Final answer:
The increase in mass of magnesium after burning is due to its combination with oxygen to form magnesium oxide, which is in accordance with the Law of Conservation of Mass. No mass is lost or created; the oxygen simply adds to the original mass of the magnesium metal.
Explanation:
The observation that a piece of magnesium gains mass when burnt can be explained without violating the Law of Conservation of Mass. This law states that mass is neither created nor destroyed in a chemical reaction. When magnesium burns in the presence of oxygen, a chemical change occurs, resulting in the formation of magnesium oxide, a white crumbly powder. The reaction can be depicted by the word equation magnesium + oxygen → magnesium oxide. The increase in mass from 51 grams to 53 grams is due to the addition of oxygen from the air. When magnesium (Mg) reacts with oxygen (O₂), the oxygen atoms combine with magnesium to form magnesium oxide (MgO), thus accounting for the increase in mass.
For example, if you heat 10.0 grams of calcium carbonate (CaCO₃) and produce 4.4 g of carbon dioxide (CO₂) and 5.6 g of calcium oxide (CaO), the total mass of the products equals the original mass of the reactants, which is in agreement with the Conservation of Mass. Similarly, in the magnesium burning experiment, the mass of the magnesium and the oxygen that combined during burning would equal the mass of the magnesium oxide produced, if all the reactants and products could be contained and measured.
The average atomic weight of copper, which has two naturally occurring isotopes, is 63.5. One of the isotopes has an atomic weight of 62.9 amu and constitutes 69.1% of the copper isotopes. The other isotope has an abundance of 30.9%. The atomic weight (amu) of the second isotope is ________ amu.
Answer: The atomic weight of the second isotope is 64.81 amu.
Explanation:
Average atomic mass of an element is defined as the sum of atomic masses of each isotope each multiplied by their natural fractional abundance
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex] .....(1)
We are given:
Let the mass of isotope 2 be 'x'
Mass of isotope 1 = 62.9 amu
Percentage abundance of isotope 1 = 69.1 %
Fractional abundance of isotope 1 = 0.691
Mass of isotope 2 = 'x'
Percentage abundance of isotope 2 = 30.9%
Fractional abundance of isotope 2 = 0.309
Average atomic mass of copper = 63.5 amu
Putting values in equation 1, we get:
[tex]\text{Average atomic mass of copper}=[(62.9\times 0.691)+(x\times 0.309)][/tex]
[tex]x=64.81amu[/tex]
Hence, the atomic weight of second isotope will be 64.81 amu.
By using the weighted average formula and the given details, we can determine the atomic weight of the second naturally occurring isotope of copper approximately to be 64.93 amu.
Explanation:The atomic weight of an isotope is calculated by adding the products of the abundance percentages and the atomic weights of each isotope. For copper (Cu), one of its isotopes is Copper-63, which makes up 69.1% of naturally occurring copper, and it has an atomic weight of 62.9 amu. Therefore, we can calculate the atomic weight of the second isotope using the given average atomic weight of copper (63.5 amu) as follows:
Average atomic weight = (abundance of Copper-63 * atomic weight of Copper-63) + (abundance of Copper-65 * atomic weight of Copper-65)
63.5 = (0.691*62.9) + (0.309 * x)
After we calculate the first part, we subtract it from 63.5 to solve for the atomic weight (x) of the second isotope, known as Copper-65.
Therefore, the atomic weight of the second isotope of copper is approximately 64.93 amu.
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What is the conjugate acid in this reaction? HC2H3O2(aq)+H2O(l)⇌H3O++C2H3O−2(aq) View Available Hint(s) What is the conjugate acid in this reaction? HC2H3O2 H2O H3O+ C2H3O−2
Hey there!
HC2H3O2(aq) + H2O(l) → H3O⁺(aq) + C2H3O2⁻(aq)
↓ ↓ ↓ ↓
acid base acid base
If we consider the only forward reaction H3O⁺is the conjugate acid of the base H2O . For reversse reaction CH3COOH is the conjugate acid of the base CH3COO⁻.
Hope this helps!
In the given reaction, after water (H2O) accepts a proton (H+) from acetic acid (HC2H3O2), it forms H3O+ (hydronium ion), which is the conjugate acid.
Explanation:In this reaction, the conjugate acid is the species that forms after a base has accepted a proton. So, here, the base is H2O and it accepts a proton, H+, from HC2H3O2 to become H3O+. Thus, in the reaction HC2H3O2(aq) + H2O(l) ⇌ H3O+ + C2H3O−2(aq), the conjugate acid that forms is H3O+ (hydronium ion).
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"A chiral alkyne A with molecular formula C6H10 is reduced with H2 and Lindlar catalyst to B having the R configuration at its stereogenic center. What are the structures of A and B? Be sure to answer all parts."
Answer:Kindly find the structure of A and B in attachment.
Explanation:
Lindlars catalyst is basically a palladium metal based catalyst which reduces specifically the alkynes to just alkenes and only reduce them in cis manner.
The palladium catalyst is poisoned by lead or quinoline so that the reduction reaction can be stopped at a point when alkynes are just reduced to alkenes because palladium in presence of hydrogen reduces the alkyne completely to alkane.
So lindlars catalys is very specific in its action and even in terms of stereochemistry as it only reduces alkynes to alkenes and the stereochemistry of reduced alkyne is cis.
The general formula of lindlars catalyst is : 5%Pd-CaCO3,Pb(OCOCH3)2 and quinoline.
A chiral alkyne would have all 4 different substituents present at the carbon next to triple bond.
since the reduction of chiral alkyne A would be done using Lindlars catalyst hence the reduced product formed would have Cis -stereochemistry.
The structures of A&B are drawn in attachments. Kindly find in attachment.
The chiral alkyne is 2-hexyne and when it's reduced using H2 and Lindlar catalyst, it forms cis-2-hexene which has the R configuration at its stereogenic center.
Explanation:The molecule A is a chiral alkyne with the molecular formula C6H10. This indicates that this alkyne molecule contains a chiral center. The most likely structure is 2-hexyne (CH3CH2C≡CCH2CH3), which features a triple bond between the 2nd and 3rd carbons.
When reduced with H2 and Lindlar catalyst, the molecule A is converted into molecule B. The Lindlar catalyst is used specifically to reduce alkynes to cis-alkenes. Thus, molecule B would be cis-2-hexene (CH3CH2CH=CHCH2CH3) as this is the alkene version with a chiral center.
It is specified that molecule B has the R configuration at its stereogenic center. The R or S configuration can be determined by the Cahn-Ingold-Prelog (CIP) priority rules. In the case of cis-2-hexene, assuming the hydrogen atom is bonded to the chiral carbon, a clockwise (R) configuration can be achieved.
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To determine the concentration of X in an unknown solution, 1.00 mL of 8.48 mM S was added to 3.00 mL of the unknown X solution and the mixture was diluted to 10.0 mL. After chromatographic separation, this solution gave peak areas of 5473 and 4851 for X and S, respectively. Determine the concentration of S in the 10.0 mL solution.
The concentration of S in the 10.0 mL solution is calculated by considering the initial 8.48 mM concentration and the dilution factor due to the increase in volume to 10.0 mL. The final concentration of S is found to be 0.848 mM.
Explanation:To determine the concentration of S in the 10.0 mL solution, you must take into account the initial concentration of S and how it changes with dilution. Initially, 1.00 mL of 8.48 mM (millimolar) S was added to the solution. The dilution can be calculated using the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume after dilution.
The initial volume of S (V1) is 1.00 mL and the final volume of the diluted solution (V2) is 10.0 mL. Therefore, using the initial concentration of S (C1) as 8.48 mM, we can solve for the final concentration (C2) as follows:
C1V1 = C2V2
(8.48 mM)(1.00 mL) = (C2)(10.0 mL)
Solving for C2 gives us:
C2 = (8.48 mM)(1.00 mL) / (10.0 mL)
C2 = 0.848 mM
The final concentration of S in the 10.0 mL solution is 0.848 mM (millimolar).
A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be: 119 mg of compound A and 97 mg of compound B. Both compounds were recrystallized and weighed again. After recrystallization, the mass of compound A was 83 mg and the mass of compound B was 79 mg. Calculate the percent recovery from recrystallization for both compounds.
Answer:
The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.
Explanation:
Mass of compound A in a mixture = 119 mg
Mass of compound A after re-crystallization = 83 mg
Percent recovery from re-crystallization :
[tex]\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100[/tex]
Percent recovery of compound A:
[tex]\frac{83 mg}{119 mg}\times 100=69.74\%[/tex]
Mass of compound B in a mixture = 97 mg
Mass of compound B after re-crystallization = 79 mg
Percent recovery of compound B:
[tex]\frac{79 mg}{97 mg}\times 100=81.44\%[/tex]
Determine the molar solubility ( ???? ) of Zn(CN)2 in a solution with a pH=1.33 . Ignore activities. The ????sp for Zn(CN)2 is 3.0×10−16 . The ????a for HCN is 6.2×10−10 .
The molar solubility of Zn(CN)2 in a solution with a given pH and Ksp can be calculated using principles of acid-base equilibria and solubility. However, without exact initial concentrations, a complete calculation can't be provided. Similar applications of these principles are seen in the examples of Cadmium Sulfide (CdS) or mercury chloride mentioned previously.
Explanation:The question involves the determination of the molar solubility of Zinc Cyanide (Zn(CN)2) in a solution with a given pH. The main principle involved is the understanding of acid-base equilibria and solubility. The pH value provided implies an [H3O+] = 10^-1.33. The Ka for Hydrogen Cyanide (HCN) is given which can be used to calculate [CN-]. After these concentrations are calculated, they can be utilized to find the molar solubility of Zn(CN)2 using the given Ksp.
Based on the information provided, some calculations similar to those mentioned but applied to Zn(CN)2 will have to be performed. However, we do not have certain required values like the exact initial concentration of the solution. Therefore, a complete solution can't be provided
However, similar stoichiometry-based calculations are applied to other salts' equilibria for determining molar solubility like Cadmium Sulfide (CdS) and Dissolution stoichiometry of mercury chloride in the provided data. These applications demonstrate how such problems are typically approached and solved.
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Knowing that the protease shows highest activity between pH 4-6, what could be the identity of the base residue acting as a in the reaction? Explain and write the expected mechanism for the reaction. b. Assuming that the optimal pH for the reaction strictly depends exclusively on the protonation state of the active site residues at which pH do you expect maximal enzymatic activity? Explain your rationale. c. What is the fraction of deprotonated Cys and protonated B at the calculated optimum pH? d. How these fractions are changing at 1 pH unit below the optimal pH? Which fraction of the enzyme will be active at this pH?
hey there!:
A) Knowing theatre the protease is showing the highest activity at pH 4-6, implies that the amino acid that amino acid that it is acting in is an amino acid with a basic side chain. Therefore, the residues can be any one of the three basic amino acids being histidine, arginine or lysine , having basic side chains at neutral pH.
b) The mechanism of reaction of cysteine proteases is as follows:
First step in the reaction is the deprotonation of a thiol in the cysteine proteases's active site by an adjacent amino acid with a basic side chain, which might be a histidine residue. This is followed by a nucleophilic attack by the anionic sulfur of the deprotonated cysteine on the substrate carbonyl carbon.
Here, a part of the substrate is released with an amine terminus, restoring the His into a deprotonated form, thus forming a thioester intermediate, forming a link between the carboxy-terminal of the substrate and cysteine, resulting in thiol formation. Thus the name thiol proteases. The thioester bond is then hydrolyzed into a carboxylic acid moiety while again forming the free enzyme.
C) cysteine proteases have a pka of 8-9 but when they are deprotonated by a His residue, their pka would come down to 6-8, which would be their optimal pH for functioning. This is because there is a deprotonation of the thiol group , later restoring the HIS deprotonated form and then formation of a thioester bond. This thioester bond when hydrolysed will a carboxylate moeity , which is responsible for bringing the pH down towards a more acidic side.
d) at the optimal pH , the fraction of deprotonated cysteine and protonated B will be equal which will change with the change in pH.
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Compare suspension and emulsion polymerizations. Describe the microscopic environment of each process shows where the monomer, initiator and additives are located in the initial stages of the polymerization and after 25% conversion is achieved. (Draw clear pictures and label them) B. What are the advantages of producing polyethene using Ziegler Natta catalyst? C. Why is step-growth polymerization not applicable to ethene molecules?
Answer:
The main difference between suspension and emulsion polymerization is that suspension polymerization requires a dispersing medium, monomer(s), stabilizing agents and initiators whereas emulsion polymerization requires water, monomer and a surfactant.
Explanation:
True or False Aluminum does not perform well at elevated temperatures
Answer:
False
Explanation:
Which of the following four fatty acids has the lowest melting point? 1. сњењењењенгсоон г. сњењењењењењењењењеоон 3. CH3CH2CH2CH2CH2CH2CH2COOH CH3CH-CHCH2CH2COOH a. 1 b. 2
Answer: Option (1) is the correct answer.
Explanation:
Melting point is defined as the point at which a solid state of a substance starts to melt and it converts into liquid state.
Also, it is known that more is the number of linear carbon atoms attached to each other, more is the melting point of a substance. Whereas more is the branching present in a compound least will be its melting point.
For example, [tex]CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COOH[/tex] has only 5 carbon atoms attached to carboxylic acid group. Whereas [tex]CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}COOH[/tex] has 9 carbon atoms attached to carboxylic acid group.
Therefore, compound [tex]CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COOH[/tex] will have low melting point.
Thus, we can conclude that out of the given options, fatty acid [tex]CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}COOH[/tex] has the lowest melting point.
Liquid A has a vapor pressure of 264 torr at 20∘C, and liquid B has a vapor pressure of 96.5 torr at the same temperature. If 5.50 moles of liquid A and 8.50 moles of liquid B are combined to form an ideal solution, what is the total vapor pressure (in torr) above the solution at 20.0∘C?
Answer: 161.8 torr
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]
where, x = mole fraction
[tex]p^0[/tex] = pressure in the pure state
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_1+p_2[/tex][tex]p_{total}=x_Ap_A^0+x_BP_B^0[/tex]
[tex]x_{A}=\frac{\text {moles of A}}{\text {moles of A+moles of B}}=\frac{5.50}{5.50+8.50}=0.39[/tex],
[tex]x_{B}=\frac{\text {moles of B}}{\text {moles of A+moles of B}}=\frac{8.50}{5.50+8.50}=0.61[/tex],
[tex]p_{A}^0=264torr[/tex]
[tex]p_{B}^0=96.5torr[/tex]
[tex]p_{total}=0.39\times 264+0.61\times 96.5=161.8torr[/tex]
The total vapor pressure above the solution is 161.8 torr.
At 20°C, an aqueous solution that is 24.0% by mass in ammonium chloride has a density of 1.0674 g/mL. What is the molarity of ammonium chloride in the solution? The formula weight of NH4Cl is 53.50 g/mol.
Answer:
Molarity = 4.79 M
Explanation:
Mass percentage -
Mass percentage of A is given as , the mass of the substance A by mass of the total solution multiplied by 100.
i.e.
mass % A = mass of A / mass of solution * 100
Given,
24% by mass of ammonium chloride,
so,
100 g solution contains , 24 g of ammonium chloride,
mass of solution = 100g
and mass of the solute , i.e. , ammonium chloride = 24 g .
Hence,
Moles -
Moles are calculated as the given mass divided by the molecular mass.
i.e. ,
moles = ( mass / molecular mass )
Given,
The molecular mass of ammonium chloride is 53.50 g /mol
moles of ammonium chloride = 24 g / 53.50 g /mol
moles of ammonium chloride = 0.449 mol
Density -
Density of a substance is given as the mass divided by the volume ,
Density = mass / volume ,
Volume = mass / density
Given ,
Density of ammonium chloride = 1.0764 g /mL
Calculated above , mass of solution = 100 g
volume of solution = 100 g / 1.0764 g/ mL
volume of solution = 93.69 mL
Since , 1 ml = 1/1000 L
volume of solution = 93.69 /1000 L
volume of solution = 0.09369 L
Molarity -
Molarity of a solution is given by the moles of solute per liter of the solution
Hence,
M = moles of solute / volume of solution (in L)
As calculated above,
moles of ammonium chloride = 0.449 mol
volume of solution = 0.09368 L
Putting in the above formula
Molarity = 0.449 mol / 0/09368 L
Molarity = 4.79 M
The molarity of ammonium chloride in the solution is 4.79 M
What is the molarity of a solution?The number of moles of solute dissolved in one liter of solution is the molarity (M) of the solution.
Calculation of molarity:
Given,
The mass of the solution is 24.0%
Density is 1.0674 g/ml
The weight of NH4Cl is 53.50 g/mol
Step 1: Convert Mass % into gram
Considering the mass of solution = 100g
24% of 100g
Mass is 24 gram
Step 2: Calculate the mole compound
Moles = mass divided from molecular mass
The molecular mass of ammonium chloride is 53.50 g /mol
Thus,
[tex]\bold{Moles = \dfrac{24}{53.50} = 0.449 mol}[/tex]
Step 3: Calculating the volume
[tex]\bold{Volume = \dfrac{mass}{density} }[/tex]
[tex]\bold{Volume = \dfrac{ 100}{1.0674 g/m} = 93.69 ml}[/tex]
[tex]\bold{Volume\; of \;solution= \dfrac{ 93.69}{1000 L} = 0.09369 L}[/tex]
Step 4: Now, Molarity of the compound is
[tex]\bold{Molarity = \dfrac{moles\; of\; solute}{volume\; of \;solution (in L)}}[/tex]
Mole of ammonium chloride = 0.449 mol
Volume of solution = 0.09368 L
By formula,
[tex]\bold{Molarity = \dfrac{0.449 mol}{0.09368 L} = 4.79 m}[/tex]
Thus, the Molarity of ammonium chloride is = 4.79 m.
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One way in which the useful metal copper is produced is by dissolving the mineral azurite, which contains copper(II) carbonate, in concentrated sulfuric acid. The sulfuric acid reacts with the copper(II) carbonate to produce a blue solution of copper(II) sulfate. Scrap iron is then added to this solution, and pure copper metal precipitates out because of the following chemical reaction: Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq) Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 250.mL copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Be sure your answer has the correct number of significant digits.
Answer:
Molar concentration of [tex]CuSO_4[/tex] solution = 0.0056 moles / liter
Explanation:
Looking at the chemical reaction we realize that the precipitate, formed by adding the iron powder, is cooper.
Then for finding the number of moles of precipitated copper:
[tex] number.of.moles= \frac{mass (g)}{molecular.mass (g/mole)}[/tex]
[tex]number.of.moles.of.solid.copper = \frac{0.089}{63.5} =1.4*10^{3}[/tex]
From the chemical reaction we deduce that [tex]1.4*10^{3}[/tex] moles of Cu equals to [tex]1.4*10^{3}[/tex] moles of [tex]CuSO_4[/tex] in the initial solution.
So molar concentration is defined as:
[tex]molar.concentration = \frac{number.of.moles}{solution.volume (liters)}[/tex]
[tex]molar.concentration.of.CuSO_4.solution= \frac{1.4*10^{3}}{0.250}= 5.6*10^{3} moles/liter = 0.0056 moles/liter[/tex]
Answer:
[tex]M=5.6x10^{-3}M[/tex]
Explanation:
Hello,
In this case, we first must consider the given already-balanced chemical reaction to realize that 89 mg of copper were recovered, moreover we can relate such mass with the employed moles of copper (II) sulfate via this reaction's stoichiometry as follows:
[tex]n_{CuSO_4}=89mg Cu*\frac{1gCu}{1000mgCu} *\frac{1molCu}{63.546gCu} *\frac{1molCuSO_4}{1molCu} \\n_{CuSO_4}=1.4x10^{-3}molCuSO_4[/tex]
Now, if we state the molarity (mol/L) as the required concentration, we apply its mathematical definition as shown below:
[tex]M=\frac{n_{CuSO_4}}{V_{sln}} =\frac{1.4x10^{-3}molCuSO_4}{0.250L} \\M=5.6x10^{-3}M[/tex]
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If two protons and two neutrons are removed from the nucleus of an oxygen-16 atom, a nucleus of which element remains? Express your answer as an isotope (e.g., as 31H).
Answer:
[tex]_{6}^{12}\text{C}[/tex]
Explanation:
A particle with two protons and two neutrons is a helium nucleus.
Your unbalanced nuclear equation is:
[tex]_{8}^{16}\text{O} \longrightarrow \, _{x}^{y}\text{Z} + \, _{2}^{4}\text{He}[/tex]
The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the equation.
Then
8 = x + 2, so x = 8 - 2 = 6
16 = y + 4, so y = 16 - 4 =12
Element 6 is carbon, so the nuclear equation becomes
[tex]_{\ 8}^{16}\text{O} \longrightarrow \, _{\ 6}^{12}\text{C} + \, _{2}^{4}\text{He}[/tex]
For the aqueous solution containing 75 mg of compound C in 1.0 mL of water, what will be the total amount of the solute C that will be extracted after seven (7) extractions using a certain solvent D if 0.90 mL of a certain organic solvent D was used for each extraction. Given Ksolvent D/water = 1.8.
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
The melting point of phenol is 40.5∘C and that of toluene is −95∘C. What is the best explanation for this difference? Select the correct answer below: a. The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot. b. Phenol has only one hydrogen on the −OH group available to form hydrogen bonds, so the hydrogen bond is stronger. c. In toluene, the hydrogen bond is spread over all three hydrogens on the methyl group, so the interaction is weaker overall. d. Phenol has a higher molecular mass than toluene. e. None of the above.
Answer:
None of the above
Explanation:
The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot.
Phenol has only one hydrogen on the −OH group available to form hydrogen bonds, so the hydrogen bond is stronger. In toluene, the hydrogen bond is spread over all three hydrogens on the methyl group, so the interaction is weaker overall.
Phenol has a higher molecular mass than toluene.
Answer:
a. The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot.
Explanation:
Hello,
We firs must consider that the hydroxyl functional group is present in phenol as a highly polar section into its structure. Thus, phenol molecules are strongly associated by the presence of hydrogen bonds which toluene does not have due to its apolarity.
Consequently, since associating interactions are present in the phenol but absent in the toluene, more energy must be supplied to the phenol to melt it down, that is why phenol's melting point is higher than toluene's that one.
Best energy
If 42.7 mL of a 0.208 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution? Ca(OH)2(aq) + 2HCl CaCl2(aq) + 2H2O(l) g
Answer: The mass of calcium hydroxide that are present in the solution is 0.3256 grams.
Explanation:
To calculate the moles of a solute, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Volume of hydrochloric acid = 42.7mL = 0.0427 L (Conversion factor: 1 L = 1000 mL)
Molarity of the solution = 0.208 moles/ L
Putting values in above equation, we get:
[tex]0.208mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0427L}\\\\\text{Moles of hydrochloric acid}=0.0088mol[/tex]
For the given chemical reaction:
[tex]Ca(OH)_2(aq.)+2HCl\rightarrow CaCl_2(aq.)+2H_2O(l)[/tex]
By Stoichiometry of the reaction:
2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.
So, 0.0088 moles of hydrochloric acid will react with = [tex]\frac{1}{2}\times 0.0088=0.0044mol[/tex] of calcium hydroxide.
To calculate the mass of calcium hydroxide, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of calcium hydroxide = 0.0044 moles
Molar mass of calcium hydroxide = 74 g/mol
Putting values in above equation, we get:
[tex]0.0044mol=\frac{\text{Mass of calcium hydroxide}}{74g/mol}\\\\\text{Mass of calcium hydroxide}=0.3256g[/tex]
Hence, the mass of calcium hydroxide that are present in the solution is 0.3256 grams.
0.325 grams of Ca(OH)₂ must be in the solution of HCl to neutralize a solution.
How we calculate moles from molarity?Molarity of any solution is defined as:
M = n/V, where
n = no. of moles
V = volume
Given chemical reaction is:
Ca(OH)₂(aq) + 2HCl → CaCl₂(aq) + 2H₂O(l)
From the stoichiometry of the reaction, it is clear that:
2 moles of HCl = react with 1 mole of Ca(OH)₂
1 mole of HCl = react with 1/2 mole of Ca(OH)₂
Moles of HCl will be calculated by using the molarity formula and given data as:
Given concentration of HCl = 0.208M
Given volume of HCl = 42.7mL = 0.0427L
Moles of HCl = 0.208 × 0.0427 = 0.0088 moles
0.0088 moles of HCl = react with 0.0088 × 1/2 = 0.0044moles of Ca(OH)₂
Mass of Ca(OH)₂ will be calculated by using the formula:
n = W/M, where
W = required mass
M = molar mass = 72g/mole
W = 0.0044mole × 72g/mole = 0.325 grams
Hence, 0.325 grams of Ca(OH)₂ will be in the solution.
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Be sure to answer all parts. Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas: 2NO(g) + O2(g) → 2NO2(g)
In one experiment, 0.857 mol of NO is mixed with 0.498 mol of O2.
Determine which of the two reactants is the limiting reactant. Calculate also the number of moles of NO2 produced. Limiting reactant: Moles of NO2 produced: moles
Answer: NO is the limiting reagent in the given reaction and 0.857 moles of [tex]NO_2[/tex] will be produced.
Explanation:
Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.
Excess reagent is defined as the reagent which is present in large amount.
For the given chemical reaction:
[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]
We are given:
Moles of NO = 0.857 mol
Moles of oxygen = 0.498 mol
By stoichiometry of the reaction:
If 2 moles of NO reacts with 1 mole of oxygen gas.
So, 0.857 moles of NO will react with = [tex]\frac{1}{2}\times 0.857=0.4285mol[/tex] of [tex]O_2[/tex]
As, the given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, NO is considered as the limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
If 2 moles of NO produces 2 moles of nitrogen dioxide gas.
So, 0.857 moles of NO will produce = [tex]\frac{2}{2}\times 0.857=0.857mol[/tex] of [tex]NO_2[/tex]
Hence, NO is the limiting reagent in the given reaction and 0.857 moles of [tex]NO_2[/tex] will be produced.
Determine the balanced chemical equation for this reaction.C8H18(g) O2(g)→CO2(g) H2O(g)Enter the coefficients for each compound in order, separated by commas. For example, 1,2,3,4 would indicate one mole of C8H18, two moles of O2, three moles of CO2, and four moles of H2O.
Answer:
2C₈H₁₈(g) + 25O₂(g)→16CO₂(g) + 18H₂O
Explanation:
To balance an equation, the moles of one element on one side of the equation should be the same as those on the other side of the equation. This is because (as a law of thermodynamics), in a chemical reaction, the matter is not destroyed nor created - atoms are only rearranged.
Answer: The coefficients are 2, 17, 8 and 18
Explanation:
A balanced chemical equation always follow law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.
For the given chemical reaction, the balanced equation follows:
[tex]2C_8H_{18}(g)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]
By Stoichiometry of the reaction:
2 moles of octane reacts with 25 moles of oxygen gas to produce 16 moles of carbon dioxide gas and 18 moles of water vapor.
Hence, the coefficients are 2, 25, 16 and 18
What is the entropy change of the system when 17.5 g of liquid benzene (C6H6) evaporates at the normal boiling point? The normal boiling point of benzene is 80.1°C and ΔH vap is 30.7 kJ/mol.
Answer : The entropy change of the system is, 19.5 J/K
Solution :
First we have to calculate the moles of benzene.
[tex]\text{Moles of }C_6H_6=\frac{\text{Mass of }C_6H_6}{\text{Molar mass of }C_6H_6}=\frac{17.5g}{78.11g/mole}=0.224moles[/tex]
Now we have to calculate the entropy change of the system.
Formula used :
[tex]\Delta S=\frac{n\times \Delta H_{vap}}{T_b}[/tex]
where,
[tex]\Delta S[/tex] = entropy change of the system = ?
[tex]\Delta H[/tex] = enthalpy of vaporization = 30.7 kJ/mole
n = number of moles of benzene = 0.224 mole
[tex]T_b[/tex] = normal boiling point of benzene = [tex]80.1^oC=273+80.1=353.1K[/tex]
Now put all the given values in the above formula, we get the entropy change of the system.
[tex]\Delta S=\frac{0.224mole\times (30.7KJ/mole)}{353.1K}=0.0195kJ/K=0.0195\times 1000=19.5J/K[/tex]
Therefore, the entropy change of the system is, 19.5 J/K
What is dynamic equilibrium? What is dynamic equilibrium? Dynamic equilibrium in a chemical reaction is the condition in which the rate of the forward reaction is lower than the rate of the reverse reaction. Dynamic equilibrium in a chemical reaction is the condition in which the rate of the forward reaction is higher than the rate of the reverse reaction. Dynamic equilibrium in a chemical reaction is the condition in which the rate of the forward reaction equals the rate of the reverse reaction.
Answer:
The correct answer is: Dynamic equilibrium in a chemical reaction is the condition in which the rate of the forward reaction equals the rate of the reverse reaction.
Explanation:
Dynamic equilibrium is a chemical equilibrium between froward reaction and backward or reverse reaction where rate of reaction going forwards is equal to the rate of reaction going backward (reverse).
Some other properties of dynamic equilibrium are:
Chemical equilibrium are attained is closed system. The macroscopic remains constant like: volume, pressure, energy etc. The concentration of the reactants and products remain constant.They are not always equal.Answer:
B: At dynamic equilibrium, the rate of the forward reaction is higher than the rate of the reverse reaction.
Write the symbol for each of the following ions: (a) the ion with a 3+ charge, 28 electrons, and a mass number of 71 (b) the ion with 36 electrons, 35 protons, and 45 neutrons (c) the ion with 86 electrons, 142 neutrons, and a 4+ charge (d) the ion with a 2+ charge, atomic number 38, and mass number 87
Answer:
[tex]\boxed{\math{_{31}^{71}\text{Ga}^{3+}}}\boxed{\math{_{35}^{80}\text{Br}^{-}}} \boxed{\math{_{90}^{232}\text{Th}^{4+}}} \boxed{\math{_{38}^{87}\text{Sr}^{2+}}}[/tex]
Explanation:
(a)
If the ion has 28 electrons and a 3+ charge, it normally has 31 electrons. It must also have 31 protons. Atom 31 is gallium.
[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{31}^{71}\textbf{Ga}^{3+}}}[/tex]
(b)
If the ion has 35 protons, it normally has 35 electrons. It has 36 electrons, so it has a negative charge. Atom 35 is bromine.
A = p + n = 35 + 45 = 80, so the isotopic mass number is 80
\text{The symbol for the ion is }\boxed{\mathbf{_{35}^{80}\textbf{Br}^{-}}}
(c)
If the ion has 86 electrons and a charge of 4+, it normally has 90 electrons. It must also have 90 protons. Atom 90 is thorium.
A = p + n = 90 + 142 = 232, so the isotopic mass number is 232.
[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{90}^{232}\textbf{Th}^{4+}}}[/tex]
(d)
If the ion has 38 protons, it is strontium.
[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{38}^{87}\textbf{Sr}^{2+}}}[/tex]
The symbol of ion will be:[tex]^{71}_{31}Ga^{+3}[/tex]
The symbol of ion will be:[tex]^{80}_{35}Br^{-1}[/tex]
The symbol of ion will be:[tex]^{232}_{90}Th^{+4}[/tex]
The symbol of ion will be: [tex]^{87}_{38}Sr^{+2}[/tex]
Explanation:
The symbol of an ion of an element is written as:
[tex]^A_ZX^C[/tex]
Where:
A = Mass number of the atom
Z = Atomic number of the atom
X = Symbol the atom
C = Charge on the atom
Given:
a) The ion with a 3+ charge, 28 electrons, and a mass number of 71.
b) The ion with 36 electrons, 35 protons, and 45 neutrons
c) The ion with 86 electrons, 142 neutrons, and a 4+ charge
d) The ion with a 2+ charge, atomic number 38, and mass number 87
To find:
The symbols of given ions
Solution:
a)
Charge on the ion = C = 3+
Number of electrons in the ion = 28
Number of electrons in ion's parent atom = 28 +3 = 31
Number of electrons in parent atom = Number of proton in ion/ paarent atom
= 31 = 31
Atomic number = Number of protons
Atomic number of the ion = Z = 31
Mass number of ion = A = 71
The element with atomic number 31 is gallium with the chemical symbol Ga, so the symbol of ion will be:
[tex]^{71}_{31}Ga^{+3}[/tex]
b)
Charge on the ion = C = ?
Number of electrons in the ion = 36
Number of protons in the ion = 35
Number of neutrons = 45
Number of electrons in parent atom = Number of protons in ion/ parent atom
= 35 = 35
Atomic number = Number of protons
Atomic number of the ion = Z = 35
Number of electrons in ion's parent atom = Number of protons in the ion = 35
Charge on the ion = C = 36 - 35 = 1-
Mass number of ion = Number of protons + Mass of neutrons
A = 35 + 45 = 80
The element with atomic number 35 is bromine with the chemical symbol Br, so the symbol of ion will be:
[tex]^{80}_{35}Br^{-1}[/tex]
c)
Charge on the ion = C = 4+
Number of electrons in the ion = 86
Number of electrons in ion's parent atom = 86+4 = 90
Number of electrons in parent atom = Number of protons in ion/ parent atom
= 90= 90
Atomic number = Number of protons
Atomic number of the ion = Z = 90
Number of neutrons = 142
Mass number of ion = Number of protons + Mass of neutrons
A = 90+ 142 = 232
The element with atomic number 90 is thorium with the chemical symbol Th, so the symbol of ion will be:
[tex]^{232}_{90}Th^{+4}[/tex]
d)
Charge on the ion = C = 2+
Atomic number of the ion = Z = 38
Mass number of ion = A = 87
The element with atomic number 38 is strontium with chemical symbol Sr, so the symbol of ion will be:
[tex]^{87}_{38}Sr^{+2}[/tex]
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A certain heat engine operates between 800 K and 300 K. (a) What is the maximum efficiency of the engine? (b) Calculate the maximum work that can be done by for each 1.0 k) of hea a reversible process for each 1.0 kJ supplied by the hot source? t supplied by the hot source. (c) How much heat is discharged into the cold sink in
Answer :
(a) The maximum efficiency of the engine is, 62.5 %
(b) The maximum work done is, 0.625 KJ.
(c) The heat discharge into the cold sink is, 0.375 KJ.
Explanation : Given,
Temperature of hot body [tex]T_h[/tex] = 800 K
Temperature of cold body [tex]T_c[/tex] = 300 K
(a) First we have to calculate the maximum efficiency of the engine.
Formula used for efficiency of the engine.
[tex]\eta =1-\frac{T_c}{T_h}[/tex]
Now put all the given values in this formula, we get :
[tex]\eta =1-\frac{300K}{800K}[/tex]
[tex]\eta =0.625\times 100=62.5\%[/tex]
(b) Now we have to calculate the maximum work done.
Formula used :
[tex]\eta =\frac{Q_h-Q_c}{Q_h}=\frac{w}{Q_h}[/tex]
where,
[tex]Q_h[/tex] = heat supplied by hot source = 1 KJ
[tex]Q_c[/tex] = heat supplied by hot source
w = work done = ?
Now put all the given values in this formula, we get :
[tex]\eta =\frac{w}{Q_h}[/tex]
[tex]0.625=\frac{w}{1KJ}[/tex]
[tex]w=0.625KJ[/tex]
(c) Now we have to calculate the heat discharge into the cold sink.
Formula used :
[tex]w=Q_h-Q_c[/tex]
[tex]Q_c=Q_h-w[/tex]
[tex]Q_c=1-0.625[/tex]
[tex]Q_c=0.375KJ[/tex]
Therefore, (a) The maximum efficiency of the engine is, 62.5 %
(b) The maximum work done is, 0.625 KJ.
(c) The heat discharge into the cold sink is, 0.375 KJ.
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/°C. The temperature increases by 1.126°C. Calculate the heat given off by the burning Mg, in kJ/g and in kJ/mol.
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter (heat capacity of 3024 J/°C), causing a temperature increase of 1.126°C. The heat given off by the burning Mg is -24.76 kJ/g and -601.9 kJ/mol.
When a sample of magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity (C) of 3024 J/°C, the temperature increases by 1.126°C (ΔT). We can calculate the heat absorbed by the calorimeter (Qc) using the following expression.
[tex]Qc = C \times \Delta T = \frac{3024J}{\° C} \times 1.126 \° C \times \frac{1kJ}{1000J} = 3.405 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the calorimeter and the heat released by the reaction (Qr) is zero.
[tex]Qc + Qr = 0\\\\Qr = -Qc = -3.405 kJ[/tex]
3.405 kJ are released by the combustion of 0.1375 g of Mg. The heat released per gram of Mg is:
[tex]\frac{-3.405kJ}{0.1375g} = -24.76 kJ/g[/tex]
Finally, we will convert -24.67 kJ/g to kJ/mol using the molar mass of Mg (24.31 g/mol).
[tex]\frac{-24.76kJ}{g} \times \frac{24.31g}{mol} = -601.9 kJ/mol[/tex]
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter (heat capacity of 3024 J/°C), causing a temperature increase of 1.126°C. The heat given off by the burning Mg is -24.76 kJ/g and -601.9 kJ/mol.
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Tooth enamel ( Ca5(PO4)3(OH) ) reacts with acid according to the reaction below: Ca_5(PO_4)_3(OH) (s) + H^+(aq) \leftrightarrow 5Ca^{2+}(aq) + 3HPO_4^{2-}(aq) + OH^-(aq) C a 5 ( P O 4 ) 3 ( O H ) ( s ) + H + ( a q ) ↔ 5 C a 2 + ( a q ) + 3 H P O 4 2 − ( a q ) + O H − ( a q ) What happens if we add more acid?
Final answer:
Adding more acid to tooth enamel causes increased solubility of enamel and may lead to cavities. The added acid reacts with hydroxide ions in the enamel, driving the reaction forward. Fluoride in dental products can help protect enamel by reducing solubility in acid.
Explanation:
When more acid, which in chemical terms is a source of H+ ions, is added to the solution where tooth enamel, chemically known as calcium hydroxyapatite Ca5(PO4)3(OH), is present, it results in the reaction shifting to the right according to Le Chatelier's principle. This process increases the solubility of the tooth enamel in acid, leading to enamel dissolution and potentially to the formation of dental cavities. The reaction with acid produces 5Ca2+(aq), 3HPO42-(aq), and OH-(aq). Hydroxide ions (OH-) react with added H+ ions to form water (2H2O), driving the reaction forward and increasing enamel solubility. To mitigate this effect, toothpastes and mouth rinses may contain fluoride compounds, replacing the strong base hydroxide in the enamel with the weaker base fluoride. This renders the enamel more resistant to acid attack by reducing the extent to which the equilibrium shifts upon acid addition.
The reaction 2NO(g)+Cl2(g)→2NOCl(g) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of 21 torr/min , what is the rate of change of the total pressure of the vessel
Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.
Explanation : Given,
[tex]\frac{d[NO]}{dt}[/tex] =21 torr/min
The balanced chemical reaction is,
[tex]2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)[/tex]
The rate of disappearance of [tex]NO[/tex] = [tex]-\frac{1}{2}\frac{d[NO]}{dt}[/tex]
The rate of disappearance of [tex]Cl_2[/tex] = [tex]-\frac{d[Cl_2]}{dt}[/tex]
The rate of formation of [tex]NOCl[/tex] = [tex]\frac{1}{2}\frac{d[NOCl]}{dt}[/tex]
As we know that,
[tex]\frac{d[NO]}{dt}[/tex] =21 torr/min
So,
[tex]-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}[/tex]
[tex]\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min[/tex]
And,
[tex]\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}[/tex]
[tex]\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min[/tex]
Now we have to calculate the rate change.
Rate change = Reactant rate - Product rate
Rate change = (21 + 10.5) - 21 = 10.5 torr/min
Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.
The rate of change of the total pressure of the vessel is 10.5 torr/min
The given reaction is expressed as:
[tex]\mathbf {2O_{(g)} + Cl_{2(g)} \to 2NOCl_{(g)}}}[/tex]
From chemical kinetics, the average rate (r) can be expressed as:
[tex]\mathbf{r = -\dfrac{1}{2}\dfrac{d[NO]}{dt}= -\dfrac{d[Cl_2]}{dt}=\dfrac{1}{2}\dfrac{d[NOCl]}{dt} }[/tex]
where;
the negative sign (-) indicates the rate of disappearance of the substances.∴
rate of disappearance of NO [tex]\mathbf{= -\dfrac{1}{2} \dfrac{d[NO]}{dt}}[/tex] rate of disappearance of Cl₂ = [tex]\mathbf{\dfrac{-d[Cl_2]}{dt}}[/tex] rate of appearance of NOCl = [tex]\mathbf{\dfrac{1}{2} \dfrac{d[NOCl]}{dt}}[/tex]We are being told that the partial pressure of NO is decreasing at 21 torr/min
i.e.
[tex]\mathbf{\dfrac{d[NO]}{dt}}[/tex] = 21 torr/minand we know that:
[tex]\mathbf{\dfrac{-d[Cl_2]}{dt}= -\dfrac{1}{2} \dfrac{d[NO]}{dt}}}[/tex]
∴
[tex]\mathbf{\dfrac{-d[Cl_2]}{dt}= -\dfrac{1}{2}(21 \ torr/min) }}[/tex] [tex]\mathbf{\dfrac{d[Cl_2]}{dt}= 10.5 \ torr/min }}[/tex]Similarly;
[tex]\mathbf{-\dfrac{1}{2} \dfrac{d[NOCl]}{dt} = \mathbf{-\dfrac{1}{2} \dfrac{d[NO]}{dt}}}[/tex] [tex]\mathbf{\dfrac{d[NOCl]}{dt} = \mathbf{ \dfrac{d[NO]}{dt}}}[/tex] [tex]\mathbf{\dfrac{d[NOCl]}{dt} =21 \ torr/min}}[/tex]Now, we need to determine the rate of change of the total pressure at which these substances are decreasing;
Rate change = rate of reactant - rate of product.
[tex]\mathbf{Rate \ change =} \mathbf{\mathbf{ \dfrac{d[NO]}{dt}} +\dfrac{d[Cl_2]}{dt} - \dfrac{d[NOCl]}{dt} }[/tex]
[tex]\mathbf{Rate \ change =} \mathbf{(21 \ torr/min) +(10.5 \ torr/min) -( 21 \ torr/min})[/tex]
Rate change = 10.5 torr/min
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Calculate the number of vacancies per cubic meter for some metal, M, at 773°C. The energy for vacancy formation is 0.97 eV/atom, while the density and atomic weight for this metal are 7.81 g/cm3 (at 773°C) and 67.81 g/mol, respectively.
Answer:
0.112eV/atom
Explanation:
since
p=m/v
then
pv=pv
7.81*0.97=67.81*V
V=7.58/67.81
V= 0.112eV/atom
A solution contains two isomers, n-propyl alcohol and isopropyl alcohol, at 25°C. The total vapor pressure is 38.6 torr. What are the mole fractions of each alcohol in the liquid and in the vapor phase? The vapor pressures are 21.0 torr for n-propyl alcohol and 45.2 torr for isopropyl alcohol.
Answer:
Mole fraction of alcohols in liquid phase [tex]x_1=0.2727\& x_2=0.7273[/tex].
Mole fraction of alcohols in vapor phase [tex]y_1=0.1468\& y_2=0.8516[/tex].
Explanation:
The total vapor pressure of the solution = p =38.6 Torr
Partial vapor pressure of the n-propyl alcohol =[tex]p^{o}_1=21.0 Torr[/tex]
Partial vapor pressure of the isopropyl alcohol =[tex]p^{o}_2=45.2 Torr[/tex]
[tex]p=x_1\times p^{o}_1+x_2\times p^{o}_2[/tex] (Raoult's Law)
[tex]p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2[/tex]
[tex]38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr[/tex]
[tex]x_1=0.2727[/tex]
[tex]x_2=1-0.2727=0.7273[/tex]
[tex]x_1\& x_2[/tex] is mole fraction in liquid phase.
Mole fraction of components in vapor phase [tex]y_1\& y_2[/tex]
[tex]p_1=y_1\times p[/tex] (Dalton's law of partial pressure)
[tex]y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}[/tex]
[tex]y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468[/tex]
[tex]y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}[/tex]
[tex]y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516[/tex]
Mole fraction of alcohols in vapor phase [tex]y_1=0.1468\& y_2=0.8516[/tex]
A chemist carefully measures the amount of heat needed to raise the temperature of a 894.0g sample of a pure substance from −5.8°C to 17.5°C . The experiment shows that 4.90kJ of heat are needed. What can the chemist report for the specific heat capacity of the substance? Round your answer to 3 significant digits.
An 894.0 g sample with a specific heat capacity of 2.35 × 10⁻⁴ kJ/g.° C, increases its temperature from -5.8 °C to 17.5 °C when absorbing 4.90 kJ of heat.
A chemist has a sample with a mass of 894.0 g. When it absorbs 4.90 kJ of heat its temperature increases from -5.8 °C to 17.5 °C. The chemist can calculate the specific heat capacity of the substance using the following expression.
[tex]Q = c \times m \times \Delta T[/tex]
where,
c: specific heat capacity of the substancem: mass of the sampleΔT: change in the temperature of the sample[tex]Q = c \times m \times \Delta T\\\\c = \frac{Q}{m \times \Delta T} = \frac{4.90 kJ}{894.0 g \times (17.5 \° C - (-5.8 \° C))} = 2.35 \times 10^{-4} kJ/g\° C[/tex]
An 894.0 g sample with a specific heat capacity of 2.35 × 10⁻⁴ kJ/g.° C, increases its temperature from -5.8 °C to 17.5 °C when absorbing 4.90 kJ of heat.
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The specific heat capacity of the substance is approximately 0.234 J/g°C. This was found by applying the formula for specific heat capacity (q = mcΔT) and doing the necessary calculations with the given values.
Explanation:To determine the specific heat capacity of the substance, we can use the formula q = mcΔT, where q is the heat energy absorbed, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. Given that q equals 4.90 kJ (or 4900 J to match the unit of specific heat capacity), m equals 894.0g and ΔT is 17.5°C - (-5.8°C) = 23.3°C, we can rearrange the formula to solve for c: c = q / (mΔT).
Substituting the given values, we find c = 4900 J / (894.0g * 23.3°C), yielding a specific heat capacity of approximately 0.234 J/g°C to three significant figures.
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Ammonia gas is dissolved in a 1.00 ×10-4M solution of CuSO4to give an equilibrium concentration of [NH3] = 1.60 ×10-3M. Calculate the equilibrium concentration of Cu2+(aq) ions in the solution
Answer:
Explanation:
Mixing Ammonia gas into a solution of Copper(II) Sulfate will give Ammonium Sulfate and a precipitate of Copper(II) Hydroxide (Cu(OH)₂). The Ksp of Cu(OH)₂ is published => 2.2 x 10⁻²². Such gives a solubility* of the Cu(OH)₂ to be ~1.77 x 10⁻⁷M => [Cu⁺²] ~1.77 x 10⁻⁷M and [OH⁻] = 2(1.77 x 10⁻⁷)M = 3.53 x 10⁻⁷M. The reaction of Ammonium Hydroxide and Copper(II) Sulfate will generate 1 x 10⁻⁴ mole Cu(OH)₂ as a precipitate but only 1.77 x 10⁻⁷ mole of the hydroxide will remain in 1 Liter of solution b/c of extreme limited solubility.
*Solubility of 1:2 ionization ratio salts = CubeRt(Ksp/4).