Answer:
ok, here is your answer
Explanation:
a. For the given thermal decomposition reaction:
2NaHCO3(s) ≪−⋯Na2CO3(s) + CO2(g) + H2O(g)
The equilibrium constant expression Kp can be defined as:
Kp = (PNa2CO3 x PCO2 x PH2O)² / (PNaHCO3)²
Where,
PNa2CO3, PCO2, PH2O, and PNaHCO3 are the partial pressures of Na2CO3, CO2, H2O, and NaHCO3, respectively.
b. For the given thermal decomposition reaction:
2CaSO4(s) <−⋯2CaO(s) + 2SO2(g) + O2(g)
The equilibrium constant expression Kp can be defined as:
Kp = (PCaO² x PSO2² x PO2) / (PCaSO4)²
Where,
PCaO, PSO2, PO2, and PCaSO4 are the partial pressures of CaO, SO2, O2, and CaSO4, respectively.
Therefore, these are the equilibrium constant expressions for the given thermal decomposition reactions.
mark me as brainliestWhat volume (ml) of potassium hydroxide will it take to reach the equivalence point in a 16.3 ml aliquot of triprotic acid?
To reach the equivalence point in a 16.3 ml aliquot of triprotic acid 48.9 ml of 0.1 M potassium hydroxide is needed.
An aliquot is a portion of a sample that is taken for the analysis or testing . It is also defined as the sub measured volume of the original sample.
To calculate the volume of potassium hydroxide needed to reach the equivalence point in a 16.3 ml aliquot of triprotic acid, we need to know the concentration of the potassium hydroxide solution and the molarity of the triprotic acid.
Assuming that the triprotic acid is fully ionized and that the potassium hydroxide is a strong base, we can use the following balanced chemical equation:
[tex]\rm 3 HX + KOH \rightarrow KX + 3 H_2O[/tex]
where HX is the triprotic acid and KX is the potassium salt of the acid.
At the equivalence point, the moles of acid and base are equal, so we can use the following equation to calculate the volume of potassium hydroxide needed:
moles of acid = moles of base
Molarity of acid × volume of acid = Molarity of base × volume of base
Since, the triprotic acid has three acidic protons, its molarity is three times the concentration of the 16.3 ml aliquot.
Let's assume that the concentration of the aliquot is 0.1 M, then the molarity of the triprotic acid is 0.3 M.
Let's also assume that the concentration of the potassium hydroxide solution is 0.1 M.
Using the equation above, we can solve for the volume of potassium hydroxide needed:
0.3 M × 16.3 ml = 0.1 M × volume of potassium hydroxide
volume of potassium hydroxide = 48.9 ml
Therefore, it would take 48.9 ml of 0.1 M potassium hydroxide to reach the equivalence point in a 16.3 ml aliquot of 0.3 M triprotic acid.
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What is the name of the ionic compound of lihso4
Answer:
Lithium hydrogen sulfate
The Ksp of magnesium hydroxide, Mg(OH)2, is
5.61×10−12 Calculate the molar solubility, s, of this compound.
The molar solubility (s) of magnesium hydroxide Mg(OH)₂, is approximately 1.19 × 10⁻4 mol/L.
The balanced equation for the dissociation of Mg(OH)2 is:
Mg(OH)₂ ↔ Mg₂+ + 2OH⁻
Hence the solubility product constant expression (Ksp) for magnesium hydoxide can be written as:
Ksp = [Mg₂+][OH⁻]₂
the concentration of Mg²⁺ and OH⁻ ions in terms of the molar solubility (s) as follows:
[Mg²⁺] = s
[OH⁻] = 2s
Now, substituting these values into the Ksp expression, we get:
Ksp = (s)(2s)²
Ksp = 4s³
Therefore, 4s³ = 5.61×10⁻¹²
s³ = (5.61 × 10⁻¹²) / 4
s³ = 1.4025 × 10⁻¹²
Taking the cube root we get:
s = (1.4025 × 10⁻¹²)¹/³
s ≈ 1.19 × 10⁻⁴ mol/L
Therefore, the molar solubility (s) of magnesium hydroxide (Mg(OH)₂) is approximately 1.19 × 10⁻⁴ mol/L.
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At -57 oC the equilibrium constant for the reaction:
2 HI(g) H2(g) + I2(g)
is KP = 7.21e-12. If the initial pressure of HI is 0.00506 atm, what are the equilibrium partial pressures of HI, H2, and I2?
p(HI) = .
p(H2) = .
p(I2) = .
Therefore, the answers are: p(HI) = 845.2 × 10⁻⁶ atmp(H2) = 422.6 × 10⁻⁶ atmp(I2) = 422.6 × 10⁻⁶ atm
Given the chemical equation:2 HI(g) ⇌ H2(g) + I2(g)At -57 oC
the equilibrium constant for the above reaction:
KP = 7.21 × 10⁻¹²
Also, the initial pressure of HI is 0.00506 atm.
The balanced chemical equation can be represented as:
H2(g) + I2(g) ⇌ 2 HI(g)
Here, 2 moles of HI will give 1 mole each of H2 and I2 at equilibrium.
If "x" is the equilibrium concentration of HI,
then the equilibrium concentrations of H2 and I2 will be "x/2" each.
Therefore, the expression for equilibrium constant (KP) can be written as:
KP = (p(HI))^2/ (p(H2) x p(I2))
Where,
p(HI) = Partial pressure of HIp(H2) = Partial pressure of H2p(I2) = Partial pressure of I2
Given:
KP = 7.21 × 10⁻¹²
p(HI) = 0.00506 atm
And, p(H2) = p(I2) = x/2
Let us assume the value of "x" to be the equilibrium concentration of HI.
Then,
KP = (p(HI))²/ (p(H2) x p(I2))7.21 × 10⁻¹²
KP = (0.00506 atm)²/ (x/2)²
or
x² = (2 × 0.00506²) / 7.21 × 10⁻¹²
or
x² = 0.7149 × 10⁶
or
x = 845.2
So, the equilibrium partial pressures of HI, H2, and I2 are:
p(HI) = 845.2 x 10⁻⁶
atmp(H2) = 422.6 x 10⁻⁶
atmp(I2) = 422.6 x 10⁻⁶atm
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Assign the stereochemistry of the carbon (r or s) at the 17 position for 17-a-estradiol and 17-b-estradiol. what is the stereochemical relationship between the two isomers?
Stereochemistry of the carbon at the 17 position for 17-a-estradiol and 17-b-estradiol:
17-a-estradiol: The hydroxyl group at the 17 position is on the R side of the molecule.17-b-estradiol: The hydroxyl group at the 17 position is on the S side of the molecule.What is the stereochemical relationship?Stereochemical relationship between the two isomers:
17-a-estradiol and 17-b-estradiol are enantiomers. This means that they are mirror images of each other, and they cannot be superimposed.
A diagram of the two isomers, with the hydroxyl group at the 17 position highlighted is attached.
The 17-a-estradiol molecule is shown on the left, and the 17-b-estradiol molecule is shown on the right. The hydroxyl group at the 17 position is on the R side of the 17-a-estradiol molecule, and it is on the S side of the 17-b-estradiol molecule.
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Water sample contains
CO2=9 MG/L
Ca2+=52 MG/L
MG2+=16mg/L
Na+= 25 mg/L
K+=5.3 ,g/L
HCO-3=110 MG/L
SO24=94mg/L
CL-=49mg/L
The hardness of the water, expressed in mg/L as CaCO3 is 196.5. The total hardness of the water is 3.93.
a) For Step 1 of the Lime-Soda softening process, what is the concentration of lime added (meq/L) for neutralizing carbonic acid?
b) What is the concentration of lime added (meq/L for calcium carbonate-hardness(meq/L)?
c) What is the concentration of soda added for calcium non-carbonate hardness?
Every equivalent of non-carbonate hardness requires one equivalent of soda. The concentration of soda added for calcium non-carbonate hardness is 2.97 meq/L.
a) For Step 1 of the Lime-Soda softening process, the concentration of lime added (meq/L) for neutralizing carbonic acid is calculated as follows:
Carbonic acid is neutralized by the addition of lime. Carbonic acid can be expressed as H2CO3. Lime reacts with carbonic acid as follows:
Ca(OH)2 + H2CO3 → CaCO3 + 2H2OCaCO3 is precipitated as a result of the reaction.
The amount of Ca(OH)2 required is determined by the amount of CO2 present in the water sample, as well as the number of equivalents of CaCO3 that may be formed.
CO2 content = 9 mg/LAs CO2 combines with Ca(OH)2, it forms
CaCO3.Ca(OH)2 + CO2 → CaCO3 + H2O
Molecular weight of CO2 = 44 gm/molNumber of equivalents in 44 gm of CO2 = 1Number of equivalents in 9 mg of CO2 = 9/44 x 1000 mg/moleq. weight of CaCO3 = 100 gm/molNumber of equivalents in 100 gm of CaCO3 = 2∴
The number of equivalents of CaCO3 in 9 mg of CO2 is (9/44 x 1000)/2
The number of equivalents of Ca(OH)2 required = Number of equivalents of CaCO3 formed
= 9/44 x 1000/2 = 102.27 meq/L
The concentration of lime added (meq/L) for neutralizing carbonic acid is 102.27
.b) For calcium carbonate-hardness(meq/L), the concentration of lime added (meq/L) is determined as follows:The equation for the reaction between lime and calcium bicarbonate is given by:
Ca(OH)2 + Ca(HCO3)2 → 2CaCO3 + 2H2OThe reaction produces two equivalents of CaCO3 per equivalent of Ca(HCO3)2.
Lime's equivalent weight is 74 gm/equivalent, and calcium's equivalent weight is 50 gm/equivalent.1 mg/L of calcium carbonate hardness as CaCO3 equals 50 mg/L as Ca2+.
As a result, the concentration of lime required is given by:
Ca(HCO3)2 content = Ca2+ (from sample) x 50 mg/LTotal hardness
= 3.93 meq/L / 50 mg/L
= 0.0786 equivalents/LCa(HCO3)2 = (total hardness - Ca2+)
= 0.0786 - 0.052 = 0.0266 meq/LConcentration of lime required to remove Ca(HCO3)2
= Ca(HCO3)2/2
= 0.0266/2
= 0.0133 meq/L
The concentration of lime added (meq/L) for calcium carbonate-hardness is 0.0133.
c) For calcium non-carbonate hardness, the concentration of soda added is calculated as follows:Soda ash reacts with MgSO4 and CaSO4 to form MgCO3 and CaCO3, respectively.
Na2CO3 + MgSO4 → MgCO3 + Na2SO4Na2CO3 + CaSO4 → CaCO3 + Na2SO4
The number of equivalents of MgSO4 and CaSO4 in the water sample are:
MgSO4
= (mg/L of Mg2+) x (1 eq/24.3 gm)CaSO4
= (mg/L of Ca2+) x (1 eq/40.1 gm)MgSO4
= 16/24.3 = 0.66 meq/LCaSO4
= 52/40.1
= 1.30 meq/L
Total equivalents of hardness
= 3.93 meq/L - (0.66 + 1.30) meq/L
= 2.97 meq/L
Soda is needed to react with the remaining hardness, which is all due to non-carbonate hardness.
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Answer:To calculate the concentration of lime added for neutralizing carbonic acid (H2CO3) in the Lime-Soda softening process,
Explanation:
Convert the given concentration of CO2 (carbon dioxide) from mg/L to meq/L.
The conversion factor for CO2 is 1 meq/L = 22.4 mg/L.
a) Concentration of lime added for neutralizing carbonic acid:
CO2 = 9 mg/L
CO2 (meq/L) = CO2 (mg/L) / 22.4
CO2 (meq/L) = 9 / 22.4
So, the concentration of lime added for neutralizing carbonic acid is approximately 0.402 meq/L.
To calculate the concentration of lime added for calcium carbonate hardness (CaCO3), we need to convert the given hardness values from mg/L to meq/L. The conversion factor for Ca2+ and Mg2+ is
1 meq/L = atomic weight (g) / valence.
b) Concentration of lime added for calcium carbonate hardness:
Ca2+ = 52 mg/L
Ca2+ (meq/L) = Ca2+ (mg/L) / (atomic weight of Ca2+ / valence of Ca2+)
Ca2+ (meq/L) = 52 / (40.08 / 2)
MG2+ = 16 mg/L
MG2+ (meq/L) = MG2+ (mg/L) / (atomic weight of Mg2+ / valence of Mg2+)
MG2+ (meq/L) = 16 / (24.31 / 2)
Total hardness (meq/L) = Ca2+ (meq/L) + MG2+ (meq/L)
Total hardness (meq/L) = Ca2+ (meq/L) + MG2+ (meq/L) + CO2 (meq/L)
Substituting the given values:
3.93 = Ca2+ (meq/L) + MG2+ (meq/L) + 0.402
We know that the hardness of the water expressed in mg/L as CaCO3 is 196.5. The molecular weight of CaCO3 is 100.09 g/mol. Therefore:
196.5 mg/L = Total hardness (meq/L) * (100.09 / 2)
Solving these equations will give us the concentration of lime added for calcium carbonate hardness.
C) For calcium non-carbonate hardness, the concentration of soda added is calculated as follows:Soda ash reacts with MgSO4 and CaSO4 to form MgCO3 and CaCO3, respectively.
Na2CO3 + MgSO4 → MgCO3 + Na2SO4Na2CO3 + CaSO4 → CaCO3 + Na2SO4
The number of equivalents of MgSO4 and CaSO4 in the water sample are:
MgSO4
= (mg/L of Mg2+) x (1 eq/24.3 gm)CaSO4
= (mg/L of Ca2+) x (1 eq/40.1 gm)MgSO4
= 16/24.3 = 0.66 meq/LCaSO4
= 52/40.1
= 1.30 meq/L
Total equivalents of hardness
= 3.93 meq/L - (0.66 + 1.30) meq/L
= 2.97 meq/L
Soda is needed to react with the remaining hardness, which is all due to non-carbonate hardness.
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a solution contains zn2 , hg2 , ag , nh4 , and ba2 . identify the precipitate after the addition of 6 m hcl, then h2s and 0.2 m hcl.
The precipitate formed after the addition of 6 M HCl, followed by H₂S and 0.2 M HCl, in the solution containing Zn²⁺, Hg₂²⁺, Ag⁺, NH₄⁺, and Ba²⁺ is AgCl.
When 6 M HCl is added to the solution, it will react with Ba²⁺ ions forming BaCl₂, which remains in solution. The other ions (Zn²⁺, Hg₂²⁺, Ag⁺, and NH₄⁺) do not form precipitates with HCl at this stage.
Next, when H₂S is added, it reacts with the remaining Zn²⁺, Hg₂²⁺, and Ag⁺ ions in the solution. Zn²⁺ and Hg₂²⁺ ions form insoluble sulfides, ZnS, and HgS, respectively, which are precipitates. However, Ag⁺ ions react with H₂S to form Ag₂S, another insoluble sulfide, resulting in the precipitation of Ag₂S.
Finally, when 0.2 M HCl is added, it does not affect the precipitates of ZnS, HgS, or Ag₂S. Therefore, the only precipitate remaining in the solution is AgCl, which is formed by the reaction between Ag⁺ ions and Cl⁻ ions from HCl.
In conclusion, the precipitate formed after the addition of 6 M HCl, followed by H₂S and 0.2 M HCl, in the given solution is AgCl.
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true or false: in polar covalent bonds the shared electrons spend more time around the larger atom making that are slightly negative
In a polar covalent bonds the shared electrons spend more time around the one atom than the other. Hence the statement is true.
In a polar covalent bond, the electrons are unequally distributed among the atoms and spend more time close to one atom than the other. Different regions of the molecule gain slightly positive (+) and slightly negative (-) charges as a result of the unequal distribution of electrons among the atoms of various components.
The electrons that the atoms share in a polar covalent connection spend more time near one nucleus than the other. The shared electrons of the atoms form a polar covalent link when they spend, on average, more time closer to one nucleus than the other nucleus.
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Ammonia is a stronger ligand than water. which is correct when concentrated aqueous ammonia solution is added to dilute aqueous copper(ii) sulfate solution?
The correct statement is that the concentrated aqueous ammonia solution will cause the formation of the copper(II) tetraammine complex by displacing water molecules.
When concentrated aqueous ammonia solution is added to dilute aqueous copper(II) sulfate solution, the following reaction occurs:
CuSO₄(aq) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)n]²⁺(aq) + SO₄²⁻(aq)
In this reaction, ammonia (NH₃) acts as a ligand and forms a complex with the copper ion (Cu²⁺). The resulting complex is called a copper(II) tetraammine complex, [Cu(NH₃)₄(H₂O)n]²⁺, where n represents the number of water molecules attached to the complex.
Ammonia is a stronger ligand than water, meaning it has a higher affinity for forming complexes with metal ions. When concentrated aqueous ammonia is added to the copper(II) sulfate solution, ammonia displaces water molecules from the coordination sphere of the copper ion and forms the copper(II) tetraammine complex.
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For the following reaction, 6.57 grams of barium hydroxide are mixed with excess sulfuric acid. The reaction yields 8.49 grams of barium sulfate. barium hydroxide (aq) + sulfuric acid (aq) ⟶ barium sulfate (s) + water (I) What is the theoretical yield of barium sulfate? grams What is the percent yield of barium sulfate? A 0.5956 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 1.2525g. What is the mass percentage of chlorine in the original compound?
The mass percentage of chlorine in the original compound is 85.61%.
For the following reaction, 6.57 grams of barium hydroxide are mixed with excess sulfuric acid.
The reaction yields 8.49 grams of barium sulfate.
barium hydroxide (aq) + sulfuric acid (aq) ⟶ barium sulfate (s) + water (l)
The given balanced chemical reaction is:
Ba(OH)2(aq) + H2SO4(aq) ⟶ BaSO4(s) + 2H2O(l)
The mole ratio between barium hydroxide and barium sulfate is 1:1.
Therefore, the amount of barium sulfate formed in the reaction would be equal to the amount of barium hydroxide taken, which can be calculated as shown below:
Mass of Ba(OH)2 = 6.57 g
Molar mass of Ba(OH)2 = 171.34 g/mol
Number of moles of Ba(OH)2 = mass of Ba(OH)2 / molar mass of Ba(OH)2
= 6.57 / 171.34= 0.0382 mol
According to the balanced chemical equation, 1 mole of Ba(OH)2 produces 1 mole of BaSO4.
Therefore, the number of moles of BaSO4 produced in the reaction would be 0.0382 moles.
The molar mass of BaSO4 is 233.39 g/mol, so the theoretical yield of BaSO4 can be calculated as shown below:
Theoretical yield of BaSO4 = number of moles of BaSO4 × molar mass of BaSO4
= 0.0382 × 233.39= 8.92 g
Therefore, the theoretical yield of BaSO4 is 8.92 grams.
The actual yield of BaSO4 is given as 8.49 grams.
So, the percent yield can be calculated using the formula:
% yield = (actual yield / theoretical yield) × 100
Substituting the values, we get:
% yield = (8.49 / 8.92) × 100= 95.07%
Therefore, the percent yield of barium sulfate is 95.07%.
A 0.5956 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate.
The mass of the resulting AgCl is found to be 1.2525g.
The reaction between chloride ion and silver nitrate can be represented as follows:
AgNO3(aq) + NaCl(aq) ⟶ AgCl(s) + NaNO3(aq)
The mole ratio between NaCl and AgCl is 1:1.
Therefore, the number of moles of AgCl formed in the reaction would be equal to the number of moles of NaCl taken, which can be calculated as shown below:
Number of moles of AgCl = mass of AgCl / molar mass of AgCl= 1.2525 / 143.32= 0.00874 mol
According to the balanced chemical equation, 1 mole of NaCl produces 1 mole of AgCl.
Therefore, the number of moles of NaCl taken would also be 0.00874 moles.
The molar mass of NaCl is 58.44 g/mol, so the mass of NaCl taken can be calculated as shown below:
Mass of NaCl = number of moles of NaCl × molar mass of NaCl= 0.00874 × 58.44= 0.510 g
Therefore, the mass percentage of chlorine in the original compound can be calculated as shown below:
Mass percentage of Cl in the original compound = (mass of Cl / mass of the sample) × 100
Mass of Cl = mass of NaCl= 0.510 g
Mass of the sample = 0.5956 g
Substituting the values, we get:
Mass percentage of Cl in the original compound = (0.510 / 0.5956) × 100= 85.61%
Therefore, the mass percentage of chlorine in the original compound is 85.61%.
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What masses of dimethylamine and dimethylammonium chloride do you need to prepare 4.00 l of ph = 12.00 buffer if the total concentration of the two components is 0.500 m?
45.1 g of DMA and 119.0 g of DMAC are required to make 4.00 L of pH 12.00 buffer with equal concentrations of DMA and DMAC at a total concentration of 0.500 M.
Buffers are used to smooth the flow of data in a computer system and prevent bottlenecks. They are used in many different contexts, such as streaming video, network communications, and file input/output operations. Buffers can also be used to improve performance by minimising the number of read and write operations that are required.
pH = pKa + log ([base]/[acid])
[base] >> [acid]
pH ≈ pKa + log [base]
[base] =[tex]10^{(pH - pKa)}[/tex]
[base] =[tex]10^{(12.00 - 10.7) }[/tex]
= 2.24 M
[acid] = 0.500 M - [base]
[acid] = 0.500 M - 2.24 M
= -1.74 M
pH = pKa + log [base]/[acid]
= 10.7 + log (0.250/0.250)
= 10.7
DMA: 45.1 g/mol
DMAC: 119.0 g/mol
moles DMA = (0.250 M) (4.00 L)
= 1.00 mol
moles DMAC = (0.250 M) (4.00 L)
= 1.00 mol
mass DMA = (1.00 mol) (45.1 g/mol)
= 45.1 g
mass DMAC = (1.00 mol) (119.0 g/mol)
= 119.0 g
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One way jewelers can distinguish the difference between a diamond and a cubic zirconia is by using the density test Data: the density of a diamond is 3.51 g/cm. Density of cubic zirconia 5,5 g/cm Each stone is put into a thick solution that has a density of 4,0 g/cm'. How do you tell which is the diamond and why? The stone that sinks to the bottom is the diamond, because the density of the diamond is less than the density of the cuble zirconia. The stone that sinks to the bottom is the diamond, because the density of the diamond is greater than the density of the thick liquid The stone that sinks to the bottom is the cubic zirconia, because the density of the cubic zirconia is less than the density of the diamond The stone that sinks to be bottom is the cubic zirconia, because the density of the cubic zirconia is greater than the density of the thick liquid
By observing which stone sinks to the bottom in the density test, one can identify the diamond as it has a higher density compared to the cubic zirconia. A density test is a physical test or measurement used to determine the density of a substance.
In the given scenario, the thick solution has a density of 4.0 g/cm³. Since the density of cubic zirconia is 5.5 g/cm³, it is denser than the liquid and will sink to the bottom when placed in the solution. On the other hand, the density of a diamond is 3.51 g/cm³, which is lower than the density of the thick liquid. Therefore, the diamond will have a buoyancy greater than the liquid and will float or remain suspended in the solution rather than sinking to the bottom.
By observing which stone sinks to the bottom, jewelers can distinguish between a diamond and cubic zirconia based on their different densities.
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Rank the following substances in order from most soluble in water to least soluble in water: methane, CH 4
; 1-hexanol, C 6
H 13
OH; magnesium chloride, MgCl 2
; and ethane, C 2
H 6
. Rank from most to least soluble in water. To rank items as equivalent, overlap them.
In order from most to least soluble in water: Magnesium chloride (MgCl2) > 1-hexanol (C6H13OH) > Methane (CH4) > Ethane (C2H6).
The given substances: methane (CH4), 1-hexanol (C6H13OH), magnesium chloride (MgCl2) and ethane (C2H6), can be ranked from most to least soluble in water as follows:1. Magnesium chloride (MgCl2): MgCl2 dissociates into Mg2+ and Cl- ions in water. Being ionic, MgCl2 is highly soluble in water.2. 1-hexanol (C6H13OH): 1-hexanol is a molecule with a hydroxyl group that can participate in hydrogen bonding with water molecules.
It is, therefore, moderately soluble in water.3. Methane (CH4): Methane is non-polar and does not have any charge on it. Water, being polar, does not interact with it to a significant extent. Methane is therefore almost insoluble in water.4. Ethane (C2H6): Ethane is non-polar and like methane does not have any charge on it. It is therefore not soluble in water or any other polar solvent.
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A chemist designs a galvanic cell that uses these two half-reactions: 2h2o 2e h2 2)h
The galvanic cell design involves the reduction of hydrogen ions to hydrogen gas at the cathode and the oxidation of hydrogen gas to hydrogen ions at the anode, resulting in a transfer of electrons and the overall reaction of 2H₂O + H₂ → 2H⁺ + 2OH⁻ + H₂.
The two half-reactions provided are:
1. 2H₂O → 2H⁺ + 2e⁻ (reduction half-reaction)
2. H₂ → 2H⁺ + 2e⁻ (oxidation half-reaction)
To design a galvanic cell, we need to combine these half-reactions in such a way that the reduction and oxidation reactions occur separately. This can be achieved by connecting the two half-cells with a salt bridge or a porous membrane.
In this case, the reduction half-reaction involves the reduction of water (H₂O) to hydrogen gas (H₂) by gaining two electrons (2e⁻). The oxidation half-reaction involves the oxidation of hydrogen gas (H₂) to hydronium ions (H⁺) by losing two electrons (2e⁻).
To construct the galvanic cell, we would typically represent the anode (oxidation) and cathode (reduction) compartments. The anode is where oxidation occurs, and the cathode is where reduction occurs. The half-cell notation for each half-reaction would look like this:
Anode (oxidation half-reaction): H₂ → 2H⁺ + 2e⁻
Cathode (reduction half-reaction): 2H₂O + 2e⁻ → 2H₂ + 2OH⁻
Overall, the balanced reaction of the galvanic cell would be:
2H₂O + H₂ → 2H⁺ + 2OH⁻ + H₂
This reaction represents the transfer of electrons from the anode to the cathode, with hydrogen ions (H⁺) being reduced to form hydrogen gas (H₂) at the cathode, and hydrogen gas (H₂) being oxidized to form hydrogen ions (H⁺) at the anode.
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6. (4 points) Consider 2,2.4-trimethvlhexane sighting along C3-C4: Draw the Newman projections of 2,2,4-trimethylhexane showing the most stable staggered conformation and the least stable eclipsed conformation.
The most stable staggered conformation of 2,2,4-trimethyl hexane minimizes steric hindrance, while the least stable eclipsed conformation maximizes steric hindrance.
To draw the Newman projections of 2,2,4-trimethyl hexane, we need to consider the relative positions of the atoms and groups along the[tex]C_{3}-C_{4}[/tex]bond. Here's how you can draw the most stable staggered conformation and the least stable eclipsed conformation:
Most Stable Staggered Conformation:
In the most stable staggered conformation, the methyl groups are positioned as far apart as possible, minimizing steric hindrance. Here's how you can draw it: (image)
In this conformation, the front carbon ([tex]C_{3}[/tex]) is represented by a dot, and the rear carbon ([tex]C_{4}[/tex]) is represented by a circle. The groups on the front carbon ([tex]C_{3}[/tex]) are shown in the axial position, while the groups on the rear carbon ([tex]C_{4}[/tex]) are shown in the equatorial position.
Least Stable Eclipsed Conformation:
In the least stable eclipsed conformation, the methyl groups are positioned directly in front of each other, leading to significant steric hindrance. Here's how you can draw it: (image)
In this conformation, the front carbon ([tex]C_{3}[/tex]) is represented by a dot, and the rear carbon ([tex]C_{4}[/tex]) is represented by a circle. The groups on both carbons are aligned, creating a higher energy conformation due to increased steric hindrance.
Remember that the most stable conformation is the one with the least steric hindrance, while the least stable conformation has the most steric hindrance.
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The experimental van't Hoff factor (i) for a 0.10 M CoCl2 solution was determined to be 2.506 at a constant temperature. What is the % dissociation for the CoCl2 in this solution at the experimental temperature? The ideal van't Hoff factor for CoCl2 would be 3.
The experimental van't Hoff factor is less than the ideal van't Hoff factor, the % dissociation is negative and is equal to -16.5%.
The van't Hoff factor (i) is used to predict the degree of dissociation of a solute in a solution and to determine the number of moles of solute dissolved in a solution. When compared to the ideal van't Hoff factor, the experimental van't Hoff factor indicates how many particles of solute have dissociated in a solution. The formula for the % dissociation of a solute in a solution is as follows:
% dissociation = (Experimental van't Hoff factor/ Ideal van't Hoff factor -1) x 100
For a 0.10 M CoCl2 solution, the ideal van't Hoff factor is 3.
However, the experimental van't Hoff factor is 2.506.
Therefore, the % dissociation of CoCl2 in the given solution can be calculated using the above formula as follows:
% dissociation = (2.506/3 - 1) x 100
= (0.835 - 1) x 100
= -0.165 x 100
= -16.5%
As the answer is negative, it implies that the experimental van't Hoff factor is less than the ideal van't Hoff factor, and as such, the degree of dissociation of CoCl2 in the given solution is less than the degree of dissociation that would have been expected at that temperature.
In summary, the % dissociation for the CoCl2 in a 0.10 M CoCl2 solution at a constant temperature can be determined by using the formula % dissociation
= (Experimental van't Hoff factor/ Ideal van't Hoff factor -1) x 100.
Since the experimental van't Hoff factor is less than the ideal van't Hoff factor, the % dissociation is negative and is equal to -16.5%.
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How can the law of conservation of mass apply to a burning log, if all that remains of it is ash?.
The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction.
In the case of a burning log, although the log appears to be reduced to ash, the mass is still conserved. When wood burns, it undergoes a chemical reaction called combustion, where it reacts with oxygen from the air to produce carbon dioxide, water vapor, and other byproducts. These byproducts include gases and smoke that are released into the atmosphere, as well as the remaining solid residue, which is ash.
Even though the log is reduced to ash, the total mass of the ash, gases, and smoke produced is equal to the initial mass of the log. The process of combustion converts the wood's carbon, hydrogen, and other elements into different compounds, but the total mass of all the reactants and products remains the same.
Therefore, the law of conservation of mass still applies, ensuring that the total mass before and after the burning of the log remains constant.
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the rate of decomposition of ph3 was studied at 861.00 °c. the rate constant was found to be 0.0575 s–1.
the concentration of P[tex]H_{3}[/tex] after 12.50 seconds is approximately 0.313 M.
To determine the concentration of P[tex]H_{3}[/tex] after 12.50 seconds, we can use the first-order rate equation:
[tex]\[ \text{Rate} = k \cdot [\text{PH_{3} }] \][/tex] P[tex]H_{3}[/tex]]
Given:
Initial concentration of P[tex]H_{3}[/tex] = 0.95 M
Rate constant (k) = 0.0745[tex]s^(-1)[/tex]
Time (t) = 12.50 s
Using the first-order rate equation, we can rearrange it to solve for the concentration of P[tex]H_{3}[/tex]at a specific time:
{ P[tex]H_{3}[/tex]} = { P[tex]H_{3}[/tex]} [tex]e^{-kt}[/tex]
Substituting the given values:
[tex]PH_{3} = 0.95 \, \text{M} \cdot e^{-(0.0745 \, \text{s}^{-1} \cdot 12.50 \, \text{s})} \][/tex]
Calculating this expression, we find:
{ P[tex]H_{3}[/tex]} = 0.313
Therefore, the concentration of P[tex]H_{3}[/tex] after 12.50 seconds is approximately 0.313 M.
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The rate of decomposition of PH3 was studied at 861.00 °C. The rate constant was found to be 0.0745 s–1. If the reaction is begun with an initial PH3 concentration of 0.95 M, what will be the concentration of PH3 after 12.50 s?
draw the structure of
1) 4 carbon branched carboxylic acid
2) 5 carbon di-alochol
The structure of the branched ether with chemical formula c4h10o is
[tex]CH_{3} -CH_{2} - CH_{2}-O-CH_{2} - CH_{2}-CH_{3}[/tex]
We have to draw the structure of branched ether. [tex]C_{4}H_{10}[/tex] is a branched ether which is also known as an alkoxy alkane or a glycol ether. It is an organic compound that is composed of four carbon atoms, ten hydrogen atoms, and one oxygen atom. Its molecular weight is 86.13 g/mol.
Its structure is linear, with a carbon backbone and an oxygen atom attached to two of the carbons in the chain. The oxygen atom is then connected to two methyl ([tex]CH_{3}[/tex]) groups, which are present on each side of the central carbon atom. The structure of the branched ether will look like this;
[tex]CH_{3} -CH_{2} - CH_{2}-O-CH_{2} - CH_{2}-CH_{3}[/tex].
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The complete question is "Draw the structure(s) of the branched ether(s) with the chemical formula c4h10o?"
indicate whether each of the following statements about the reactivity of elements is true or false.
Each of the statements about the reactivity of elements should be evaluated as follows;
The eight-element periodicity found in the Periodic Table is related to the number of electrons in the outermost energy level of the atoms that make up each element: true.Electrons in the first energy level of an atom are called valence electrons: false.For atoms important to life, if the first energy level is the outermost shell, it is stable with 2 electrons. If any other energy level is the outermost shell, it is stable with 8 electrons: true.Atoms tend to react in ways that give each atom a stable outer shell of electrons: true.Atoms with an outer shell that is almost empty are located on the right side of the Periodic Table while atoms with an outer shell that is full or almost full are located on the left side of the Periodic Table: false.Atoms in the same row of the Periodic Table tend to have the same number of valence electrons: falseAtoms with 7 valence electrons tend to be non-reactive: false.What is a chemical element?A chemical element is a pure substance that comprises atoms having the same atomic number (number of protons) in its nuclei and as such, it is the primary constituent of matter.
In Chemistry, valence electrons can be defined as the number of electrons that are present in the outermost shell of an atom of a specific chemical element.
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Complete Question:
Indicate whether each of the following statements about the reactivity of elements is true or false
1. The eight-element periodicity found in the Periodic Table is related to the number of electrons in the outermost energy level of the atoms that make up each element.
2. Electrons in the first energy level of an atom are called valence electrons.
3. For atoms important to life, if the first energy level is the outermost shell, it is stable with 2 electrons. If any other energy level is the outermost shell, it is stable with 8 electrons.
4. Atoms tend to react in ways that give each atom a stable outer shell of electrons.
5. Atoms with an outer shell that is almost empty are located on the right side of the Periodic Table while atoms with an outer shell that is full or almost full are located on the left side of the Periodic Table.
6. Atoms in the same row of the Periodic Table tend to have the same number of valence electrons.
7. Atoms with 7 valence electrons tend to be non-reactive.
Magnesium fluoride can be formed by burning magnesium in fluorine gas. With reference to its bonding, explain why magnesium fluoride has a very high melting point.
Answer:
Magnesium Flouride is a ionic compound and thus has a giant lattice structure. Its ions are held together in this lattice by strong electrostatic forces of attraction. A large amount of energy is needed to overcome the strong electrostatic forces of attraction between the Mg2+ ions and the F- ions to separate the ions. Hence Magnesium fluoride has a very high melting point.
biomass burning unlikely to account for missing source of carbonyl sulfide. geophysical research letters
Biomass burning is likely to account for missing source of carbonyl sulfide. Burning biomass has long been known to be an OCS source, especially in smouldering fires.
The most prevalent sulphur gas in the atmosphere, carbonyl sulphide (OCS or COS), is significant as a source of stratospheric aerosols as well as a tracer for gross primary production. The activities required to maintain the OCS budget balance are not entirely taken into account by the surface fluxes and atmospheric sinks of OCS that are currently estimated.
DMS is a consequence of phytoplanktonic activity, and COS and CS2 are largely formed in the ocean through photochemical reactions involving organosulfur compounds. The significant rise in ocean COS emissions throughout the deglaciation means that all three sulphur gases have seen increased emissions. COS is a precursor to background stratospheric sulphate aerosol, which has an adverse effect on the chemistry of the stratosphere and a net radiative effect.
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For the following reaction K=30.0 2 A(aq)−>B(aq)+C(aq) The initial concentration of A is 2.4M (with no B or C ). What is the equilibrium concentration of B in M ?
The equilibrium concentration of B is 0.589 M.
The reaction considered is 2 A(aq) → B(aq) + C(aq) with K = 30.0 and an initial concentration of A being 2.4M (no B or C).
The equilibrium concentration of B in M can be calculated as follows:Let the initial concentration of A be [A]₀.
The concentration of A at equilibrium would be [A]₀ - 2x M, where x is the concentration of A that reacted to give B and C.
The concentrations of B and C would be x M, as both are produced in a 1:1 molar ratio with respect to A. Using the equilibrium constant expression:K = [B][C]/[A]²
Substituting the above expressions for
[B], [C], and [A]:K
= (x)(x)/([A]₀ - 2x)²K
= x²/([A]₀ - 2x)²Since K
= 30.0 and [A]₀ = 2.4 M:30.0
= x²/(2.4 - 2x)²
Expanding the denominator and solving for x using the quadratic formula, we get:x = 0.589 MThe equilibrium concentration of B is thus:x = [B] = 0.589 M.
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Which of the following is an example of an extensive property of copper nugget?
An extensive property is a property that is dependent on the amount of matter present in the sample. Copper nugget is an example of an extensive property, meaning that it varies depending on the amount of copper nuggets present in the sample.
Copper is a chemical element that is found in Group 11 of the periodic table. It is a soft, malleable, and ductile metal that has a reddish-orange color. It is an excellent conductor of heat and electricity, and it is used in a variety of applications, including electrical wiring, plumbing, and roofing.What is an extensive property?Extensive properties are properties of matter that depend on the amount of matter present. Examples of extensive properties include mass, volume, and length. They are additive, which means that the value of the property increases as more matter is added to the sample. In the case of copper nuggets, the mass is an extensive property because it depends on the amount of copper nuggets present in the sample.
The mass of a copper nugget depends on its size, shape, and density. The greater the number of copper nuggets, the higher the mass of the sample. Copper nuggets can be measured using scales or balances to determine their mass, which is usually expressed in grams or kilograms.
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A diabetic is recommended to use 1 ml of insulin for every 10 g of carbohydrates consumed. the recommended daily intake of carboydrates is 300 g. a diabetic has eaten a slice of toast and has consumed 5% of their daily value of carbohydrates. how many ml of insulin should the diabetic use to maintain a proper blood sugar level after eating the piece of toast?
The diabetic should use 1.5 ml of insulin to maintain a proper blood sugar level after eating the piece of toast.
Insulin is a hormone that plays a crucial role in regulating blood sugar levels in the body, particularly in individuals with diabetes. In a person with diabetes, the body either doesn't produce enough insulin or is unable to effectively use the insulin it produces.
Insulin helps facilitate the absorption of glucose (sugar) from the bloodstream into cells, where it is used for energy or stored for later use. It acts as a "key" that unlocks cells, allowing glucose to enter.
The goal of insulin therapy in diabetes management is to keep blood sugar levels within a target range to prevent complications associated with high or low blood sugar. The dosage and timing of insulin administration are determined based on factors such as individual needs, carbohydrate intake, activity level, and blood sugar monitoring.
If the recommended daily intake of carbohydrates is 300 g, and the slice of toast represents 5% of that,
Carbohydrates in the slice = 5% of 300 g = (5/100) * 300 g = 15 g
It is recommended that the ratio of 1 ml of insulin for every 10 g of carbohydrates,
Insulin needed = (Carbohydrates in the slice) / 10 = 15 g / 10 = 1.5 ml
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Thio neutralization rebuilds the disulfide bonds by removing the _____________that was added by the permanent waving solution.
Thio neutralization rebuilds the disulfide bonds by removing the excess sulfur that was added by the permanent waving solution.
This process involves using a neutralizing solution, which helps to break down the thioglycolate and reduce its reactivity. The neutralizing solution contains an oxidizing agent, such as hydrogen peroxide, which helps to remove the excess sulfur and restore the hair's natural pH balance. By removing the excess sulfur, the disulfide bonds are able to reform, resulting in the desired curl or wave pattern.
This neutralization step is crucial in the permanent waving process to ensure the hair's integrity and prevent any further damage. In summary, thio neutralization helps rebuild the disulfide bonds by removing the excess sulfur, allowing the hair to maintain its new shape.
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Thio neutralization rebuilds the disulfide bonds by removing the excess sulfur that was added by the permanent waving solution.
This process involves using a neutralizing solution, which helps to break down the thioglycolate and reduce its reactivity. The neutralizing solution contains an oxidizing agent, such as hydrogen peroxide, which helps to remove the excess sulfur and restore the hair's natural pH balance. By removing the excess sulfur, the disulfide bonds are able to reform, resulting in the desired curl or wave pattern.
This neutralization step is crucial in the permanent waving process to ensure the hair's integrity and prevent any further damage. In summary, thio neutralization helps rebuild the disulfide bonds by removing the excess sulfur, allowing the hair to maintain its new shape.
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Thio neutralization rebuilds the disulfide bonds by removing the excess sulfur that was added by the permanent waving solution.
This process involves using a neutralizing solution, which helps to break down the thioglycolate and reduce its reactivity. The neutralizing solution contains an oxidizing agent, such as hydrogen peroxide, which helps to remove the excess sulfur and restore the hair's natural pH balance. By removing the excess sulfur, the disulfide bonds are able to reform, resulting in the desired curl or wave pattern.
This neutralization step is crucial in the permanent waving process to ensure the hair's integrity and prevent any further damage. In summary, thio neutralization helps rebuild the disulfide bonds by removing the excess sulfur, allowing the hair to maintain its new shape.
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Enter your answer in the provided box. Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction CaF 2
+H 2
SO 4
→CaSO 4
+2HF In one process, 5.85 kg of CaF 2
is treated with an excess of H 2
SO 4
and yields 2.35 kg of HF. Calculate the percent yield of HF.
The percent yield of HF can be calculated by dividing the actual yield (2.35 kg) by the theoretical yield and multiplying by 100. The theoretical yield is determined based on the stoichiometry of the reaction and the given amount of CaF2 (5.85 kg). The percent yield represents the efficiency of the reaction in producing the desired product.
The balanced chemical equation for the reaction is: CaF2 + H2SO4 -> CaSO4 + 2HF.
From the given data, we have 5.85 kg of CaF2 and the actual yield of HF is 2.35 kg.
To determine the theoretical yield of HF, we use the stoichiometry of the reaction. According to the balanced equation, 1 mole of CaF2 produces 2 moles of HF. We need to convert the mass of CaF2 to moles using its molar mass. CaF2 has a molar mass of approximately 78 g/mol.
Calculating the moles of CaF2: 5.85 kg * (1000 g/kg) / 78 g/mol = 75 moles.
Since the mole ratio between CaF2 and HF is 1:2, the theoretical yield of HF is 2 * 75 moles = 150 moles.
Converting moles of HF to mass: 150 moles * (20 g/mol) = 3000 g or 3 kg.
The percent yield is calculated as (actual yield / theoretical yield) * 100 = (2.35 kg / 3 kg) * 100 ≈ 78.3%.
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What is the conjugate base of ch3ch2sh?
1) ch3ch2s-
2) ch3ch2-
3) ch3ch2sh2
4) ch3ch2s2-
5) none of these
The conjugate base of CH₃CH₂SH (ethyl mercaptan) is CH₃CH₂S⁻. Therefore, option (1) is correct.
When CH₃CH₂SH loses a proton (H⁺), it forms CH₃CH₂S⁻, where the sulfur atom (S) has gained an extra electron to maintain charge neutrality. This is the conjugate base of CH₃CH₂SH.
Option 2) CH₃CH₂⁻ is not the correct conjugate base because it represents the removal of a proton from the ethyl group rather than the sulfur atom.
Option 3) CH₃CH₂SH₂ is the original compound and not the conjugate base.
Option 4) CH₃CH₂S₂⁻ represents the presence of two sulfur atoms and an additional negative charge, which does not reflect the conjugate base of CH₃CH₂SH.
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The rate of reaction of calcium carbonate (chalk) with hydrochloric acid likely depends on _____.
The rate of reaction of calcium carbonate (chalk) with hydrochloric acid likely depends on the concentration of hydrochloric acid, the particle size of the calcium carbonate, and the temperature.
Factors that can affect the rate of reaction between calcium carbonate and hydrochloric acid include the following: The concentration of hydrochloric acid: The greater the concentration of the acid, the more quickly the reaction will occur.
The particle size of the calcium carbonate: The smaller the particles, the more rapidly they will react because there is more surface area available for the acid to act upon.
Temperature: The higher the temperature, the quicker the reaction will proceed because particles will be moving more quickly and colliding with greater force and frequency.
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Identify the compound whose majoe IR stretching frequency signals are 1100,2900 cm −1
Acetone (CH 3
COCH 3
) Benzene Dimethyl ether (CH 3
OCH 3
) Acetaldehyde (CH 3
CHO) Identify the major IR stretching signals for Ethanol (CH 3
CH 2
OH) 3300 (broad) cm−1 3300 (sharp) cm−1 1600 cm−1 2300 cm−1 The IR spectrum of a sample contains absorptions at 1750 cm−1. To what class of organic compound does this sample most likely belong? Alkene Ketone Alkane Alcohol
The sample containing an absorption band at 1750 cm-1 belongs to the Ketone class of organic compounds.The compound whose major IR stretching frequency signals are 1100, 2900 cm-1 is Acetone (CH3COCH3).
In the given options, the compound whose major IR stretching frequency signals are 1100, 2900 cm-1 is Acetone (CH3COCH3).Acetone has the following structure:It is also known as 2-propanone. It is a colorless, volatile, and flammable liquid and has a sweetish odor. Acetone is an important solvent used in various industries and is also used as a laboratory reagent. The IR spectrum of Acetone has a strong carbonyl (C=O) absorption band at around 1700 cm-1, a strong band at 2900 cm-1 due to aliphatic C-H stretching, and a weak band at 1100 cm-1 due to C-C-C bending.To identify the major IR stretching signals for Ethanol (CH3CH2OH) are 3300 (broad) cm−1 and 1050 cm−1.
The absorption band at 3300 (broad) cm−1 is due to O-H stretching and the band at 1050 cm−1 is due to C-O stretching.To identify the class of organic compound containing an absorption band at 1750 cm-1, we have to look for the absorption band due to the carbonyl group (C=O). The carbonyl group is present in Aldehydes, Ketones, and Carboxylic acids. The absorption band due to the carbonyl group is generally in the range of 1650-1750 cm-1. Hence, the sample containing an absorption band at 1750 cm-1 belongs to the Ketone class of organic compounds.
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