diagram represents the valence electron configuration of a main-group element. This element is in group 2A According to the octet rule, this element would be expected to form a(n) with a charge of cation anion If X is in period 4, the ion formed has the same electron configuration as the noble gas The symbol for the ion is________
Answer: The symbol of the ion formed is [tex]Ca^{2+}[/tex]
Explanation:
An ion is formed when a neutral atom looses or gains electrons.
When an atom looses electrons, it results in the formation of positive ion known as cation.When an atom gains electrons, it results in the formation of negative ion known as anion.Electronic configuration is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom is determined by the atomic number of that atom.
The element present in Group 2-A and in period 4 is Calcium (Ca)
Electronic configuration of Ca atom: [tex]1s^22s^22p^63s^23p^64s^2[/tex]
This atom will loose 2 electrons to attain stable electronic configuration similar to Argon element (noble gas)
The electronic configration of [tex]Ca^{2+}\text{ ion}=1s^22s^22p^63s^23p^6[/tex]
Hence, the symbol of the ion formed is [tex]Ca^{2+}[/tex]
Answer: The octet rule says that in forming ions, main-group atoms gain or lose electrons in order to attain a noble gas electron configuration. Except for the He configuration, this means that atoms gain or lose electrons to have an octet of electrons in the outermost shell.
Explanation: If x is in period 4, the ion formed has the same electron configuration as the noble ga
Molecular orbitals formation involved in the combination of same type atomic orbitals which also have same symmetry for diatomic molecules.So the correct combinations to the formation of molecular orbitals are,
1s + 1s -----> Ï1s + Ï*1s
2s + 2s -----> Ï2s + Ï*2s
2pz + 2pz -----> Ï2pz + Ï*2pz
2py + 2py -----> Ï2py + Ï*2py
2px + 2px -----> Ï2px + Ï*2px
Answer:2py + 2py -----> Ï2py + Ï*2py
2px + 2px -----> Ï2px + Ï*2px
Explanation:
Molecular orbitals are constructed from atomic orbitals by linear combination of atomic orbitals. The 1s, 2s and 2pz orbitals overlap in an end to end manner hence they only form sigma bonding and anti bonding orbitals. The 2px and 2py orbitals overlap side by side and form pi bonding and anti bonding orbitals. Hence the answer.
Fission reactions are induced in nuclear power plants because they produce great amounts of energy. Often, a nucleus must be forced to undergo fission by being hit by another particle. A U-235 nucleus being hit with some other particle causes it to undergo fission and emit more particles. Those particles eventually hit other U-235 nuclei, which propagates the reaction. What is this type of reaction called
Answer:
Chain reaction
Explanation:
A chain reaction is a reaction that sustains itself. It has the ability to continue for a very long time without adding any more materials to the reaction system. It may be succinctly described as a self propagating reaction.
In a nuclear fission, uranium-235 is bombarded with neutrons to produce unstable uranium-236 which disintegrates to form daughter nuclei and produce more neutrons that bombard more uranium-235 and the reaction continues indefinitely.
NaCl is purified by adding HCl to a saturated solution of NaCl (317 g/L). Will pure NaCl precipitate when 30 mL of 4.5 M HCl is added to 0.15 L of saturated solution? Group of answer choices NaCl will precipitate from solution No NaCl will precipitate
The thing which will happen when NaCl is purified by adding HCl to a saturated solution of NaCl (317 g/L) is
NaCl will not precipitate.What is a Saturated Solution?This refers to the type of solution which has a lot of solute that enables the solution to dissolve.
Hence, if NaCl is purified by the addition of HCl to a saturated solution of NaCl (317 g/L) and 30 mL of 4.5 M HCl is added to 0.15 L of saturated solution, then the NaCl will not precipitate because it cannot be transformed the dissolved substance to an insoluble solid as a result of the extra HCL added.
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A sample of 0.281 gg of an unknown monoprotic acid was dissolved in 25.0 mLmL of water and titrated with 0.0950 M NaOH NaOH. The acid required 30.0 mLmL of base to reach the equivalence point.What is the molar mass of the acid?
Answer:
98.6 g/mol.
Explanation:
Equation of the reaction
HX + NaOH--> NaX + H2O
Number of moles = molar concentration × volume
= 0.095 × 0.03
= 0.00285 moles
By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.
Molar mass = mass ÷ number of moles
= 0.281 ÷ 0.00285
= 98.6 g/mol.
Which one of the following choices describes most accurately the actual, internal reaction temperature (in other words, the temperature of the reaction mixture inside the reaction vial) for the Fischer esterification experiment of 1-butanol with acetic acid to form n-butyl acetate? Select one, and explain your answer.
a) Sand bath temperature (160-180 °C)
b) Boiling point of 1-butanol (116-118 °C)
c) Boiling point of the reaction mixture (reflux temperature)
d) Boiling point of acetic acid (117 °C)
e) Boiling point of n-butyl acetate (124-126 °C)
Answer:
c) Boiling point of the reaction mixture (reflux temperature)
Explanation:
Hello,
At first, it is important to remember that esterification is an organic chemical reaction related with the neutralization of organic acids and alcohols to form esters, in this case from 1-butanol and acetic acid to n-butyl acetate as shown below:
[tex]CH_3COOH+CH3(CH_2)2CH_2OH \rightleftharpoons CH_3COOCH_2(CH_2)2CH3+H_2O[/tex]
It is shown that is a reaction which equilibrium condition is present since the n-butyl acetate is likely to come back to the acetic acid and the 1-butanol. Moreover, it is necessary to catalyze esterification with sulfuric acid and including constant heating and stirring, nonetheless, such heating induces boiling of the reacting mixture containing the acetic acid and the 1-butanol which are likely to boil. Therefore, reflux must be implemented as shown on the attached picture to prevent reactant lost which shift the reaction leftwards, diminishing n-butyl acetate yield, thus, the most accurately way to describe the actual temperature is c) boiling point of the reaction mixture (reflux temperature) since acetic acid and 1-butanol have a composition which modifies their boiling point into an only one that is the mixture's boiling point which is also related with the temperature at which the reflux is performed.
Best regards.
The option that best describes most accurately the actual, internal reaction temperature is Boiling point of the reaction mixture (reflux temperature).
What is Reflux Reactions about?It is said to be a very hard task when one is trying to monitor and control the temperature of a reaction chamber when a person do not have expensive and well equipped laboratory tools.
People often uses phase when the above is not in place as it is Phase said to be a form of melting or boiling occur at particular temperatures, and all heat exchanged are said to be done in a phase transition goes into the phase transition.
The main reason of refluxing a solution is done so as to heat a solution in a manner where one can control the outcome at a constant temperature and this is the option that is best for the Fischer esterification experiment of 1-butanol with acetic acid to create n-butyl acetate.
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A generic metal thiocyanate, M(SCN)2, has a Ksp value of 2.00×10−5. Calculate the molar solubility of the metal thiocyanate in 0.421 M KSCN. Express your answer numerically in units of mM to 4 decimal places.
Answer:
The molar solubility of the metal thiocyanate is [tex]1.127\times 10^{-4} M[/tex].
Explanation:
Concentration of potassium thiocyanate = 0.421 M
Concentration of thiocyanate ion =[tex][SCN^-]= 0.421 M[/tex]
Concentration of metal ion =[tex][M^{2+}]= ?[/tex]
The solubility product of metal thiocyanate = [tex]K_{sp}=2.00\times 10^{-5}[/tex]
[tex]M(SCN)_2\rightleftharpoons M^{2+}+2SCN^-[/tex]
S 2S
At equilbrium
S (2S+0.421)
The expression of solubility product is given by :
[tex]K_{sp}=[M^{2+}]\times [SCN^-]^2[/tex]
[tex]2.00\times 10^{-5}=S\times (2S+0.421)^2[/tex]
Solving for S:
[tex]S = 1.128\times 10^{-4} M[/tex]
[tex][M^{2+}]=\frac{2.00\times 10^{-5}}{(0.421 M)^2}=1.127\times 10^{-4} M[/tex]
The molar solubility of the metal thiocyanate is [tex]1.127\times 10^{-4} M[/tex].
A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250 M KSCNC.3.00 mL of 0.050 M HNO3Calculate the initial concentrations of all ions, after mixing, prior to the reaction occurring. The equilibrium concentration of Fe(NCS)2+ was determined using a spectrophotometer to be 3.6 x 10-5 M. Calculate the concentrations of all ions at equilibrium. Calculate the value for the equilibrium constant, K.
Explanation:
[tex]Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}[/tex]
A. 2.00 mL of 0.00250 M [tex]Fe(NO_3)_3[/tex]
Moles of ferric nitrate = n
Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)
Molarity of ferric nitrate = 0.00250 M
[tex]n=0.00250 M\times 0.002 L=0.000005 mol[/tex]
B. 5.00 mL of 0.00250 M [tex]KSCN[/tex]
Moles of KSCN = n'
Volume of KSCN = 5.00 ml = 0.005 L ( 1 mL=0.001 L)
Molarity of KSCN = 0.00250 M
[tex]n'=0.00250 M\times 0.005 L=0.0000125 mol[/tex]
C. 3.00 mL of 0.050 M [tex]HNO_3[/tex]
Moles of nitric acid = n''
Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)
Molarity of nitric acid = 0.050 M
[tex]n=0.050 M\times 0.003 L=0.00015 mol[/tex]
After mixing A, B and C together and their respective initial concentration before reaction.
After mixing A, B and C together the volume of the solution becomes = V
V = 0.002 L=0.005 L+0.003 L= 0.010 L
Concentration of ferric nitrate :
[tex][Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M[/tex]
Concentration of ferric ions :
[tex][Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M[/tex]
Concentration of nitrate ions from ferric nitrate:
[tex][NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M[/tex]
Concentration of KSCN :
[tex][KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M[/tex]
Concentration of [tex]SCN^-[/tex] ions:
[tex][SCN^-]=1\times [KSCN]=0.00125 M[/tex]
Concentration of potassium ions:
[tex][K^+]=1\times [KSCN]=0.00125 M[/tex]
Concentration of nitric acid :
[tex][HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M[/tex]
Concentration of hydrogen ion :
[tex][H^+]=1\times [HNO_3]=0.015 M[/tex]
Concentration of nitrate ions from nitric acid :
[tex][NO_3^{-}]=1\times [HNO_3]=0.0015 M[/tex]
Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M
[tex]Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}[/tex]
given concentration of [tex] Fe(NCS)^{2+}[/tex] at equilbrium = [tex]3.6\times 10^{-5} M = 0.000036 M[/tex]
initially :
0.0005 M 0.00125 M 0
At equilibrium
(0.0005-0.000036) M (0.00125-0.000036) M 0.000036 M
0.000464 M 0.001214 M 0.000036 M
The expression of an equilibrium constant will be given as;
[tex]K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}[/tex]
[tex]=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91 [/tex]
The value for the equilibrium constant is 63.91.
Gold has always been a highly prized metal, and it has been widely used from the beginning of history as a store of value. It does not rust like iron and does not become tarnished like silver. It is so chemically inert that it will not react with even the strongest concentrated acids. But it can be dissolved in aqua regia – a fresh-prepared mixture of concentrated HNO3 and HCl (1:3).
When Germany invaded Denmark in World War II, the Hungarian chemist George de Hevesy dissolved the gold Nobel Prizes of Max von Laue and James Franck in aqua regia to prevent the Nazis from stealing them. He placed the jar with the solution on a shelf in his laboratory, and after the war, precipitated the gold out of the acid and returned it to the Royal Swedish Academy of Sciences and the Nobel Foundation who recast the medals and again presented them to Laue and Franck.
The unbalanced equation for the reaction of gold with aqua regia is given below.
Add the stoichiometric coefficients to the equation to balance it.
Au(s) + HNO3(aq) + HCl(aq) → HAUCl4(aq) + NO2(g) + H20(l)
What's the function of HCL?
Answer: The balanced equation is
Au(s) + 3HNO3(aq) + 4HCl(aq) ---> HAuCl4(aq) + 3NO2(g) +3H2O(l)
The function of HCl in a solution of Aqua regia, that is used to dissolve gold is to dissolve other metals like quartz or iron stone that surround the gold.
Explanation: To balance the equation, we check the ratio of each element in the reacting side and the product side ( left and right hand side). Let their ratio be equal be adding moles to the compound of the element or the element it's self in either side of the equation.
HCl which is called hydrochloric acid, is an acid that does not react with gold, but it react with every other substance, like your skin, metals etc. it is used to clean a gold, by dipping the gold inside it, all the metals on the surface of the gold will dissolve.
When dissolving a gold in aqua regia solution, HCL is added to prepare this solution because it will help to dissolve all other substance on the surface of the gold.
How would the calculated ratio of water molecules present in Epsom salt be affected if the procedure was changed so the analysis was preformed using 2.500g of hydrated salt instead of 1.500g? Assume you still heated the sample to a constant mass. (Too High, Too Low, Unchanged) Explain
Answer:
Unchanged
Explanation:
Epsom salt is also referred to as magnesium sulphate heptahydrate with the chemical formula [tex]MgSO_4.7H_2O.[/tex]
There are 7 moles of water and 1 mole of magnesium sulphate in every 1 mole of epsom salt. The water ratio remains the same irrespective of the amount or weight of epsom salt taken.
Hence, if 2.500 g of epsom salt is taken and heated to a constant mass instead of 1.500 g, the ratio of water molecules present will be the same.
When aqueous solutions of magnesium iodide and sodium carbonate are combined, solid magnesium carbonate and a solution of sodium iodide are formed. The net ionic equation for this reaction is:______
Answer:
Mg+2 (aq) + CO3-2 (aq) ---> MgCO3 (s)
Explanation:
The net ionic equation for the reaction of magnesium iodide and sodium carbonate is Mg⁺² (aq) + CO₃⁻²(aq) ---> MgCO₃ (s)
What is an ionic equation?The ionic equations are those in which the electrolytes are dissociated into the ion of a solution.
The equations in which the written in dissociated ions of an electrolyte solution.
Thus, the ionic equation is Mg⁺² (aq) + CO₃⁻²(aq) ---> MgCO₃ (s).
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Initial alcohol is converted to compound A. Compound A undergoes three different transformations. Name compound A and each the final product in three transformations.
Explanation:
Initial alcohol that is formed from methane is methanol that can be converted to acetic acid.Methanol in presence of [tex]CO_2[/tex] and hydrogen gas gets oxidized to acetic acid with the release of water.Hence, compound A will be Acetic acid.
[tex]CH_3OH + CO_2 +H_2[/tex] → [tex]CH_3COOH + H_2O[/tex]
The acetic acid formed can be transformed into -[tex]CH_3COOH[/tex] → [tex]CH_4 + CO[/tex]The product formed is methane and carbon monoxide.
2. [tex]CH_3COOH[/tex] → [tex]CH_2CHO + H_2O[/tex]
The product formed is formaldehyde and water.
3. [tex]CH_3COOH + NaHCO_3[/tex] → [tex]CH_3COONa +CO_2 +H_2O[/tex]
The product formed is sodium acetate, carbon dioxide, and water.
Oxalic acid is a diprotic acid. If a solid material contains 53.66 percent of oxalic acid (H 2C 2O 4), by mass, then a 0.6543-g sample of that solid will require ________ mL of 0.3483 M NaOH for neutralization. 11.19 97.78 28.59 1.119 22.39
Answer: The volume of NaOH required is 22.39 mL
Explanation:
We are given:
Mass of sample = 0.6543 g
Mass percent of oxalic acid = 53.66 %
This means that 53.66 grams of oxalic acid is present in 100 grams of sample
Mass of oxalic acid in the given amount of sample = [tex]\frac{53.66}{100}\times 0.6543=0.351g[/tex]
To calculate the number of moles, we use the equation:[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of oxalic acid = 0.351 g
Molar mass of oxalic acid = 90 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of oxalic acid}=\frac{0.351g}{90g/mol}=0.0039mol[/tex]
The chemical equation for the reaction of oxalic acid and NaOH follows:
[tex]C_2H_2O_4+2NaOH\rightarrow Na_2C_2O_4+2H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of oxalic acid reacts with 2 moles of NaOH
So, 0.0039 moles of oxalic acid will react with = [tex]\frac{2}{1}\times 0.0039=0.0078mol[/tex] of NaOH
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Moles of NaOH = 0.0078 moles
Molarity of solution = 0.3483 M
Putting values in above equation, we get:
[tex]0.3483M=\frac{0.0078\times 1000}{V}\\\\V=\frac{0.0078\times 1000}{0.3483}=22.39mL[/tex]
Hence, the volume of NaOH required is 22.39 mL
To determine the volume of NaOH needed to neutralize a 0.6543-g sample of a solid containing 53.66% oxalic acid, we calculate the moles of oxalic acid in the sample and use stoichiometry to determine the moles of NaOH required for neutralization. Finally, we use the formula for molarity to find the volume of NaOH needed.
Explanation:Oxalic acid, H2C2O4, is a diprotic acid, meaning it can donate two protons (H+) in an acid-base reaction. To determine the volume of 0.3483 M NaOH needed for neutralization of a 0.6543-g sample of the solid containing 53.66% oxalic acid by mass, we need to first calculate the moles of oxalic acid in the sample.
First, we calculate the moles of H2C2O4 in the sample:
Moles of H2C2O4 = mass of H2C2O4 / molar mass of H2C2O4
Next, we use stoichiometry to calculate the moles of NaOH required for neutralization:
Moles of NaOH = 2 * moles of H2C2O4
Finally, we use the formula for molarity to determine the volume of NaOH needed:
The volume of NaOH (L) = moles of NaOH / molarity of NaOH
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Thermal energy is Question 2 options: A) solar energy, i.e. energy that comes from the sun. B) the energy stored within the structural units of chemical substances. C) energy available by virtue of an object's position. D) the energy associated with the random motion of atoms and molecules.
Answer:
D) the energy associated with the random motion of atoms and molecules.
Explanation:
Thermal energy is the manifestation of energy in the form of heat. In all materials, the atoms that make up their molecules are in continuous movement, either moving or vibrating, which implies that the atoms have a certain kinetic energy that we call heat or thermal energy. In a way, thermal energy is the internal energy of a body.
Final answer:
Thermal energy is the energy associated with the random motion of atoms and molecules, a type of kinetic energy that increases with object's temperature, making Option D the correct choice.
Explanation:
The question poses multiple options for defining thermal energy. Option A suggests solar energy, which isn't correct as solar energy is a form of radiant energy. Option B refers to the energy within chemical substances, known as chemical energy. Option C discusses energy by virtue of an object's position, commonly known as potential energy. The correct answer is Option D, which states that thermal energy is the energy associated with the random motion of atoms and molecules.
More specifically, thermal energy is related to the kinetic energy resulting from the random translational motions of particles and also includes vibrational and rotational energies within molecules. An increase in this motion, as measured by temperature, corresponds to an increase in thermal energy. Hence, thermal energy is a form of internal energy of an object, manifesting as the microscopic motion of its constituent particles.
A solution is prepared by dissolving 27.0 g of urea [(NH2)2CO], in 150.0 g of water. Calculate the boiling point of the solution. Urea is a nonelectrolyte.
Answer: The boiling point of solution is 101.56°C
Explanation:
Elevation in boiling point is defined as the difference in the boiling point of solution and boiling point of pure solution.
The equation used to calculate elevation in boiling point follows:
[tex]\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}[/tex]
To calculate the elevation in boiling point, we use the equation:
[tex]\Delta T_b=iK_bm[/tex]
Or,
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
where,
Boiling point of pure water = 100°C
i = Vant hoff factor = 1 (For non-electrolytes)
[tex]K_b[/tex] = molal boiling point elevation constant = 0.52°C/m.g
[tex]m_{solute}[/tex] = Given mass of solute (urea) = 27.0 g
[tex]M_{solute}[/tex] = Molar mass of solute (urea) = 60 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = 150.0 g
Putting values in above equation, we get:
[tex]\text{Boiling point of solution}-100=1\times 0.52^oC/m\times \frac{27\times 1000}{60\times 150}\\\\\text{Boiling point of solution}=101.56^oC[/tex]
Hence, the boiling point of solution is 101.56°C
A solution is prepared by dissolving 27.0 g of urea in 150.0 g of water has a boiling point of 101.54 °C.
The normal boiling point of water is 100 °C. However, when 27.0 g of urea is dissolved in 150.0 g of water, we expect the boiling point of the solution to be higher.
What is the boiling point elevation?Boiling point elevation is the phenomenon that occurs when the boiling point of a liquid is increased when another compound is added. It is a colligative property. We can calculate the increase in the boiling point using the following expression.
ΔT = i × Kb × m
where,
i is the Van't Hoff factor (i = 1 for nonelectrolytes).Kb is the molal boiling point constant (Kb = 0.513 °C/m for water).m is the molality.What is molality?Molality (m) is defined as the total moles of a solute contained in a kilogram of a solvent. We can calculate it using the following expression.
m = mass solute / molar mass solute × kg solvent
m = 27.0 g / (60.06 g/mol) × 0.1500 kg = 3.00 m
The boiling point elevation is:
ΔT = i × Kb × m = 1 × (0.513 °C/m) × 3.00 m = 1.54 °C
Then, the boiling point of the solution is:
T = 100 °C + 1.54 °C = 101.54 °C
A solution is prepared by dissolving 27.0 g of urea in 150.0 g of water has a boiling point of 101.54 °C.
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A cylindrical specimen of some metal alloy having an elastic modulus of 129 GPa and an original cross-sectional diameter of 4.4 mm will experience only elastic deformation when a tensile load of 1570 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm.
Explanation:
The cross-sectional area of the specimen is calculated as follows.
[tex]A_{o} = \frac{pi}{4} d^{2}[/tex]
= [tex]\frac{3.14}{4} \times (\frac{4.4}{1000})^{2}[/tex]
= [tex]1.5197 \times 10^{-5} m^{2}[/tex]
Equation of stress is as follows.
[tex]\sigma = \frac{F}{A_{o}}[/tex]
And, the equation of strain is as follows.
[tex]\epsilon = \frac{\Delta l}{l_{o}}[/tex]
Hence, the Hook's law is as follows.
E = [tex]\frac{\sigma}{\epsilon}[/tex]
E = [tex]\frac{\frac{F}{A_{o}}}{\frac{\Delta l}{l_{o}}}[/tex]
= [tex]\frac{F \times l_{o}}{A_{o} \times \Delta l}[/tex]
or, [tex]l_{o} = \frac{E \times \Delta l \times A_{o}}{F}[/tex]
= [tex]\frac{129 \times 10^{9} \times \frac{0.48}{1000} \times 1.662 \times 10^{-5}}{1570}[/tex]
= 0.6554 m
or, [tex]l_{o}[/tex] = 655.4 mm
Thus, we can conclude that the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm is 655.4 mm.
Some CH2Cl2 is placed in a sealed flask and heated to 517 K. When equilibrium is reached, the flask is found to contain CH2Cl2 (3.42×10-2 M), CH4 (3.69×10-2 M), and CCl4 (4.12×10-2 M). What is the value of the equilibrium constant for this reaction at 517 K?
Answer: The value of equilibrium constant for the given reaction at 517 K is 1.30
Explanation:
The chemical equation for the dissociation of [tex]CH_2Cl_2[/tex] follows:
[tex]2CH_2Cl_2(g)\rightleftharpoons CH_4(g)+CCl_4(g)[/tex]
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=\frac{[CH_4][CCl_4]}{[CH_2Cl_2]^2}[/tex]
We are given:
[tex][CH_4]_{eq}=3.69\times 10^{-2}M[/tex]
[tex][CCl_4]_{eq}=4.12\times 10^{-2}M[/tex]
[tex][CH_2Cl_2]_{eq}=3.42\times 10^{-2}M[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(3.69\times 10^{-2})\times (4.12\times 10^{-2})}{(3.42\times 10^{-2})^2}\\\\K_{eq}=1.30[/tex]
Hence, the value of equilibrium constant for the given reaction at 517 K is 1.30
Changes in pressure have a measurable effect: (select all that apply) Select all that apply:
in any system only in systems in which gases are involved
when there are equal numbers of moles of gas on the reactant and product sides of the equilibrium only
when the chemical reaction produces a change in the total number of gas molecules in the system
Answer:
1. only in systems in which gases are involved
2. only when the chemical reaction produces a change in the total number of gas molecules in the system
Explanation:
According to Le Chatelier's principle, pressure will only affect a system in equilibrium containing gaseous reactants and products. However, change in the pressure will only affect the gaseous system in which the total number of moles of the reactants are different from the total number of moles of the products.
Pressure changes affect equilibrium systems involving gases, particularly when the reaction results in different numbers of gas molecules on each side. Thus option A is correct.
Pressure changes in a system at equilibrium have a measurable effect primarily in gaseous systems. Here's how they work:
Pressure changes impact systems in which gases are involved.The effect is significant when there is a change in the total number of gas molecules in the system.If the number of gas moles is equal on both sides of the reaction, there is no effect on equilibrium.This occurs because increasing the pressure (by decreasing the volume) will shift the equilibrium toward the side with fewer moles of gas, while decreasing the pressure (by increasing the volume) will shift it toward the side with more moles of gas.
Complete question:
Changes in pressure have a measurable effect: (select all that apply) Select all that apply:
A. in any system only in systems in which gases are involved
B. when there are equal numbers of moles of gas on the reactant and product sides of the equilibrium only
C. when the chemical reaction produces a change in the total number of gas molecules in the system
D. None of these
Carbon dioxide and water react to form methanol and oxygen, like this: CO2(g) + H2O (g) ----> CH3OH (l) + O2 (g) At a certain temperature, a chemist finds that 8.6 L a reaction vessel containing a mixture of carbon dioxide, water, methanol, and oxygen at equilibrium has the following composition: compound amount CO2 2.25 gH2O 2.72 gCH3OH 3.82 gO2 1.98 gCalculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.
Final answer:
The concentration equilibrium constant (Kc) will not include CH₃OH because it is a pure liquid. Then, Kc = 68.
Explanation:
First, we will calculate the molar concentration (M) of each substance using the following expression.
M = mass of the substance / molar mass of the substance × volume of solution
CO₂
M = 2.25 g / 44.01 g/mol × 8.6 L = 0.0059 M
H₂O
M = 2.72 g / 18.02 g/mol × 8.6 L = 0.018 M
CH₃OH
M = 3.82 g / 32.04 g/mol × 8.6 L = 0.014 M
O₂
M = 1.98 g / 32.00 g/mol × 8.6 L = 0.0072 M
Let's consider the following reaction at equilibrium.
CO₂(g) + H₂O(g) ⇄ CH₃OH(l) + O₂(g)
The concentration equilibrium constant (Kc) will not include CH₃OH because it is a pure liquid. Then,
Kc = [O₂] / [CO₂] × [H₂O]
Kc = 0.0072 / 0.0059 × 0.018
Kc = 68
Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significant figures and include the appropriate units.
Answer: The equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]
Explanation:
We are given:
Molarity of oxalic acid solution = 0.20 M
Oxalic acid [tex](H_2C_2O_4)[/tex] is a weak acid and will dissociate 2 hydrogen ions.
The chemical equation for the first dissociation of oxalic acid follows:[tex]H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)[/tex]
Initial: 0.20
At eqllm: 0.20-x x x
The expression of first equilibrium constant equation follows:
[tex]Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}[/tex]
We know that:
[tex]Ka_1\text{ for }H_2C_2O_4=0.059[/tex]
Putting values in above equation, we get:
[tex]0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of hydronium ion = x = 0.083 M
The chemical equation for the second dissociation of oxalic acid:[tex]HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)[/tex]
Initial: 0.083
At eqllm: 0.083-y 0.083+y y
The expression of second equilibrium constant equation follows:
[tex]Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}[/tex]
We know that:
[tex]Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}[/tex]
Putting values in above equation, we get:
[tex]6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of hydronium ion = y = 0.0000639 M
Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M
Hence, the equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]
The equilibrium concentration of H3O+ in a solution of oxalic acid is linked to the acid dissociation constant Ka. Calculation of this requires specific values related to the nature of oxalic acid, a weak acid, and its behaviour in solution.
Explanation:In order to find the equilibrium concentration of H3O+ in a solution of oxalic acid, we must first understand that oxalic acid is a weak acid, and thus does not fully ionize in solution. In the ionization process of oxalic acid H2C2O4, it would donate a proton (H+) to water (H2O), forming a hydronium ion (H3O+) and a mono hydrogen oxalate ion. The equilibrium concentration of H3O+ (hydronium ions) thus corresponds to the acid dissociation constant Ka, with the equation Ka=[H3O+][HC2O4-]/[H2C2O4]. Considering the initial concentration of the oxalic acid, specific calculations will be required to find an accurate equilibrium concentration which is not possible without the value of Ka. So, the specific answer depends on the Ka value of oxalic acid.
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Calculate the volume in liters of a 0.0015/molL calcium sulfate solution that contains 25.0g of calcium sulfate CaSO4 . Be sure your answer has the correct number of significant digits.
Answer:
The volume of a 0.0015 [tex]\frac{moles}{liters}[/tex] calcium sulfate solution that contains 25.0 g of calcium sulfate CaSO₄ is 122.53 liters.
Explanation:
Molarity (M) is the number of moles of solute that are dissolved in a given volume. Molarity is expressed by:
[tex]Molarity=\frac{number of moles of solute}{Dissolution volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex]
In this case you have 25.0g of calcium sulfate CaSO₄. First of all you need to know the amount of moles that mass represents. For that you must first know the atomic masses of each element:
Ca: 40 g/molS: 32 g/molO: 16 g/molThen the molar mass of the compound calcium sulfate is:
molar mass= 40 g/mol + 32 g/mol + 4*16 g/mol= 136 g/mol
It is then possible to apply a rule of three as follows: if 136 g represents 1 mol of the compound calcium sulfate, 25 g how many moles are they?
[tex]moles=\frac{25 g*1 mole}{136 g}[/tex]
moles≅0.1838
Now you can apply a rule of three knowing the molarity of 0.0015 [tex]\frac{moles}{liters}[/tex]: if 0.0015 moles represents 1 liter of solution, 0.1838 moles how many liters are they?
[tex]volume=\frac{0.1838moles*1 liter}{0.0015moles}[/tex]
volume=122.53 liters
The volume of a 0.0015 [tex]\frac{moles}{liters}[/tex] calcium sulfate solution that contains 25.0 g of calcium sulfate CaSO₄ is 122.53 liters.
To find the volume of a 0.0015 mol/L calcium sulfate solution that contains 25.0g of calcium sulfate, first convert the mass to moles, then use the molarity to find the volume. The approximate volume is 122L.
Explanation:To calculate the volume of the solution, we first need to convert the mass of calcium sulfate in grams to moles. We do this by dividing by the molar mass of calcium sulfate, which is about 136.14 g/mol.
25.0g CaSO4 * (1 mol / 136.14 g) = approx 0.183 mol CaSO4
Then, we use the molarity of the solution to find the volume. Remember that the definition of molarity is moles of solute per liter of solution.
Volume = moles / molarity = 0.183 mol / 0.0015 mol/L = approx 122 L
So, the volume of the calcium sulfate solution is approximately 122 liters.
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Rank the following substances in order from most soluble in water to least soluble in water: methane, CH4; 2-pentanol, C5H11OH; copper(II) sulfate, CuSO4; and propane, C3H8.
The ranking from most soluble to least soluble in water is:
1. Copper(II) sulfate, CuSO4
2. 2-Pentanol, C5H11OH
3. Methane, CH4
4. Propane, C3H8
To rank the substances in order of solubility in water, we can consider the types of intermolecular forces involved. Water is a polar molecule, and substances with polar or ionic characteristics tend to be more soluble in water.
1. **Copper(II) sulfate, CuSO4:**
- This substance is an ionic compound containing copper ions (Cu^2+) and sulfate ions (SO4^2-). Ionic compounds often dissolve well in water through ion-dipole interactions. Thus, copper(II) sulfate is likely to be the most soluble.
2. **2-Pentanol, C5H11OH:**
- 2-Pentanol is a polar molecule due to the presence of the hydroxyl (OH) functional group. It can form hydrogen bonds with water molecules, making it moderately soluble in water.
3. **Methane, CH4:**
- Methane is a nonpolar molecule. It lacks a permanent dipole moment and cannot form strong interactions with water molecules. Thus, it is expected to be less soluble in water than polar substances.
4. **Propane, C3H8:**
- Similar to methane, propane is a nonpolar molecule. It also lacks a permanent dipole moment and is expected to be the least soluble in water among the given substances.
So, the ranking from most soluble to least soluble in water is:
1. Copper(II) sulfate, CuSO4
2. 2-Pentanol, C5H11OH
3. Methane, CH4
4. Propane, C3H8
A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution are monovalent (charge e) gold ions, and you've calculated that you must deposit 0.60 g of gold to reach the necessary thickness.
How much current do you need, in mA, to plate the statue in 3.5 hr?
Answer: The current needed, in mA, to plate the statue in 3.5 hr is 20 mA
Explanation:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=9.6352\times 10^4C[/tex]
[tex]Au^++e^-\rightarrow Au[/tex]
197 g of gold is deposited by = 96500 C of electricity
Thus 0.60 g of gold is deposited by =[tex]\frac{96500}{197}\times 0.60=294 C[/tex] of electricity
To calculate the current required, we use the equation:
[tex]I=\frac{q}{t}[/tex]
where,
I = current passed = ?
q = total charge = [tex]294C[/tex]
t = time required = 3.5 hrs =[tex]3.5\times 3600=12600s[/tex]
Putting values in above equation, we get:
[tex]I=\frac{294C}{12600}\\\\I=0.02A=20mA[/tex]
Hence, the current needed, in mA, to plate the statue in 3.5 hr is 20 mA
Germanium forms a substitutional solid solution with silicon. Compute the weight percent of germanium that must be added to silicon to yield an alloy that contains 3.43 × 1021 Ge atoms per cubic centimeter. The densities of pure Ge and Si are 5.32 and 2.33 g/cm3, respectively. The atomic weights for germanium and silicon are 72.64 and 28.09 g/mol, respectively.
Answer:
The answer to tnhe question is;
The weight percent of germanium to be added is 16.146 %.
Explanation:
To solve the question, we note that the formula to calculate the weight percent of an element in terms of the number of atoms per cm³ in a 2 element alloy is given by,
[tex]C_1=\frac{100}{1+\frac{N_A\rho_2}{N_1A_1} -\frac{\rho_2}{\rho1} }[/tex]
Where
N[tex]_A[/tex] = Avogadro's Number
ρ₁ = Density of alloy whose weight percent is sought
ρ₂ = density of the other alloy
N₁ = Number of atoms per cubic centimeter
A₁ = Atomic weight of the element whose weight percent is sought
Therefore
[tex]C_{Ge}=\frac{100}{1+\frac{(6.022*10^{23})*(2.33)}{(3.43*10^{21})*(72.64)} -(\frac{2.33}{5.32}) } = \frac{100}{1+5.632 -0.43797 } = 16.146 %[/tex]
[tex]C_{Ge}[/tex] = 16.146 %.
Duncan knows that it takes 36400 calcal of energy to heat a pint of water from room temperature to boiling. However, Duncan has prepared ramen noodles so many times he does not need to measure the water carefully. If he happens to heat 0.900 pintpint of room-temperature water, how many kilojoules of heat energy will have been absorbed by the water at the moment it begins to boil?
Final answer:
To calculate the amount of heat energy that will be absorbed by the water, we need to use the specific heat capacity of water and convert the volume of water to liters. By using the formula Q = mcΔT, we can calculate that 143.32 kJ of heat energy will be absorbed by the water.
Explanation:
To calculate the amount of heat energy that will be absorbed by the water, we need to use the specific heat capacity of water, which is 4184 J/kg/°C. We also need to convert the volume of water from pint to liters. There are approximately 0.473 liters in a pintpint.
First, we convert the volume of water to liters: 0.900 pint × (0.473 liters/pint) = 0.426 liters.
Next, we calculate the temperature change: boiling point (100°C) - room temperature (20°C) = 80°C.
Finally, we use the formula Q = mcΔT, where Q is the heat energy absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change: Q = (0.426 kg) × (4184 J/kg/°C) × (80°C) = 143,315.52 J. To convert this to kilojoules (kJ), divide by 1000: 143,315.52 J / 1000 = 143.32 kJ.
Identify the statement that is FALSE. Identify the statement that is FALSE. Entropy increases with dissolution. Entropy generally increases with increasing molecular complexity. For noble gasses, entropy increases with size. The entropy of a gas is greater than the entropy of a liquid. Free atoms have greater entropy than molecules.
Answer:
The statement that is FALSE.
Entropy increases with dissolution.
Entropy generally increases with increasing molecular complexity.
For noble gasses, entropy increases with size.
The entropy of a gas is greater than the entropy of a liquid.
Free atoms have greater entropy than molecules.
Explanation:
Free atoms have greater entropy than molecules.
The energy is increased in faster moving particles, the opposite happens in slower ones. The entropy increases if there is randomized distribution of the energy, the system becomes then more stable as energy is released in a cluster.
The false statement is: 'Free atoms have greater entropy than molecules'. Because complexity brings more possibilities for microstates, actual molecules usually have greater entropy than single atoms. All of the other statements provided in the original query are correct.
Explanation:In the given context, the statement that is false is: "Free atoms have greater entropy than molecules". Entropy, which is a measure of randomness or disorder in a system, is influenced by the structure of the particles (atoms or molecules) that comprise a substance.
More complex molecules, with greater number of atoms, typically have a higher entropy than solitary atoms because they have more ways to vibrate and thus more potential microstates, increasing the entropy of the system. This goes contrary to the false statement made.All the other statements are true. For instance, entropy does increase with dissolution, reflecting the increase in molecular disorder and the number of potential microstates.
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At constant temperature and volume, a sample of oxygen gas is added to a sample of nitrogen gas. The pressure of the mixture is found by adding the pressures of the two individual gases. This is an example of:
(A) Boyle's Law
(B) Charles's Law
(C) Avogadro's Law
(D) Dalton's Law
D. Dalton's Law
Explanation:
As the pressure of the gas is related to the sum of the partial pressures of the individual gases present in the mixture was explained by Dalton's law, the given system is an example of Dalton's law.
Boyle's law relates the inverse proportionality of volume and pressure of an ideal gas.
Charles's Law reveals the direct relationship of temperature and volume of an ideal gas.
Avogadro's Law states the relationship between the volume of gas and number of molecules at same pressure as well as temperature.
(D) Dalton's Law states that the pressure of the mixture is found by adding the pressures of the two individual gases.
Dalton’s Law also known as the Law of Partial Pressures states that in a mixture of non-reacting gases the total pressure exerted is equal to the sum of the partial pressures of the gases in the mixture.
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The boiling point of chloroform is 61.7 C. The enthalapy of vaporization is 31.4 kj/mol. Caculate the entropy of vaporization. Does the sign for entropy of vaporizatiou make sense
Answer:
Δ S = 93.8 J/mol-K
Explanation:
Given,
Boiling point of chloroform = 61.7 °C
= 273 + 61.7 = 334.7 K.
Enthalapy of vapourization = 31.4 kJ/mol.
Using Gibbs free energy equation
Δ G = Δ H - T (ΔS)
at equilibrium (when the liquid is boiling), Δ G = 0
so, 0 = ΔH - T (Δ S)
T (Δ S) = Δ H
and ΔS = ΔH / T
Δ S = (31400 J/mol.) / 334.7 K
Δ S = 93.8 J/mol-K
Final answer:
The molar entropy of vaporization (ΔvapS) of acetone can be calculated using the formula ΔvapS = ΔvapH / T. By converting the boiling point to Kelvin and substituting the values, we find ΔvapS to be 88.4 J/mol·K. The positive sign of the entropy change is expected as vaporization increases disorder.
Explanation:
The student's question pertains to the calculation of the molar entropy of vaporization (ΔvapS) of acetone at its normal boiling point. Given the normal boiling point of acetone is 56°C, and its molar enthalpy of vaporization (ΔvapH) is 29.1 kJ/mol, we can calculate the molar entropy of vaporization using the equation:
ΔvapS = ΔvapH / T
Where T is the absolute temperature in Kelvin (K). To obtain T, convert the boiling point from Celsius to Kelvin:
T = 56°C + 273.15 = 329.15 K
Now substituting the values into the equation:
ΔvapS = 29.1 kJ/mol / 329.15 K
ΔvapS = 0.0884 kJ/mol·K or 88.4 J/mol·K
The sign for entropy of vaporization is positive, which makes sense since the entropy of the system is expected to increase when a liquid turns into a gas due to the increase in disorder.
) Any gas exerts a pressure against its surroundings. For example, if you put more gas in a balloon, it exerts more pressure and expands the balloon. Describe what causes the pressure in a gas, in terms of the fact that molecules of gas are moving.
Explanation:
As we know that molecules of a gas move far away from each other because they are held by weak Vander waal forces. So, when a balloon is filled by a gas then molecules of a gas strike at its walls leading to more number of collisions.
This will also lead to more expansion of the balloon. As we know that pressure is the force exerted n per unit area of a substance or object.
Thus, more is the kinetic energy of the molecules of the gas more will be the force exerted by them. Hence, more will be the pressure exerted.
An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg of an ideal gas at 700 kPa and 59°C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank.
Answer:
[tex]P_2=350\ kPa[/tex]
[tex]T_2=59^{\circ}\ C[/tex]
Explanation:
Given that
mass , m = 4 kg
Initial pressure ,[tex]P_1=700\ kPa[/tex]
Initial temperature ,[tex]T_1=59^{\circ}\ C[/tex]
The volume of rigid tanks are same
[tex]V_1=V[/tex]
[tex]V_2=2 V[/tex]
Let's take final temperature[tex] =T_2[/tex]
Given that tank is insulated that is why heat transfer in the tank will be zero.
By using energy balance
[tex]E_{in}-E_{out}=\Delta U[/tex]
[tex]\Delta U[/tex]= Change in the internal energy of the gas
[tex]0 = m C_V(T_2-T_1[/tex]) ( Cv=Specific heat capacity at constant volume)
[tex]0 = T_2-T_1[/tex]
Therefore [tex]T_1=T_2[/tex]
[tex]T_2=59^{\circ}\ C[/tex]
We know that ideal gas equation for gas
P V = m R T
P=pressure ,V=Volume ,m=mass ,R= gas constant ,T=temperature
By using mass conservation
[tex]m=\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}[/tex]
Now by putting the values in the above equation
[tex]\dfrac{700\times V}{RT_1}=\dfrac{P_2\times 2V}{RT_1}[/tex]
[tex]P_2=\dfrac{700}{2}\ kPa[/tex]
[tex]P_2=350\ kPa[/tex]
Therefore the final volume will be 350 kPa and temperature will be 59°C.
The final temperature in the tank is approximately 332.15K. When the volume doubles, the pressure is halved, yielding a final pressure of approximately 350 kPa.
Explanation:In this problem, your task is to determine the final temperature and pressure in a tank after an ideal gas is allowed to expand. Given the system in question is both insulated (adiabatic) and rigid, we can infer that neither heat (Q) nor work (W) is done, as indicated by the equation AEint=Q-W = 0.
Furthermore, as the internal energy does not change, this implies that the temperature will remain constant. So, the final temperature in the tank is the same as the initial, 59°C. We must convert this into Kelvin, as the equation of state of the ideal gas requires temperature to be in Kelvin. The conversion is as follows: T(K) = T(°C) + 273.15. Therefore, the final temperature is approximately 332.15K.
On the other hand, according to the ideal gas law parameters in this problem, when the volume doubles (since the partition is removed), the pressure is halved. This is reflected by the formula P = nRT/V, where 'n' is the amount of the gas, 'R' is the ideal gas constant, 'T' is the temperature and 'V' is the volume. Therefore, the final pressure is the initial pressure divided by 2, yielding a final pressure of approximately 350 kPa.
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