Explanation:
First, we will calculate the feed rate of alum as follows.
[tex]\frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}[/tex]
= 228000 g/day
Converting this amount into g/min as follows.
[tex]\frac{228000 g}{1 day} \times \frac{1 day}{1440 min}[/tex]
= 158 g/min
Now, the chemical equation will be as follows.
[tex]Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O[/tex]
[tex]\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}[/tex]
= 0.151 mmol [tex]mmol SO^{2-}_{4}/L[/tex]
[tex]\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}[/tex]
= 15.15 mg [tex]CaCO_{3}[/tex]/L
For precipitate:
[tex]Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O[/tex]
[tex]\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}[/tex]
= 7.88 [tex]Al(OH)_{3}/L[/tex]
[tex]\frac{7.88 mg Al(OH)_{3}}{1 L} \times \frac{3800 m^{3}}{1 day} \times \frac{1000 L}{1 m^{3}} \times \frac{1 kg}{10^{6} mg}[/tex]
= 29.9 [tex]Al(OH)_{3}/day[/tex]
Find the net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere. Express your answer in terms of the radius R and the total charge Q.
Answer:
Explanation:
The final net force will be in the Z- direction. Let's find out the z component of the force on the differential volume of charge is:
df = dqEcosθz
[tex]E = \frac{1}{4\pi epsilon} \frac{Qr}{R^{3} }[/tex]
dq = ρdV = [tex]\frac{3Q}{4\pi R^{3} }[/tex][tex]r^{2}[/tex]dr.sinθdθdΦ
integrate it over half ball,
[tex]F_{z} = \int\limits^._V {df_{x}dV} =\frac{1}{4\pi epsilon } \frac{Q}{R^{3} } \frac{3Q}{4\pi R^{3} }\int\limits^R_0 {\int\limits^\frac{\pi }{2} _{0} {\int\limits^\frac{\pi }{2} _0 {r^{3} } \, dr } \, } \,[/tex].sinθcosθdθdΦ.( these are part of the integral, i was unable to write it in equation format).
= [tex]\frac{3Q^{2} }{32\pi epsilonR^{2} } \int\limits^\frac{\pi }{2} _b {} \,[/tex] sinθcosθdθ
= [tex]\frac{3Q^{2} }{64\pi epsilon R^{2} }[/tex]
[tex]F = \frac{3Q^{2} }{64\pi epsilon R^{2} } z[/tex]
The net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere is (kQ²)/(2R²), where k is the electrostatic constant, Q is the total charge on the sphere, and R is the radius of the sphere.
Explanation:The net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere can be found by considering the electric field at the surface of the sphere. Since the charge distribution is spherically symmetric, the electric field at the surface of the sphere will only have a radial component. Using Gauss's law, we can determine that the electric field at the surface is given by E = kQ/R², where k is the electrostatic constant, Q is the total charge on the sphere, and R is the radius of the sphere.
To find the force, we can multiply the electric field by the charge on the northern hemisphere. The charge on the northern hemisphere can be calculated as half the total charge on the sphere, Q/2. Therefore, the net force is given by F = (kQ²)/(2R²).
The value of Δ G ° ' for the conversion of 3-phosphoglycerate to 2-phosphoglycerate (2PG) is + 4.40 kJ/mol . If the concentration of 3-phosphoglycerate at equilibrium is 2.45 mM , what is the concentration of 2-phosphoglycerate? Assume a temperature of 25.0 ° C .
Answer:The concentration of 2-phosphoglycerate is 0.415 mM
Explanation:
[tex]3-phosphoglycerate\rightleftharpoons 2-phosphoglycerate[/tex]
Relation between standard Gibbs free energy and equilibrium constant follows:
[tex]\Delta G^o=-RT\ln K[/tex]
where,
[tex]\Delta G^o[/tex] = Standard Gibbs free energy = +4.40 kJ/mol = 4400 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = [tex]25^0C=(25+273)K=298 K[/tex]
Putting values in above equation, we get:
[tex]4400J/mol=-(8.314J/Kmol)\times 298K\times \ln K[/tex]
[tex]\ln K=-1.776[/tex]
[tex]K=0.169[/tex]
[tex]K=\frac{ 2-phosphoglycerate}{3-phosphoglycerate}[/tex]
[tex]0.169=\frac{ 2-phosphoglycerate}{2.45mM}[/tex]
[tex]2-phosphoglycerate}=0.415mM[/tex]
Thus the concentration of 2-phosphoglycerate is 0.415 mM
Which of the following statements about entropy is true? a) Entropy is a measure of system multiplicity. b) The standard unit of entropy is kcal/mol. c) Entropy cannot be visualized in terms of disorder. d) In general, the entropy of a protein increases during folding
Answer:
The correct answer is d.
Incorrect Answers:
Options a, b and c are incorrect answers because during folding of protein, surface area decreases and so there are less water molecules involved.The water molecules are free.The protein folding decreases entropy
Calculate the number of O atoms in 0.150 g of CaSO4 · 2H2O.
Answer:
There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.
Explanation:
Mass of calcium sulfate crystal = m = 0.150 g
Molar mass of calcium sulfate crystal = M = 172 g/mol
Moles of magnesium nitrate = n
[tex]n=\frac{m}{M}[/tex]
[tex]n=\frac{0.150 g}{172 g/mol}=0.0008721 mol[/tex]
1 mole of calcium sulfate crystal has 6 moles of oxygen atoms. Then 0.004446 moles calcium sulfate crystal will have :
[tex]6\times 0.0008721 mol=0.0052326mol\approx 0.00523 mol[/tex]
There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.
In 0.150 g of CaSO₄·2H₂O, there are 3.15 × 10²¹ O atoms.
We want to calculate the number of oxygen atoms in 0.150 g of CaSO₄·2H₂O. We will need a series of conversion factors.
What is a conversion factor?A conversion factor is an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.
We will need to consider the following conversion factors:
The molar mass of CaSO₄·2H₂O is 172.17 g/mol.There are 6.02 × 10²³ molecules of CaSO₄·2H₂O in 1 mole (Avogadro's number).There are 6 oxygen atoms in 1 molecule of CaSO₄·2H₂O.[tex]0.150 g CaSO_4.2H_2O \times \frac{1molCaSO_4.2H_2O}{172.17gCaSO_4.2H_2O} \times \frac{6.02\times 10^{23}moleculesCaSO_4.2H_2O }{1molCaSO_4.2H_2O} \times \frac{6\ O\ atoms}{1moleculeCaSO_4.2H_2O} = \\ = 3.15 \times 10^{21} O\ atoms[/tex]
In 0.150 g of CaSO₄·2H₂O, there are 3.15 × 10²¹ O atoms.
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A 0.4 M buffer solution was prepared with acetic acid and sodium acetate. At pH 5.5, what are the concentrations of acetic acid and acetate ion? The pKa of acetic acid is 4.76. Round the answers to two decimal places. State the units.
Answer: The concentration of acetic acid and sodium acetate (acetate ion) is 0.06 M and 0.34 M respectively
Explanation:
We are given:
Concentration of buffer solution having acetic acid and sodium acetate = 0.4 M
Let the concentration of acetic acid be x M
So, the concentration of sodium acetate will be = (0.4 - x) M
To calculate the concentration of acid for given pH, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})[/tex]
where,
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acetic acid = 4.76
[tex][CH_3COONa]=0.4-x[/tex]
[tex][CH_3COOH]=x[/tex]
pH = 5.5
Putting values in above equation, we get:
[tex]5.5=4.76+\log(0.4-x}{x})\\\\x=0.062M[/tex]
So, concentration of acetic acid = x = 0.06 M
Concentration of sodium acetate = (0.4 - x) = (0.4 - 0.06) = 0.34 M
Hence, the concentration of acetic acid and sodium acetate (acetate ion) is 0.06 M and 0.34 M respectively
What is the molar mass of a gas if it takes 7.3 min to effuse through a small hole and 6.0 min for the same amount of N2 to effuse through the same hole gas
Answer:
Molar mass of unknown gas = 41.45 g/mol
Explanation:
Time taken for a gas to effuse/diffuse ∝√(Molar Mass of the gas)
Let T₁ be the time taken for the Unknown gas to effuse. T₁ = 7.3 min = 438s
Let T₂ be the time taken for the Nitrogen gas to effuse. T₁ = 6 min = 360s
(T₁/T₂) = √(Molar mass of unknown gas/Molar mass of Nitrogen)
Molar mass of Nitrogen = 2×14 = 28 g/mol
(438/360) = √(Molar Mass of unknown gas/28)
√(Molar mass of unknown gas/28) = 1.21667
molar mass of unknown gas/28 = 1.21667² = 1.4803
Molar mass of unknown gas = 1.4803 × 28 = 41.45 g/mol
Hope this helps!
Why could the Bohr model not predict line spectra for atoms other than hydrogen?
he could not preidct it bud
rate = k[AB]2 and k = 0.20 L/mol·s. If the initial concentration of AB is 1.50 M, what is [AB] after 14.0 s?
Answer : The concentration of AB after 14.0 s is, 0.29 M
Explanation :
The expression used for second order kinetics is:
[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]
where,
k = rate constant = [tex]0.20M^{-1}s^{-1}[/tex]
t = time = 14.0 s
[tex][A_t][/tex] = final concentration = ?
[tex][A_o][/tex] = initial concentration = 1.50 M
Now put all the given values in the above expression, we get:
[tex]0.20\times 14.0=\frac{1}{[A_t]}-\frac{1}{1.50}[/tex]
[tex][A_t]=0.29M[/tex]
Therefore, the concentration of AB after 14.0 s is, 0.29 M
For the following soluble (strong electrolyte) species, represent the process for the solid compound dissolving in water: aluminum nitrate, iron (II) chloride, potassium sulfide, magnesium acetate, ammonium phosphate. Ex :MgBr2(s) Mg+2(aq) + 2Br-(aq)
Answer:
Al(NO₃)₃ → Al³⁺ (aq) + 3NO₃⁻ (aq)
FeCl₂ → Fe²⁺ (aq) + 2Cl⁻ (aq)
K₂S → 2K⁺ (aq) + S⁻²(aq)
Mg(CH3COO)₂ → Mg²⁺ (aq) + 2CH3COO⁻ (aq)
(NH₄)₃PO₄ → 3NH₄⁺ (aq) + PO₄⁻³(aq)
Explanation:
Al(NO₃)₃ → Aluminum nitrate
FeCl₂ → Iron (II) chloride
K₂S → Potassium sulfide
Mg(CH3COO)₂ → Magnesium acetate
(NH₄)₃PO₄ → Ammonium phosphate
Shells constructed from seawater incorporate the 18O/16O ratio of seawater during their lifetime within their CaCO3 shell walls, providing a paleothermometer that is used to estimate the temperatures of ancient seas.
Answer:
Hello, the above question is not complete, nonetheless let us check somethings out.
Explanation:
Paleothermometer definition is from two words, that is "Paleo" which means something that is old and ''thermometer" which is an instrument for measuring temperature. So, if we add this up, Paleothermometer is an instrument for measuring "old" temperature, that is temperature. One of the Paleothermometer that is been used is the δ18O which is the one in the question that has isotopic ratio of 18O/16O, and it deals with the measurement of 18O to 16O. The others include Alkenones Paleothermometer, Mg/Ca Paleothermometer, Leaf physiognomy and so on.
If the values of the isotopic ratio that is 18O/16O ratio is low, then the temperature is high. To Calculate the 18O/16O ratio for ancient ocean then we will be using the equation below;
δ18O = (z - 1) × 1000. Where z= [(18O/16O)/( 18O/16O)sm. And sm= standard mean.
Be sure to answer all parts. Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with superheated steam: C(s) + H2O(g) → CO(g) + H2(g) ΔH o rxn = 129.7 kJ (a) Combine the reaction above with the following two to write an overall reaction for the production of methane: CO(g) + H2O(g) → CO2(g) + H2(g) ΔH o rxn = −41 kJ CO(g) + 3H2(g) → CH4(g) + H2O(g) ΔH o rxn = −206 kJ In the overall reaction, include the physical states of each product and reactant. (b) Calculate ΔH o rxn for this overall change. 12.03 kJ (c) Using the value in (b) and calculating ΔH o rxn for the combustion of methane, find the total heat for gasifying 6.28 kg of coal and burning the methane formed. Assume water forms as a gas and the molar mass of coal is 12.00 g/mol.
Final answer:
The overall reaction for converting coal to methane is 2C(s) + 2H2O(g) → CH4(g) + CO2(g). The ΔH°rxn for the overall process is -117.3 kJ. Using the molar mass of coal and the heat of combustion for methane, the total heat for gasifying 6.28 kg of coal and burning the resulting methane is calculated.
Explanation:
To answer the student's question regarding the production of methane from coal via gasification, we combine the given reactions to find the overall reaction:
C(s) + H2O(g) → CO(g) + H2(g) ΔH°rxn = 129.7 kJCO(g) + H2O(g) → CO2(g) + H2(g) ΔH°rxn = −41 kJCO(g) + 3H2(g) → CH4(g) + H2O(g) ΔH°rxn = −206 kJThe H2O(g) and CO(g) appear on both sides of the above equations and can be cancelled out when added together. The overall reaction is:
2C(s) + 2H2O(g) → CH4(g) + CO2(g)
To calculate ΔH°rxn for the overall reaction, we sum the enthalpies of the individual steps:
ΔH°rxn (overall) = 129.7 kJ + (-41 kJ) + (-206 kJ) = -117.3 kJ
Using the heat of combustion for methane, 890.4 kJ/mol, and the assumption that coal is approximated as pure carbon (12.00 g/mol), we can calculate the total heat for gasifying 6.28 kg of coal and then burning the methane produced:
Moles of C in 6.28 kg = (6280 g) / (12.00 g/mol) = 523.33 mol
Total heat from gasification = 523.33 mol × -117.3 kJ/mol = -61411.3 kJ
Total heat from combustion = 523.33 mol × 890.4 kJ/mol = 465774.9 kJ
The total heat for the entire process is the sum of heat from gasification and combustion.
Calculate the energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 the centers of which are separated by a distance of 7.5 nm
The energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 can be calculated using Coulomb's law.
Explanation:The energy of attraction between a cation and an anion can be calculated using Coulomb's law. Coulomb's law states that the energy of attraction is directly proportional to the product of the charges and inversely proportional to the distance between them. In this case, the cation has a valence of +1 and the anion has a valence of -3. The distance between their centers is given as 7.5 nm. The equation to calculate the energy of attraction is:
E = k * (q1 * q2) / r
Where E is the energy of attraction, k is Coulomb's constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the cation and anion respectively, and r is the distance between their centers.
Substituting the values:
E = (9 x 10^9 Nm^2/C^2) * ((+1) * (-3)) / (7.5 x 10^-9 m)
Simplifying the equation gives:
E = -36 Nm
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Suppose in the lab you measure the solid NaOH and dissolve it into 100.0 mL of water. You then measure 0.2000 g of KHP (KC8H5O4, 204.22 g/mol) and place it in a clean, dry 100-mL beaker, and then dissolve the KHP in about 25 mL of water and add a couple of drops of phenolphthalein indicator. You titrate this with your NaOH(aq) solution and find that the titration requires 9.53 mL of NaOH(aq).
a. What is the concentration of your NaOH(aq) solution?
b. Determine the number of moles of NaOH(aq) that would be required to titrate 250.00 mL of your Kool-Aid solution.
Answer:
Explanation:
Equation of the reaction:
NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)
A.
Number of moles = mass/molar mass
= 0.20/204.22
= 9.79 x 10-4 mol KHP
By stoichiometry, 1 mole of NaOH reacts with 1 mole of KHP
= 6.267 x 10-4 mol NaOH
Molar concentration = number of moles/volume
= 9.79 x 10-4/0.00953
= 0.103 M NaOH
B.
Number of moles of kool aid = mass/molar mass
= 3.607/342
= 0.01055 mol
Concentration = 0.01055/0.005
= 2.1 M
C1 * V1 = C2 * V2
Concentration of NaOH = 2.1 * 0.005/0.01079
= 0.977 M
Number of moles of NaOH = 0.0105 mol.
Give full and condensed electron configurations, partial orbital diagrams showing valence electrons, and the number of inner electrons for the following elements:
(a) Ni (Z = 28) (b) Sr (Z = 38) (c) Po (Z = 84)
Answer:
As is in the attachment.
Explanation:
The condensed electronic configuration is written in the short form by expressing in terms of the elements of the noble gases.
The attached file is the explanation of the answers.
The mass percent composition of an organic compound showed that it contained 40.0% C, 6.7% H and 53.3% O. A solution of 0.673 g of the solid in 28.1 g of the solvent diphenyl gave a freezing point depression of 1.6 Celsius. Calculate the molecular formula of the solid. (Kf for diphenyl is 8.00°C/m.)
Answer:
C₄H₈O₄ is the molecular formula of the solid.
Explanation:
Let's apply the freezing point depression to solve this:
ΔT = Kf . m
where Δt = Freezing T° pure solvent - Freezing T° of solution
Kf, the cryoscopic constant
m = molalilty, moles of solute in 1kg of solvent.
We must determine the molecular weight to know the molecular formula of the solid
Let's replace the data.
1.6°C = 8 °C/m . m
We can determine molality, by this:
1.6 °C / 8°C/m = 0.2 m
mol of solute / 1kg of solvent = 0.2
Let's convert the mass of solvent from g to kg, to determine the moles of solute.
mol of solute / 0.0281 kg = 0.2 mol/kg
28.1 g . 1kg / 1000 g = 0.0281 kg
mol of solute = 0.0281 kg . 0.2 mol/kg → 0.00562 moles
Molar mass (g/mol) → 0.673 g / 0.00562 mol = 120 g/mol
Now, we can apply the percent composition.
100 g of compound have ___ 40 g C ___6.7 g H ___ 53.3 g O
120 g of compound must have:
(120 . 40) / 100 = 48 g of C
(120 . 6.7) / 100 = 8 g of H
(120 . 53.3) / 100 = 64 g of O
Let's convert the mass of each elements to moles
48 g . 1 mol/12 g = 4 C
8 g . 1 mol /1g = 8 H
64 g . 1 mol / 16g = 4 O
To determine the molecular formula of the solid compound, calculate the empirical formula using the mass percent composition. Then, divide the molar mass of the compound by the molar mass of the empirical formula to find the number of empirical formula units in the compound.
Explanation:To determine the molecular formula of the solid compound, we first need to calculate its empirical formula. We can assume a 100g sample of the compound, so we have 40g of C, 6.7g of H, and 53.3g of O. Converting the grams to moles, we find that we have approximately 3.33 moles of C, 6.65 moles of H, and 3.33 moles of O.
Next, we need to find the smallest whole number ratio between the elements. The ratio between C, H, and O is approximately 1:2:1. So, the empirical formula of the compound is CH2O.
To find the molecular formula, we need to know the molar mass of the compound. The molar mass of the empirical formula CH2O is approximately 30 g/mol. To determine the molecular formula, we need to divide the molar mass of the compound by the molar mass of the empirical formula. Assuming the molar mass of the compound is a multiple of 30 g/mol, we can divide it by 30 to find the number of empirical formula units in the compound.
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In order to prepare a solution of 600 mg O2/L of COD, how much glucose should be dissolved in distilled water?
Answer:
1.126 grams
Explanation:
Given that:
Standard Solution = 600 mg O₂/L
The molecular weight of C₆H₁₂O₆ (glucose) = 180.156 g/mol
The mass of O₂ in 1 mole of C₆H₁₂O₆ can be determined as:
C₆H₁₂O₆ = 6 × 16 g ( of one oxygen)
= 96 g
∴ 96 g of O₂ is available in 180.156 gram of C₆H₁₂O₆
Thus C₆H₁₂O₆ required for giving 600 mg = 0.60 g of O₂
∴ [tex]\frac{180.156}{96}*0.6[/tex]
= 1.876625 × 0.6
= 1.125975 g
≅ 1.126 grams
Hence, 1.126 grams of C₆H₁₂O₆ (glucose) will be added to one liter of distilled water in order to get 600mg O₂/L.
At a certain temperature this reaction follows second-order kinetics with a rate constant of : Suppose a vessel contains at a concentration of . Calculate the concentration of in the vessel seconds later. You may assume no other reaction is important. Round your answer to significant digits.
To calculate the concentration of a second-order reaction after a certain time, use the half-life equation and substitute the given values.
Explanation:A second-order reaction follows the equation relating the half-life of the reaction to its rate constant and initial concentration:
t1/2 = 1 / (k * [A]₀)
To calculate the concentration of A in the vessel after a certain number of seconds, you can substitute the given values into this equation. For example, if t1/2 = 18 min, k = 0.0576 L mol-1 min-1, and [A]₀ = 0.200 mol L-1, you can solve for [A].
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The concentration in the vessel seconds later is ≈ 0.179 M which will be required for 0.200 M and a rate constant of 5.76 × 10-² L/mol/min over 10 minutes.
To solve this question, we can use the integrated rate law for second-order reactions:
[tex]\frac{1}{[A]t} = \frac{1}{[A]_0} + kt[/tex]
The initial concentration, [tex][A]_0[/tex], is 0.200 M.The rate constant, k, is 5.76 × 10-2 L/mol/min.The time elapsed, t, is 10.0 min.Substitute these values into the integrated rate law equation:
[tex]\frac{1}{[A]t} = \frac{1}{0.200} + (5.76 \times 10^{-2} L/mol/min)(10.0 min)[/tex]
Calculate the right-hand side:
[tex]\frac{1}{[A]t} = 5.00 + 0.576 \\\\\frac{1}{[A]t} = 5.576[/tex]
So, [tex][A]t = 1 / 5.576 \approx 0.179 M[/tex]
Therefore, the concentration of butadiene remaining after 10.0 min is approximately 0.179 M.
What is the electron capacity of the nth energy level? What is the capacity of the fourth energy level?
Answer: The number of electrons present in the fourth energy level are 32
Explanation:
To calculate the number of electrons present in a particular energy level, we use the equation:
[tex]\text{Number of electrons}=2n^2[/tex]
where,
n = principle quantum number
Calculating the number of electrons for fourth energy level:
n = 4
Putting values in above equation, we get:
[tex]\text{Number of electrons}=2(4)^2\\\\\text{Number of electrons}=32[/tex]
Hence, the number of electrons present in the fourth energy level are 32
se the following key to classify each of the elements below in its elemental form: A. Discrete atoms ... C. Metallic lattice B. Molecules ... D. Extended, three-dimensional network 1. Calcium 2. Helium ... 3. Sulfur 4. Potassium ...
Explanation:
the correct match can as follows:
1. helium ⇒ discrete atoms (helium is an inert gas)
2. oxygen ⇒ molecules (oxygen exists in molecular form as O2)
3 Magnesium ⇒ matalic lattice ( Magnesium is a metal FCC crystal Structure :) )
4 Aluminum⇒ covalent network (since it is situated in the middle of the group and posses amphoteric properties too)
Final answer:
To classify the elements: Calcium and Potassium form metallic lattices (category C), Helium exists as discrete atoms (category A), and Sulfur forms molecules (category B). This is due to how the atoms or molecules arrange themselves in solids, often driven by forces between electrons and nuclei.
Explanation:
To classify each element in its elemental form based on the given key:
Calcium (Ca) would fall into category C, Metallic lattice, as it is a metal and metals typically form extended, repeating three-dimensional patterns.
Helium (He) is correctly classified as A, Discrete atoms, because it exists as individual atoms and does not form bonds easily.
Sulfur (S) in its most common form consists of eight-atom rings, making it B, Molecules.
Potassium (K), like calcium, would also be categorized as C, Metallic lattice, due to its metal characteristics, forming a crystal lattice.
Atoms arrange themselves in these structures based on the net attractive forces between electrons and atomic nuclei, and the structural arrangement significantly affects the properties of the material, such as malleability and ductility for metals. Pure metals are crystalline solids with metal atoms packed in a repeating pattern known as a unit cell.
The proposed mechanism for the reaction ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq) is
1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq) FAST
2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq) FAST
3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq) SLOW
What is the overall equation? (Type your answer using the format [NH4]+ for NH4+. Use the lowest possible coefficients. Enter 0 if necessary. Do not leave any box blank.)
(aq) + I -(aq) Cl -(aq) + (aq) (b) Identify the intermediates, if any.
Answer:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq).
Explanation:
1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq)
2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq)
3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq)
Adding all the 3 equations together gives and it gives:
ClO-(aq) + H2O(l) + I-(aq) + HClO(aq) + OH -(aq) + HIO(aq)
---> HClO(aq) + OH-(aq) + HIO(aq) + Cl -(aq) + H2O(l) + IO-(aq)
Deleting the same species on both sides of the equation gives:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)
The overall equation:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)
The overall chemical reaction is ClO- (aq) + I- (aq) --> IO- (aq) + Cl- (aq). The intermediates, compounds produced and then consumed in later reaction steps, are HClO and HIO.
Explanation:The overall reaction occurring is the sum of the provided stepwise reactions, where intermediates, or species that are formed in one step and consumed in another, are cancelled out. The chemical species that are intermediates in this case are HClO and HIO.
Adding all three reactions together and cancelling out the intermediates, we get:
ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq)
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What is the molality of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175? (The molar mass of KCl = 74.55 g/mol and the molar mass of H2O is 18.02 g/mol.)
Final answer:
The molality (m) of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175, can be calculated using the formula molality (m) = (moles of solute) / (kilograms of solvent). First, calculate the moles of KCl by multiplying the mole fraction by the mass of water and dividing by the molar mass of KCl. Then, calculate the kilograms of water by dividing the mass of water by 1000. Finally, substitute these values into the molality formula.
Explanation:
The molality (m) of an aqueous KCl solution with a mole fraction (XKCl) of 0.175 can be calculated using the following formula:
molality (m) = (moles of solute) / (kilograms of solvent)
To find the molality, we need to convert the mole fraction into moles of KCl and kilograms of water. The molar mass of KCl is 74.55 g/mol and the molar mass of H2O is 18.02 g/mol.
First, we calculate the moles of KCl:
moles of KCl = XKCl * (mass of water) / (molar mass of KCl)
Next, we calculate the kilograms of water:
kilograms of water = (mass of water) / 1000
Finally, we substitute these values into the formula:
molality (m) = moles of KCl / kilograms of water
The mole fraction of a non electrolyte (MM 101.1 g/mol) in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. Calculate the molarity of the solution.
Answer:
Molarity for the solution is 1.05 mol/L
Explanation:
Mole fraction of solute = 0.0194
Solution's density = 1.0627 g/mL
We must know that sum of mole fraction = 1
Mole fraction of solute + Mole fraction of solvent = 1
0.0194 + Mole fraction of solvent = 1
Mole fraction of solvent = 1 - 0.0194 → 0.9806
Molarity is mol of solute in 1L of solution, so we have to determine solution's volume in L
With molar mass we can determine the mass of solute and solvent and then, the solution's mass
0.0194 mol . 101.1 g/ mol = 1.96 g of non electrolyte solute
0.9806 mol . 18 g/mol = 17.65 g of water
Mass of solution = mass of solute + mass of solvent
1.96 g + 17.65 g = 19.6 g (mass of solution)
Solution's density = Solution's mass / Solution's volume
1.0627 g/mL = 19.6 g / Solution's volume
Solution's volume = 19.6 g / 1.0627 g/mol →18.4 mL
Let's convert the mass from mL to L
18.4mL . 1L / 1000 mL = 0.0184 L
We have the moles of solute, so let's determine molarity
mol/L → 0.0194 mol / 0.0184 L = 1.05 M
When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth. If a 103200 kg submarine takes on 2100 kg of water to maintain depth at 1000 feet, where the density of seawater is approximately 1033 kg/m3, what is the total displacement (volume) of the submarine in m3 (Report your answer to 4 sig figs without a written unit)?
Answer:
[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.
Explanation:
Mass of water carried by submarine at 1000 ft depth = m = 2100 kg
The density of seawater at 1000 ft depth = d = [tex]1033 kg/m^3[/tex]
Volume of the water displaced = V= ?
Total displacement of the submarine = Volume of the water displaced = V
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3[/tex]
[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.
The reaction of hydrochloric acid with potassium permanganate is described above. 526.64 g of hydrogen chloride is reacted with 229.19 g of potassium permanganate. Assuming the reaction goes to completion, calculate the mass of each product produced. g of manganese(II) chloride g of water g of chlorine g of potassium chloride
Explanation:
Equation of the reaction:
2KMnO4(aq) + 16HCl(aq) --> 2MnCl2(aq) + 2KCl(aq) + 5Cl2(g) + 8H2O(aq)
To calculate the limiting reagent, we need to calculate the number of moles of the reactants :
KMnO4:
Molar mass = (39 + 55 + (16*4))
= 158 g/mol
Number of moles = mass/molar mass
= 229.19/158
= 1.4506 mol
HCl:
Molar mass = 1 + 35.5
= 36.5 g/mol
Number of moles = 526.64/36.5
= 14.428 mol
By stoichiometry, 2 moles of KMnO4 reacted with 16 moles of HCl
The limiting reagent :
14.428 moles of HCl * 2 moles of KMnO4/16moles of HCl
= 1.8035 moles of KMnO4 is required to react with 14.428 moles of HCl
But there's 1.4506 moles of KMnO4. Therefore, KMnO4 is the limiting reagent.
Mass of the products:
KCl:
2 moles of KMnO4 will produce 2 moles of KCl
Moles of KCl = 1 * 1.4506 mol
= 1.4506 mol
Molar mass = 39 + 35.5 = 74.5 g/mol
Mass of KCl = 74.5 * 1.4506
= 108.07 g
MnCl2:
2 moles of KMnO4 will produce 2 moles of MnCl2
Number of moles of MnCl2 = 1 * 1.4506
= 1.4506 mol
Molar mass = 55 + (35.5*2)
= 126 g/mol
Mass of MnCl2= 1.4506 * 126
= 182.78 g
Cl2:
2 moles of KMnO4 will produce 5 moles of Cl2
Number of moles of Cl2 = 5/2 * 1.4506
= 3.6265 mol
Molar mass of Cl2 = 35.5*2
= 71 g/mol
Mass of Cl2 = 71* 3.6265
= 257.4815 g
H2O:
2 moles of KMnO4 will produce 8 moles of H2O
Number of moles of H2O = 8/2 * 1.4506
= 5.80 mol
Molar mass of H2O =( 1*2) + 16
= 18 g/mol
Mass of H2O = 18*5.80
= 104.44 g
Answer:
(1) mass of KCl =108.025g
(2) mass of MnCl2 =182.7 g
(3) mass of Cl2 =257.37 g
(4) mass of H20(water) =104.4g
Explanation:
we start with determining the limiting factor
molar mass of KMnO4 = 158g/mol
molar mass of HCl = 36.5g/mol
hence,
number of moles of KMnO4= mass /molar mass
number of moles of KMnO4= 229.19/158 = 1.45moles
number of moles of HCl = 526.64/36.5 = 14.43 moles
we chose the lowest number of moles from the reactants as the limiting factor
hence the limiting factor is KMnO4.
calculating the mass of the products:
for KCl
2moles of KMnO4 reacts to give 2moles of KCl
there for KCl contains 1.45moles
(1) mass of KCl = number of moles of KCl x molar mass of KCl = 1.45 x 74.5 = 108.025g
(2) from mole ratio 2 mole of KMnO4 gave 2 moles of MnCl2
therefore 1.45moles KMnO4 will give 1.45 MnCl2
mass of MnCl2 = 1.45moles x molar mass MnCl2 = 1.45 x 126 = 182.7 g
(3) from mole ratio
2moles KMnO4 gave 5moles Cl2
1.45 moles will give (5 x1.45)/2 moles of Cl2
mass of Cl2 = number of moles of Cl2 x molar mass Cl2 = 3.625 x 71 = 257.37 g
(4)from mole ratio
2moles KMnO4 gave 8moles water(H2O)
1.45 moles will give (1.45x 8)/2 moles of H20
mass of H2O produced = number of moles of H20 x the molar mass of H20 = 5.8 x 18 = 104.4g
One of the hydrates of MnSO4 is manganese(II) sulfate tetrahydrate . A 71.6 gram sample of MnSO4 4 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?
Answer:
48.32 g of anhydrous MnSO4.
Explanation:
Equation of dehydration reaction:
MnSO4 •4H2O --> MnSO4 + 4H2O
Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)
= 223 g/mol
Mass of MnSO4 • 4H2O = 71.6 g
Number of moles = mass/molar mass
= 71.6/223
= 0.32 mol.
By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4
Number of moles of MnSO4 = 0.32 mol.
Molar mass = 55 + 32 + (4*16)
= 151 g/mol.
Mass = 151 * 0.32
= 48.32 g of anhydrous MnSO4.
A saline solution contains 0.770 gg of NaClNaCl (molar mass = 58.55 g/molg/mol) in 133 mLmL of solution. Calculate the concentration of NaClNaCl in this solution, in units of molarity.
Answer:
0.104 M
Explanation:
A saline solution contains 0.770 g of NaCl (molar mass = 58.55 g/mol) in 133 mL.
The molar mass of the solute (NaCl) is 58.55 g/mol. The moles corresponding to 0.770 g are:
0.770 g × (1 mol/55.85 g) = 0.0138 mol
The volume of solution is 133 mL. In liters,
133 mL × (1 L/1000 mL) = 0.133 L
The molarity of NaCl is:
M = moles of solute / liters of solution
M = 0.0138 mol / 0.133 L
M = 0.104 M
Write the full ground-state electron configuration for each:
(a) Cl (b) Si (c) Sr
Answer:
(a) Cl
[tex]1s^22s^22p^63s^23p^5[/tex]
(b) Si
[tex]1s^22s^22p^63s^23p^2[/tex]
(c) Sr
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]
Explanation:
(a) Cl
Atomic number = 17
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^5[/tex]
(b) Si
Atomic number = 14
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^2[/tex]
(c) Sr
Atomic number- 38
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]
What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?
Answer: E = 2.455 x 10^5 N/C
Explanation:
q1 = 1.2x10^-7C
q2 = 6.2x10^-8C
Electric field, E = kQ/r²
where k = 9.0x10^9
since the location is (27 - 5)cm from q1
hence electric field, E1 = k*q1/r²
E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C
for q2:
E1 = k*q2/r²
E2 at 5cm
E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C
Hence, the total electric field at 5cm position is
E = E1 + E2
E = 22314.05 + 223200 = 245514.05 N/C
E = 2.455 x 10^5 N/C
Final answer:
The electric field at a point 5.0 cm from a negative charge is calculated using the formula E = kQ/r² with the direction toward the charge.
Explanation:
q1 = 1.2x10⁻⁷ C
q2 = 6.2x10⁻⁸C
Electric field, E = kQ/r²
where k = 9.0x10⁹
since the location is (27 - 5)cm from q1
hence electric field, E1 = k×q1/r²
E1= (9x10⁹ x 1.2x10⁻⁷)/(0.22)² = 22314.05 N/C
for q2:
E1 = k×q2/r²
E2 at 5cm
E2 = (9x10⁹ x 6.2x10⁻⁸)/(0.05)² = 223200 N/C
Hence, the total electric field at 5cm position is
E = E1 + E2
E = 22314.05 + 223200 = 245514.05 N/C
E = 2.455 x 10⁵ N/C
Classify each amino acid by the chemical properties of its side chain (R group) at pH 7·Select the amino acid that fits best in each category. Each amino acid will be used only once. 1. This amino acid has a positively charged R group: Select answer 2. This amino acid has a negatively charged R group! Select answer 3. This amino acid has a neutral polar R group: Sect answer 4 This amino acid has a nonpolar aliphatic R tryptophan aspartate valine arginine Select answer 5. This amino acid has an aromatic R group:
Answer:
. 1. This amino acid has a positively charged R group: ARGININE
2. This amino acid has a negatively charged R group: ASPARTATE
3. This amino acid has a neutral polar R group: NONE
4. This amino acid has a nonpolar aliphatic R: VALINE
5. This amino acid has an aromatic R group: TRYTOPHAN
Explanation:
1) Arginine contains an extra amino group bearing a positive charge, in its chain which imparts basic properties to it
2) Aspartate contains an extra carboxyl group with a dissociable protron. Once the Protron is dissociated, it carries an extra negative charge in its side chain (R)
3) NONE of the amino acids given belong to this group because amino acids with neutral polar R groups contain functional groups that form hydrogen bonds with water. But, this is not the case with tryptophan aspartate valine or arginine
4) Valine has a R group that is hydrocarbon in nature and thus hydrophobic.
5) Trytophan has a benzene ring in its side chain
Amino acids are classified by the properties of their R groups at pH 7 to be arginine (positively charged), aspartate (negatively charged), valine (neutral polar and nonpolar aliphatic) and tryptophan (aromatic).
Explanation:The chemical properties of an amino acid's side chain, or R group can be identified with their characteristics at pH 7. Here are each of the amino acids classified by the properties of their R groups at pH 7:
The amino acid with a positively charged R group is arginine. The amino acid with a negatively charged R group is aspartate. Valine has a neutral polar R group. The amino acid with a nonpolar aliphatic R group is valine . The amino acid with an aromatic R group is tryptophan.Learn more about Amino Acids here:https://brainly.com/question/31872499
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Dehydrohalogenation of 1-chloro-1-methylcyclopropane affords two alkenes (A and B) as products.
Explain why A is the major product despite the fact that it contains the less substituted double bond.
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A