Answer:
V1= 4.44ml
Explanation:
C1= 5.24M V2=?, C2= 5.29, V2= 4.4ml
Applying dilution formula
C1V1=C2V2
5.24×V1 = 5.29×4.4
V1 =4.44ml
Iron (II) is measured in abandoned mine drainage using a phenanthroline method (Standard Methods for the examination of water and wastewater, method 3500-Fe). In 9. solution and diluted to 100.0 ml. The solution turns deep red and is measured at 510 nm. A series of standards from 0.50 -2.00 ug/ml iron(II) are prepared and their absorbance measured. The data for these solutions is given below. Using this data calculate the concentration of iron(II) in the original mine drainage samples. Fe Conc (ug/ml) Absorbance 0.00 0.50 1.000.134 1.50 2.00 0.002 0.045 0.096 0.196 Mine drainage 0.115 0.120
Answer:
Mine drainage Sample 1: [tex][Fe(II)] sample 1 = \frac{121 \mu gFe(II)}{1ml} = 121 \frac{\mu g}{ml}[/tex]
Mine drainage Sample 2: [tex][Fe(II)] sample 2 = \frac{127 \mu gFe(II)}{1ml} = 127 \frac{\mu g}{ml}[/tex]
Explanation:
An alkyne with the molecular formula C6H10 was treated with ozone followed by water to produce only one type of carboxylic acid. Draw the structure of the starting alkyne and the product of ozonolysis.
The starting alkyne is 3-hexyne, and after the ozonolysis process and treatment with water, the resulted product is hexanoic acid.
Explanation:In the given question, an alkyne with the molecular formula C6H10 is treated with ozone followed by water, to produce one type of carboxylic acid via the process of ozonolysis. The starting alkyne for this reaction probably is 3-hexyne. During ozonolysis, the alkyne triple bond is cleaved, forming two equivalent aldehydes. But these aldehydes are further oxidized into carboxylic acids when water is used as second reagent. The produced product, hence, is hexanoic acid.
Remember, in the process of ozonolysis, alkynes are oxidized and cleaved into smaller molecules. These molecules usually have carbonyl groups such as aldehydes and ketones. But in this scenario, because of the hydration step, we end up with a carboxylic acid.
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Electrospray ionization and atmospheric pressure chemical ionization are two methods that are used to introduce the eluate from the liquid chromatography column into a mass spectrometer. Which ionization method typically requires analyte ions to be in solution prior to reaching the interface between the column in liquid chromatography and the mass spectrometer?
Answer:
Electrospray ionization
Explanation:
Electrospray ionization Is a soft ionization technique used in producing ions from macromolecules, it is employed especially in spectrometry where high voltage is used to create an aerosol from a liquid.It can be employed in Knowing molecular weights of molecules and biological macromolecules such as Peptides and proteins. Therefore, electrospray ionization is the method typically requires analyte ions to be in solution prior to reaching the interface between the column in liquid chromatography and the mass spectrometer
1. What values are needed to determine the energy of an electron in a many‑electron atom?
O n
O ????
O m????
O ms
2. What information is most important in determining the size of an orbital?
O n
O ????
O m????
O ms
3. What information is needed to determine the orientation of an orbital?
O n
O ????
O m????
O ms
4. What information is needed to determine the general shape of an orbital?
O n
O ????
O m????
O ms
Answer: 1. n
2. n
3. [tex]m_s[/tex]
4. l
Explanation:
Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1).For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as [tex]m_s[/tex] . The value of this quantum number ranges from -l to +l.
Spin Quantum number : It describes the direction of electron spin. This is represented as s.
To determine the energy of an electron in a many-electron atom, the principal and angular momentum quantum numbers are needed (n and l). The principal quantum number (n) primarily determines the size of an orbital, while the magnetic quantum number (ml) is needed for its orientation. The angular momentum quantum number (l) is responsible for the general shape of an orbital.
To determine the energy of an electron in a many-electron atom, the following quantum numbers are needed:
n (principal quantum number) determines the general range for the value of energy and the probable distances that the electron can be from the nucleus.I (angular momentum quantum number) helps to determine subshell energy levels.The most important information in determining the size of an orbital is:
n (principal quantum number), which largely determines the energy and size of the orbital.To determine the orientation of an orbital, the quantum number needed is:
ml (magnetic quantum number), which describes the orientation of the orbital in space.The quantum number needed to determine the general shape of an orbital is: I (angular momentum quantum number), which describes the shape or type of the orbital.
Each electron also has a spin quantum number, ms, which determines the spin of an electron and can have a value of either +1/2 or -1/2.
Be sure to answer all parts.
(a) How many atoms are directly bonded to the central atom in a trigonal planar molecule?
i. two
ii. three
iii. six
iv. eight
(b) How many atoms are directly bonded to the central atom in a trigonal bipyramidal molecule?
i. three
ii. four
iii. five
iv. six
(c) How many atoms are directly bonded to the central atom in an octahedral molecule?
i. three
ii. four
iii. six
iv. eight
Answer:
a) ii
b)iii
c)iii
Explanation:
three atoms directly bonded then only it is possible to achieve trigonal planar
trigonal bipyramidal means five atoms should attach to central atom
for octahedral six atoms must directly connected to central atom
t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be
Complete Question:
t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be
I + S ⇄ I*S
W + S ⇄ W*S
W*S + I*S ⇄ TBA * S * S
TBA * S ⇄ TBA + S
Derive a rate law assuming:
(a) The surface reaction is rate limiting
(b) The adsorption of isobutene is limiting
(c) The reaction follows Eley-rideal Kinetics
I*S+W ⇄ TBA * S
and surface reaction is limiting
(d) Isobutene (I) and water (W) are adsorbed on different sites
I + S₁ ⇄ I*S₁
W + S₂ ⇄ W*S₂
TBA is not on the surface, and the surface reaction is rate-limiting
[tex][Ans: r'_{TBA}=-r'_1=\frac{k[C_1C_w-C_{TBA/K_C}]}{(1+K_WC_W)(1+K_1C_1)} ][/tex]
(e) What generalizations can you make by comparing rate laws derived from part (a) through (d)?
Answer and explanation:
The mechanism for the production of t-butyl alcohol is as follows:
the reaction and rate law for the adsorption of isobutene over the amberlyst-15 is as follows:
I + S ⇄ I * S [tex]-r_{ADI} = k_I(C_1C_v-\frac{C_{I.S}}{K_I} )[/tex]
where [tex]C_V[/tex] is the concentration of vacant site
[tex]K_I[/tex] is the equilibrium constant of the adsorption
[tex]k_I[/tex] is the rate constant for forward
[tex]C_I,C_{I.S}[/tex] are concentration of isobutene and site filled with isobutene
The reaction and rate law for the adsorption of water (W) over the amberlyst-15 catalyst catalyst is as follows
W + S ⇄ W.S [tex]-r_{ADW} = k_W(C_WC_V-\frac{C_{W.S}}{K_W} )[/tex]
The reaction and rate law for the surface reaction on the catalyst is as follows
W.S + I.S ⇄ TBA . S + Sn [tex]-r_s = k_s(C_{W.S}C_{I.S}-\frac{C_{TBA.S}C_V}{K_s} )[/tex]
The reaction and rate law for the desorption of TBA from catalyst is as follows
TBA . S ⇄ TBA + S [tex]-r_{D TBA} = k_{DTBA}(C_{TBA.S}-\frac{C_{TBA}C_V}{K_{DTBA}} )[/tex]
the attached image below gives the remaining steps
TBA, an octane enhancer, is produced by the hydration of isobutene over an Amberlyst-15 catalyst, resulting in a multiphase mixture that can be made miscible with the help of cosolvents or excess TBA.
Explanation:The production of t-Butyl Alcohol (TBA) involves a liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The question provides background information on the reaction mechanism and mentions the use of cosolvents or excess TBA to achieve reasonable miscibility in the multiphase mixture. This process results in a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can aid in achieving miscibility. Essentially, the production of TBA serves as a safer alternative to lead additives in gasoline, enhancing octane levels.
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12. The Action of Phospholipases The venom of the Eastern diamondback rattler and the Indian cobra contains phospholipase A2, which catalyzes the hydrolysis of fatty acids at the C-2 position of glycerophospholipids. The phospholipid breakdown product of this reaction is lysolecithin (lecithin is phosphatidylcholine). At high concentrations, this and other lysophospholipids act as detergents, dissolving the membranes of erythrocytes and lysing the cells. Extensive hemolysis may be life- threatening. (a) All detergents are amphipathic. What are the hydrophilic and hydrophobic portions of lysolecithin
Answer:
Explanation:
Compounds derived from Phosphatidyl choline are referred to as Lysolecithins. These compounds are amphiphatic in nature and have both hydrophilic and hydrophobic regions.
Choline, glycerol, and phosphate represent the hydrophyilic portion of lysolecithins,while the fatty acid on C1 is hydrophobic in nature.
Glycerol, choline and phosphate act as a polar tail whereas the fatty acid acts as a long polar head which represents an amphipathic molecule like detergents.
What amount of solid NaOH must be added to 1.0 L of a 0.14 M H2CO3 solution to produce a solution with [H+]= 3.5×10−11 M ? There is no significant volume change as the result of the addition of the solid.
To produce a solution with [H+]= 3.5×10−11 M, 0.14 mol of solid NaOH must be added to 1.0 L of a 0.14 M H2CO3 solution.
Explanation:To calculate the amount of solid NaOH needed, we need to use the balanced chemical equation for the reaction between NaOH and H2CO3. The equation is:
NaOH + H2CO3 -> Na2CO3 + H2O
From the equation, we can see that the ratio between NaOH and H2CO3 is 1:1. Therefore, the amount of solid NaOH needed is equal to the molarity of H2CO3 multiplied by the volume, which is:
Amount of solid NaOH = 0.14 M * 1.0 L = 0.14 mol NaOH
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A sample of gas contains 0.1100 mol of N2(g) and 0.3300 mol of H2(g) and occupies a volume of 20.5 L. The following reaction takes place: N2(g) 3 H2(g) 2 NH3(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant. g
Final answer:
To find the volume of the sample after the given reaction, Avogadro's law is used to show that the volume of gases at constant temperature and pressure is directly proportional to the number of moles. The balanced equation indicates that two times the volume of NH3 gas is produced from one volume of N2 gas. The final volume of the sample is calculated to be 10.25 L after the reaction.
Explanation:
To determine the volume of the sample after the reaction between N2 and H2, we must first understand the stoichiometry of the reaction. The balanced chemical equation for the synthesis of ammonia is:
N2(g) + 3H2(g) → 2NH3(g)
According to Avogadro's law, at constant temperature and pressure, equal volumes of gases contain an equal number of molecules. Thus, in our reaction, one volume of N2 reacts with three volumes of H2 to produce two volumes of NH3. In this case, we begin with 0.1100 mol of N2 and 0.3300 mol of H2. Since H2 is present in excess (needed only 3 times the amount of N2), all of the N2 will be consumed first.
The stoichiometry tells us that 1 mol of N2 reacts with 3 mol of H2 to form 2 mol of NH3. So 0.1100 mol of N2 would produce 0.2200 mol of NH3. Since we are assuming the temperature and pressure are constant, we can use the ratio of moles to determine the change in volume. Initially, there are 0.1100 + 0.3300 = 0.4400 mol of gases and after reaction we have 0.2200 mol of NH3.
The initial volume is 20.5 L. With the moles reducing from 0.4400 to 0.2200, the volume occupied by the gases after the reaction is expected to be half of the initial volume, given the relationship V1/n1 = V2/n2 where V is volume and n is the number of moles. Therefore, the volume of the sample after the reaction is 10.25 L.
Final answer:
By applying Avogadro's law, we determine that all of the initial N2 and H2 reacts to form NH3. Because equal volumes of any gas at the same temperature and pressure have the same number of molecules, we understand that the volume of NH3 produced will be equivalent to the volume of N2 reactant consumed plus three times this volume due to 3H2 reacting. Therefore, the final volume should be less than the initial volume since 2 moles of NH3 will be produced from 4 moles of combined reactants.
Explanation:
Understanding Reaction Volumes and Avogadro's Law:
Based on the balanced chemical equation N2(g) + 3H2(g) → 2NH3(g), we can apply Avogadro's law to determine the change in volume when the reaction occurs at constant temperature and pressure.
At the onset, we have 0.1100 mol of N2(g) and 0.3300 mol of H2(g) occupying a combined volume of 20.5 L. Assuming the reaction goes to completion, we start by identifying the limiting reactant, the reactant that will be completely consumed in the reaction. The stoichiometry of the reaction indicates that for every 1 mol of N2, 3 mol of H2 are required.
For 0.1100 mol of N2, we would need 0.3300 mol of H2 to fully react based on the stoichiometry of 1:3. Since we have exactly 0.3300 mol of H2, it means N2 is the limiting reactant and all of it will be converted into NH3.
Now, according to Avogadro's law, if one mole of N2 gas occupies a certain volume and reacts to form two moles of NH3 at the same conditions of temperature and pressure, the volume of NH3 produced will double that of the N2 because two moles of NH3 contain double the number of molecules compared to one mole of N2. However, since H2 also takes up volume and we have three moles of H2 reacting for every mole of N2, the total volume initially is the volume of N2 plus three times the volume of N2 (which is the volume of H2).
The stoichiometry tells us that 1 mol of N2 reacts with 3 mol of H2 to form 2 mol of NH3. So 0.1100 mol of N2 would produce 0.2200 mol of NH3. Since we are assuming the temperature and pressure are constant, we can use the ratio of moles to determine the change in volume. Initially, there are 0.1100 + 0.3300 = 0.4400 mol of gases and after reaction we have 0.2200 mol of NH3.
The initial volume is 20.5 L. With the moles reducing from 0.4400 to 0.2200, the volume occupied by the gases after the reaction is expected to be half of the initial volume, given the relationship V1/n1 = V2/n2 where V is volume and n is the number of moles. Therefore, the volume of the sample after the reaction is 10.25 L.
1. Choose the alkyl halide(s) from the following list of C6H13Br isomers that meet each criterion below. 1) 1-bromohexane 2) 3-bromo-3-methylpentane 3) 1-bromo-2,2-dimethylbutane 4) 3-bromo-2-methylpentane 5) 2-bromo-3-methylpentane a) the compound(s) that can exist as enantiomers b) the compound(s) that can exist as diastereomers c) the compound that gives the fastest SN2 reaction with sodium methoxide d) the compound that is least reactive to sodium methoxide in methanol e) the compound(s) that undergo an SN1 reaction to give rearranged products f) the compound that gives the fastest SN1 reaction
Answer:
See explanation
Explanation:
the compound that gives the fastest SN2 reaction with sodium methoxide- 1-bromohexane
the compound that gives the fastest SN1 reaction- 3-bromo-3-methylpentane
the compound(s) that undergo an SN1 reaction to give rearranged products- 1-bromo-2,2-dimethylbutane
the compound that is least reactive to sodium methoxide in methanol -
3-bromo-3-methylpentane
the compound(s) that can exist as diastereomers - 3-bromo-3-methylpentane
the compound(s) that can exist as enantiomers- 3-bromo-2-methylpentane
Given that the rate of diffusion of nickel in iron is very much greater in the liquid state than in the solid state, what effect should this have on the ease of obtaining an equilibrium microstructure (i.e., one that is homogeneous) when an alloy containing the peritectic composition 4.5 percent nickel is cooled through the peritectic temperature
Answer:
Explanation:
The rate of diffusion of nickle in is higher in liquid state than solid state which is affect the ease of equilibrium of microstructure.
When peritectic composition of 4.5% nickle is called from peritectic temperature due to high rate of diffusion of nickle into iron we get fine microstructure containing more nickle atoms in the iron.
Due to high rate of diffusion more no. Of microstructure created from where new grain is generated. So the microstructure will get equilibrium soon after cooling.
Answer:
Explanation:
The rate of diffusion of nickle in is higher in liquid state than solid state which is affect the ease of equilibrium of micro structure.
When peritectic composition of 4.5% nickle is called from peritectic temperature due to high rate of diffusion of nickle into iron we get fine micro structure containing more nickle atoms in the iron.
Due to high rate of diffusion more no. Of micro structure created from where new grain is generated. So the micro structure will get equilibrium soon after cooling.
Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests and the reasoning illustrated in the introduction and the Week 2 Worksheet. Be sure to include equations for all the reactions which occur at each step in the scheme. Record the scheme in the data section of the lab notebook. The scheme MUST be submitted at the start of Lab Period 2. It will be reviewed by the TA at the start of the lab period, before you begin testing your scheme. Ag+, Fe 3+ Cu2+, Ni2+
Answer:
SEPARATION SCHEME FOR CATIONS
GIVEN CATIONS : [tex]Ag^{+} \ , Fe^{3+} , Cu^{2+}, Ni^{2+}[/tex]
Step 1: Add [tex]6mol/dm^3[/tex] of [tex]HCl[/tex] to the mixture solution
Result : This would cause a precipitate of [tex]AgCl[/tex] to be formed
Reaction : [tex]Ag^{+} _{(aq)} + Cl^{-} _{(aq)} ---------> AgCl(ppt)[/tex]
Step 2 : Next is to remove the precipitate and add [tex]H_2S[/tex] to the remaining
solution in the presence of [tex]0.2 \ mol/dm^3[/tex] of HCl
Result : This would cause a precipitate of [tex]CuS[/tex] to be formed
Reaction : [tex]Cu^{2+}_{(aq)} + S^{2-}_{(aq)} ------> Cu_2S(ppt)[/tex]
Step 3: Next remove the precipitate then add [tex]6 \ mol/dm^3[/tex] of aqueous
[tex]NH_3 (NH_3 \cdot H_2 O)[/tex] , process the solution in a centrifuge,when the
process is done then sort out the precipitate from the solution
Now this precipitate is [tex]Fe(OH)_3[/tex] and the remaining solution
contains [tex](Ni (NH_3)_6)[/tex]
Next take out the precipitate to a different beaker and add HCl
to it this will dissolve it, then add a drop of [tex]NH_4SCN[/tex] this will
form a precipitate [tex]Fe(SCN)_{6}^{3-}[/tex] which will have the color of
blood indicating the presence of [tex]Fe^{3+}[/tex]
Reaction : [tex]F^{3+}_{(aq)} + 30H^-_{(aq)} --------->Fe(OH)_3_{(aq)}[/tex]
[tex]Fe (OH)_{(s)} _3 + 3H^{+}_{aq} -------> Fe^{3+}_{aq} + 3H_2O_{(l)}[/tex]
[tex]Fe^{3+} + 6SCN^{-} -----> Fe(SCN)_6 ^{3-}[/tex]
Now the remaining mixture contains [tex]Ni^{2+}[/tex]
Explanation:
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) + 2Cl- (aq) ⇄ CuCl2- (aq) K = 8.7 x 104(a) Calculate the solubility of CuCl in pure water. (Ignore CuCl2- formation for part a).(b) Calculate the solubility of CuCl in 0.100 M NaCl solution.
Answer: (a) The solubility of CuCl in pure water is [tex]1.1 \times 10^{-3} M[/tex].
(b) The solubility of CuCl in 0.1 M NaCl is [tex]9.5 \times 10^{-3} M[/tex].
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.
[tex]CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)[/tex]
Initial: 0 0
Change: +x +x
Equilibm: x x
[tex]K_{sp} = 1.2 \times 10^{-6}[/tex]
And, equilibrium expression is as follows.
[tex]K_{sp} = [Cu^{+}][Cl^{-}][/tex]
[tex]1.2 \times 10^{-6} = x \times x[/tex]
x = [tex]1.1 \times 10^{-3} M[/tex]
Hence, the solubility of CuCl in pure water is [tex]1.1 \times 10^{-3} M[/tex].
(b) When NaCl is 0.1 M,
[tex]CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)[/tex], [tex]K_{sp} = 1.2 \times 10^{-6}[/tex]
[tex]Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)[/tex], [tex]K = 8.7 \times 10^{4}[/tex]
Net equation: [tex]CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)[/tex]
[tex]K' = K_{sp} \times K[/tex]
= 0.1044
So for, [tex]CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)[/tex]
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = [tex]\frac{CuCl_{2}}{Cl^{-}}[/tex]
0.1044 = [tex]\frac{x}{0.1 - x}[/tex]
x = [tex]9.5 \times 10^{-3} M[/tex]
Therefore, the solubility of CuCl in 0.1 M NaCl is [tex]9.5 \times 10^{-3} M[/tex].
Final answer:
The solubility of CuCl in pure water and in 0.100 M NaCl solution both turn out to be 1.1 x 10^-3 M when formation of CuCl2- is ignored, determined using the provided Ksp.
Explanation:
To calculate the solubility of CuCl in pure water, we can use the solubility product constant (Ksp). The dissociation of CuCl can be represented as:
CuCl(s) ⇌ Cu+(aq) + Cl-(aq)
Let the solubility of CuCl be 's' moles per liter. At equilibrium, the concentration of Cu+ and Cl- ions will both be 's' M. So, the Ksp expression for CuCl is:
Ksp = [Cu+][Cl-] = (s)(s) = s²
Giving us:
s = √(1.2 x 10^-6) = 1.1 x 10^-3 M
This is the solubility of CuCl in pure water.
For the solubility of CuCl in 0.100 M NaCl solution, we have to consider the common ion effect. In this case, Cl- is the common ion. With the presence of 0.100 M NaCl, the concentration of Cl- ions at equilibrium will be higher. However, since we ignore the formation of CuCl2-, the solubility will still be governed by the Ksp of CuCl. The calculation will remain the same as in pure water, due to the assumption of ignoring CuCl2- formation:
s = 1.1 x 10^-3 M
Therefore, the presence of NaCl does not affect the solubility under the given conditions for this specific case.
How many moles of NH3 are required to produce 12 moles of NH4Cl
Answer:
16 moles
Explanation:
16 moles of NH₃ are required to produce 12 moles of NH₄Cl
What is Stoichiometry ?Stoichiometry helps us use the balanced chemical equation to measures quantitative relationships and it is to calculate the amount of products and reactants that are given in a reaction.
What is Balanced Chemical Equation ?The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
First we have to write the balanced chemical equation.
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
Here 8 moles of NH₃ produce 3 moles of Cl₂ to form 6 moles of NH₄Cl.
So,
12 mole NH₄Cl × [tex]\frac{8\ \text{mole}\ NH_3}{6\ \text{mole}\ NH_4Cl}[/tex]
= 16 moles NH₃
Thus, we can say that 16 moles of NH₃ are required to produce 12 moles of NH₄Cl.
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Your teacher needs a 3.0M solution acid, but only has a 12.0M stock solution of sulfuric acid in the chemical store room. Calculate and describe the steps the teacher needs to take in order to make 100mL of the 3.0M solution of sulfuric acid.
Answer:
The calculations are in the explanation below.
The steps are:
1. Using a graduated pipette, accurately take 25mL of the 12.0M stock solution.2. Pour the 25mL of stock solution into a 100 mL volumetric flask3. Add distilled water up to the mark4. Cap the flask with the stopper5. Stirr by gently rotating the flask.Explanation:
To make 100 mililiter of the 3.0M solution of sulfuric acid, first you must calculate the volume of the 12.0M stock solution that contains the same number of moles as the diluted solution.
For that, you use the dilution formula:
number of moles = C₁V₁ = C₂V212.0M×V₁ = 3.0M × 100mLV₁ = 3.0M × 100mL/12.0M = 25mLThen, the steps are:
1. Using a graduated pipette, accurately take 25mL of the 12.0M stock solution.
2. Pour the 25mL of stock solution into a 100 mL volumetric flask
3. Add distilled water up to the mark
4. Cap the flask with the stopper
5. Stirr by gently rotating the flask.
Convert 10 liter into m3.
10 litres = 0.01 cubic meter
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the ratio of the ionic radius of the cation to the ionic radius of the anion is 0.840, what is the ionic radius of the anion
Answer:
the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
Explanation:
From the diagram shown below :
The anion [tex]Cl^-[/tex] is located at the corners
The cation [tex]Cs^+[/tex] is located at the body center
The Body diagonal length = [tex]\sqrt{3 \ a }[/tex]
∴ [tex]2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a[/tex]
Given that :
[tex]\frac{r^+}{r^-} =0.84[/tex] (i.e the ratio of the ionic radius of the cation to the ionic radius of
the anion )
[tex]0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a[/tex]
Also ; a = 664 pm
Then :
[tex]r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm[/tex]
Therefore, the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
The ionic radius of the anion [tex]r^-=312.52pm[/tex]
Primitive cubic structure:The anion is placed on the corners and the cation is placed on the frame center.
The Body diagonal length = [tex]\sqrt{3a}[/tex]
[tex]2r^++2r^-=\sqrt{3a} \\\\r^++r^-=\sqrt{3}/2a }[/tex]
Given:
Ratio= 0.840
[tex]\frac{r^+}{r^-}=0.840[/tex]
[tex]0.84r^-+r^-=\sqrt{3}/2a \\\\1.84r^-=3/2a\\\\r^-=\sqrt{3}/2*1.84a[/tex]
Also ; a = 664 pm
Then : [tex]r^-[/tex] =312.52 pm
Therefore, the ionic radius of the anion = 312.52 pm
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besides the ocean where else is water on Earth? Is it always in liquid form?
Answer:
water is everywhere, its in the air and the clouds, in the earth too!
it isnt always liquid form. like when it snows or hails.
Explanation:
what volume in ML of 12.0M HCL is needed to contain 3.00 moles of HCL?
Answer:
V= 250ml
Explanation:
From n= CV
3= 12×V
V= 0.25L= 250ml
At a certain temperature the vapor pressure of pure thiophene is measured to be . Suppose a solution is prepared by mixing of thiophene and of acetyl bromide . Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal.
Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. However the methodology will remain the same.
First we calculate the moles of thiophene and heptane, using their molar mass:
137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene 111 g heptane ÷ 100 g/mol = 1.11 moles heptaneTotal number of moles = 1.63 + 1.11 = 2.74 moles
The mole fraction of thiophene is:
1.63 / 2.74 = 0.59Finally, the partial pressure of thiophene vapor is:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
Partial Pressure = 0.59 * 0.60 atmPp = 0.35 atmFinal answer:
The partial pressure of thiophene vapor above an ideal solution can be determined by calculating the mole fraction of thiophene and then applying Raoult's Law, which involves multiplying the mole fraction by the vapor pressure of pure thiophene.
Explanation:
To calculate the partial pressure of thiophene vapor above the solution, we can utilize Raoult's Law, which pertains to ideal solutions. According to Raoult's Law, the partial pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution (mole fraction). To find the mole fraction of thiophene, we will sum up the total moles of both thiophene and acetyl bromide and use their respective amounts to calculate the mole fractions for each.
First, we calculate the mole fraction of thiophene (Xthiophene) in the solution:
Xthiophene = moles of thiophene / (moles of thiophene + moles of acetyl bromide)Then, we apply Raoult's Law to find the partial pressure of thiophene vapor (Pthiophene):Pthiophene = Xthiophene × vapor pressure of pure thiopheneThe partial pressure of thiophene vapor can therefore be obtained, ensuring we keep consistent units and consider significant digits as per the given values in the question.
A 60.0 g aluminum block, initially at 55.00 °C, is submerged into an unknown mass of water at 293.15 K in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 25.00 °C. What is the approximate mass of the water? The specific heat of water is 4.18 J/g . °C. The specific heat of aluminum is 0.897 J/g . °C.
Answer:
The approximate mass of the water is 80kg
Explanation: Heat lost=heat gained
M1c1(Ʃ)=M2c2(Ʃ)
M1 is mass of aluminum
M2 is the mass of water
C1 is specific heat capacity of aluminum
C2 is specific heat capacity of water
Ʃ is change in temperature.
60 x0.897 x(55-25)=M2 x 4.18 x (25-20.15)
1614.6=20.27M2
M2=79.65
M2=80kg
Write a balanced equation for the combination reaction described, using the smallest possible integer coefficients. When carbon monoxide combines with oxygen , carbon dioxide is formed.
Answer:
2 CO(g) + O₂(g) → 2 CO₂(g)
Explanation:
Let's consider the combination reaction between carbon monoxide and oxygen to form carbon dioxide.
CO(g) + O₂(g) → CO₂(g)
We have an odd number of atoms of C on each side, so we multiply CO(g) and CO₂(g) by 2.
2 CO(g) + O₂(g) → 2 CO₂(g)
Now, the equation is balanced.
The combination reaction where carbon monoxide combines with oxygen to form carbon dioxide can be represented by the balanced chemical equation: 2CO + O₂ → 2CO₂.
Explanation:The balanced equation for the combination reaction where carbon monoxide (CO) combines with oxygen (O₂) to form carbon dioxide (CO₂) can be represented as follows:
2CO + O₂ → 2CO₂
In this reaction, two molecules of carbon monoxide react with one molecule of oxygen to produce two molecules of carbon dioxide. It's important to remember that in balancing chemical equations, one must ensure the same number of each type of atom is represented on both the reactant and product side of the equation.
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A mixture of krypton and neon gas is compressed from a volume of 96.0L to a volume of 68.0L , while the pressure is held constant at 15.0atm . Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.
Answer:
-42556.5 J
Explanation:
From gas law,
The work done by gas is given as,
W = PΔV......................... Equation 1
Where W = Work done on the gas mixture, P = Pressure of the gas mixture, ΔV = Change in volume of the gas mixture.
Given: P = 15 atm = (15×101325) N/m² = 1519875 N/m², ΔV = 68-96 = -28 L = -28/1000 = -0.028 m³
Substitute into equation 2
W = 1519875(-0.028)
W = -42556.5 J
Hence the work done by the gas mixture = -42556.5 J
The work performed on the gas mixture of krypton and neon when compressed at a constant pressure of 15.0 atm from a volume of 96.0L to 68.0L is -4.26 x 10⁴ joules.
Explanation:The subject question asks to compute the work done on a gas mixture comprising krypton and neon as it's compressed from a volume of 96.0L to 68.0L, with a constant pressure of 15.0 atm. The calculation of work done on gas under constant pressure relies on the formula W = -P ΔV, where W is work, P represents pressure, and ΔV is the change in volume. In this problem, P = 15.0 atm, which we will convert to pascals (1 atm = 1.01 × 10⁵ Pa). The change in volume ΔV = Vf - Vi = 68.0L - 96.0L = -28.0L, which we will convert to cubic meters (1L = 1.0 × 10-³ m³). Applying these values, W = -P ΔV = -(15.0atm) * (-28.0L) = 420 L*atm. To convert this to joules, we can use the conversion factor 101.3 J = 1 L*atm, which results in a final answer of -4.26 x 10⁴ J. The result is negative as it indicates the work is done on the gases.
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How many moles of nitrogen are there in a 16,500 mL sample of nitrogen at STP?
Final answer:
To find the number of moles of nitrogen in a 16,500 mL sample at STP, convert the volume to liters and divide by the molar volume of 22.4 L/mol, resulting in approximately 0.7366 moles of nitrogen.
Explanation:
To calculate the number of moles of nitrogen in a 16,500 mL sample of nitrogen at STP (Standard Temperature and Pressure), we should use the molar volume of a gas at STP, which is 22.4 L/mol. This means that at STP, 1 mole of any gas occupies 22.4 liters.
First, we need to convert the volume of nitrogen from milliliters to liters:
16,500 mL × (1 L / 1,000 mL) = 16.5 L
Then, we divide the volume of nitrogen by the molar volume:
16.5 L / 22.4 L/mol = 0.7366 moles
Therefore, the sample contains approximately 0.7366 moles of nitrogen gas.
We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of
sodium chloroacetate (ClCH2COONa) in 1.0 L of water (pKa = 2,865).
to. Calculate the pH
yes. Calculate the pH using the formal forms (activities). Have on
counts the contribution of the protons (section a) in the calculation of the ionic strength.
C. Find the pH of a mixture prepared by dissolving the following compounds
in a final volume of 1L: 0.08 moles of ClCH2COOH, 0.04 moles of
ClCH2COONa, 0.05 moles of HNO3 and 0.06 moles of NaOH
Answer:
A. pH using molar concentrations = 2.56
B. pH using activities = 2.46
C. pH of mixture = 2.56
Explanation:
A. pH using molar concentrations
ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺
HA + H₂O ⇌ A⁻ + H₃O⁺
We have a solution of 0.08 mol HA and 0.04 mol A⁻
We can use the Henderson-Hasselbalch equation to calculate the pH.
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}[/tex]
B. pH using activities
(i) Calculate [H⁺]
pH = -log[H⁺]
[tex]\text{[H$^{+}$]} = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73 \times 10^{-3}\text{ mol/L}[/tex]
(ii) Calculate the ionic strength of the solution
We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.
The formula for ionic strength is
[tex]I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} + 0.00273\times(+1)^{2}\right]\\\\= \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041[/tex]
(iii) Calculate the activity coefficients
[tex]\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79[/tex]
(iv) Calculate the initial activity of A⁻
a = γc = 0.79 × 0.04= 0.032
(v) Calculate the pH
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\[/tex]
C. Calculate the pH of the mixture
The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.
The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.
The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.
The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.
(i) Calculate the ionic strength
[tex]I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} + 0.05\times(-1)^{2}\right]\\\\= \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10[/tex]
(ii) Calculate the activity coefficients
[tex]\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69[/tex]
(iii) Calculate the initial activity of A⁻:
a = γc = 0.69 × 0.05= 0.034
(iv) Calculate the pH
[tex]\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}[/tex]
Determine the primary structure of an octapeptide from the following data: acid-catalyzed hydrolysis gives 2 arg, leu, lys, met, phe, ser, and tyr. carboxypeptidase a releases ser. edman's reagent releases leu. treatment with cyanogen bromide forms two peptides with the following amino acid compositions: 1. arg, phe, ser 2. arg, leu, lys, met, tyr trypsin-catalyzed hydrolysis forms the following two amino acids and two peptides: 1. arg 2. ser 3. arg, met, phe 4. leu, lys, tyr
A Sentence that I Used:
The position of Met seems incorrect. The amino acid sequences of all proteins begin with Met because it is the amino acid that is attached to the anticodon for the AUG start codon. It looks like the student might have worked in a backward direction for transcription.
A student placed 11.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 40.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution
Answer:
There is 0.92 g of glucose in 100 mL of the final solution.
Explanation:
Initially, 11.5 g of glucose is added to the volumetric flask
Water is then added to 100 mL Mark,
The flask was then shaken until the solution was uniform.
The shaking of the mixture makes the concentration of glucose to become uniform all through the solution.
At this point, the concentration of this solution in g/mL is (11.5/100) = 0.115 g/mL
A 40.0 mL sample of this glucose solution was diluted to 0.500 L.
40.0 mL of the already mixed solution is then diluted to 0.500 L.
The mass of glucose in 40.0 mL of the mixed solution with concentration 0.115 g/mL is then given as
Mass = (conc in g/mL) × (volume) = 0.115 × 40 = 4.6 g
So, this mass is then diluted to 0.500 L mark.
New concentration = (mass)/(conc In mL) = (4.6/500) = 0.0092 g/mL
How many grams of glucose are in 100. mL of the final solution
Mass = (conc in g/mL) × (Volume in mL) = 0.0092 × 100 = 0.92 g
Hope this Helps!!!
Answer:
0.459 gram
Explanation:
Find the attachment
Sodium hydrogen carbonate (NaHCO3) , also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid (HC1) , which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HC1 through this reaction:
HC1(aq) + NaHCO3 (aq) → NaC1 (aq) + H2O (I) + CO2 (g)
The CO2 gas produced is what makes you burp after drinking the solution.
Suppose the fluid in the stomach of a man suffering from indigestion can be considered to be 50. mL of a 0.034 M HC1 solution. What mass of NaHCO 3 would he need to ingest to neutralize this much HC1 ?
Be sure your answer has the correct number of significant digits.
Answer:
Mass of NaHCO₃ required for the neutralization = 0.143 g
Explanation:
The reaction between HCl and NaHCO₃ is given as
HCl + NaHCO₃ → NaCl + H₂O + CO₂
The fluid in the stomach of a man suffering from indigestion is considered to be 50 mL of a 0.034 M HCl solution. The mass of NaHCO₃ he would need to ingest to neutralize this much HCl.
We forst need to calculate the number of moles of HCl in 50.0 mL of 0.034 M HCl
(Number of moles) = (Conc in mol/L) × (Volume in L)
Conc in mol/L = 0.034 M
Volume in L = (50/1000) = 0.05 L
Number of moles of HCl = 0.034 × 0.05 = 0.0017 moles
From the stoichiometric balance of the reaction,
1 mole of HCl requires 1 mole of NaHCO₃
0.0017 moles of HCl will require 0.0017 moles of NaHCO₃.
So, we can then.calculate the mass of NaHCO₃ required for this neutralization.
Mass = (Number of moles) × (Molar mass)
Molar mass of NaHCO₃ = 84.007 g/mol
Mass of NaHCO₃ required for this neutralization = 0.0017 × 84.007 = 0.1428119 g = 0.143 g to 3 s.f because calculated values usually have 1 more significant figure than the given parameters for the calculations.
Hope this Helps!!!
The hydroxyl radical, a fragment of water vapor known as the "atmospheric cleanser", with a chemical formula of OH, has a globally averaged number density of ~1×10^6 molecules cm-3 that is fairly constant with altitude. Determine the OH mixing ratio (in ppt) at the surface, where total pressure P ~1 atm, temperature T ~282K, and also at 10 km altitude, where P ~0.260 atm,
Answer:
Check the explanation
Explanation:
The hydroxyl radical, •OH, is the hydroxide ion (OH−) when in neutral form of the Hydroxyl radicals they are extremely reactive (easily becoming hydroxy groups) and as a result are short-lived; however, they form a significant part of radical chemistry. The hydroxyl radical composition is also highly reactive towards oxidative reactions.
Kindly check the attached image below to see the step by step explanation to the question above.
Aldol condensation of 2,5-heptanedione yields a mixture of two enone products in a 9:1 ratio. Treatment of the minor product with aqueous NaOH converts it into the major product; the interconversion proceeds as follows: Hydroxide ion adds to the double bond, forming enolate ion 1; Proton transfer occurs, yielding tetrahedral intermediate 2; Ring opening occurs, yielding enolate ion 3; Protonation of enolate ion 3 occurs, yielding 2,5-heptanedione; Deprotonation at C-6 occurs, yielding enolate ion 5; Enolate ion 5 attacks C-2, yielding tetrahedral intermediate 6; Protonation occurs to yield aldol addition product 7; Dehydration yields the more stable product.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The diagram of the mechanism of this reaction is shown on the second uploaded image
The structure of the enolate Ion 1 is shown on the third uploaded image
Explanation:
The aldol condensation of 2,5-heptanedione involves several steps of proton transfer and structural changes to yield stable products. These steps showcase the dynamic nature of chemical reactions and examples of sequential proton transfers seen in polyprotic acids.
Explanation:The aldol condensation of 2,5-heptanedione yielding a mixture of enone products is a chain of reactions involving the transfer of protons. Iterating the process you described, it starts with hydroxide ion adding to the double bond to form intermediate enolate ion 1, followed by proton transfer resulting in tetrahedral intermediate 2. The ring opening process then forms another enolate ion (3), which after protonation, yields the initial 2,5-heptanedione. A deprotonation at C-6 subsequently yields enolate ion 5, which attacks C-2 to generate tetrahedral intermediate 6. Protonation at this stage will yield aldol addition product 7. Dehydration of this product will finally yield the more stable product. This sequence is an example of sequential proton transfers often seen in polyprotic acids.
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