Answer: [tex]CrF_3[/tex] has greater impact on the freezing point depression of ice.
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f-T_f^0[/tex] = Depression in freezing point
i= vant hoff factor
[tex]K_f[/tex] = freezing point constant
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
a) i = 4 for [tex]CrF_3[/tex] as it dissociates to give 4 ions in water.
[tex]CrF_3\rightarrow Cr^{3+}+3F^-[/tex]
b) i = 3 for [tex]CaF_2[/tex] as it dissociates to give 3 ions in water.
[tex]CaF_2\rightarrow Ca^{2+}+2F^-[/tex]
As the vant hoff factor is higher for [tex]CrF_3[/tex] , it has greater impact on the freezing point depression of ice.
Final answer:
CrF3 would have a greater impact on the freezing point depression of ice than CaF2 because it dissociates into more particles, four compared to CaF2 which dissociates into three.
Explanation:
In determining which substance would have a greater impact on the freezing point depression of ice, we need to consider the number of particles each substance releases into solution when dissolved. CrF3 will ionize to form one Cr3+ ion and three F- ions, for a total of four particles. Conversely, CaF2 will dissociate to form one Ca2+ and two F- ions, resulting in three particles. According to colligative properties, CrF3 will therefore have a greater effect on freezing point depression than CaF2 because it produces a larger number of dissociated particles.
Draw a Lewis structure for C2H3Cl. Show all unshared electron pairs. None of the atoms bears a formal charge, and all atoms have octets (except for hydrogen atoms, which have duets).
Answer:
See picture below
Explanation:
To do this, we first need to know how are the bonds in this molecule are. To do so, let's calculate the number of insaturations in the molecule:
n°I = C+1 - (H-N+X)/2
These numbers indicate if the molecule has double bond, triple bond, ring, cyclo, among other options:
n°I = 3 - (3+1)/2
n°I = 1
As it's a very small molecule, we can assume this molecule only have a double bond, and it's an alkene.
So the lewis structure, shows the electrons and the bonding, and also shows the unshared electron pairs, depending on how much electron have each molecule.
In the case of carbon:
[C] = [He] 2s2 2p2 ----> 4 electrons
[H] = 1s1 ----> 1 electron
[Cl] = [Ne] 3s2 3p5 ----> 7 electrons.
Therefore, we also know that Carbon has a double bond, so, the main molecule would have something like this:
C = C
so next to the carbons, we can put two hydrogens and in the other carbon, the chlorine and the remaining hydrogen.
See picture below for structure:
The Lewis structure of C₂H₃Cl is attached in the image below.
Lewis structures are also known as Lewis dot structures or electron dot structures. They are diagrams that represent the arrangement of atoms and valence electrons in a molecule or ion.
In a Lewis structure, the symbol of each atom is used to represent the nucleus and inner-shell electrons, while dots or lines are used to represent the valence electrons. Valence electrons are the outermost electrons involved in bonding and determining the chemical properties of an atom. The image is attached below.
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Which of the following pure compounds will exhibit hydrogen bonding?
a.) CH3CH2OHb.) CH2Oc.) C2H4d.) C2H2e.) CH3OCH3f.) CH3NH2g.) C3H8h.) NH3
Explanation:
For a compound to show hydrogen bonding it is necessary that the hydrogen atom of the compound should be attached to more electronegative atom like fluorine, oxygen or nitrogen.
For example, [tex]CH_{3}CH_{2}OH[/tex], [tex]CH_{3}NH_{2}[/tex] and [tex]NH_{3}[/tex] all these compounds contain an electronegative atom attached to hydrogen atom.
Therefore, these pure compounds will exhibit hydrogen bonding.
Thus, we can conclude that out of the given options [tex]CH_{3}CH_{2}OH[/tex], [tex]CH_{3}NH_{2}[/tex] and [tex]NH_{3}[/tex] are the pure compounds which will exhibit hydrogen bonding.
What are dipoles, and what is the difference between a natural dipole and an induced dipole?
Answer:
See explanation below.
Explanation:
Dipoles are molecules that have partial charges. It happens because of the difference in electronegativity of the elements. This property is the tendency that the atom has to take the electron to it, so, in the covalent bond, the shared pair of electrons is easily found at the more electronegativity atom, and so, it has a partial negative charge, and the other, a partial positive charge. This is a natural dipole.
If the difference of electronegativity is 0, or extremely close to 0, then the molecule is nonpolar, and so the molecule doesn't have partial charges. But, to be joined together and form the substance, the partial charge must be induced, so it's an induced dipole.
Final answer:
An electric dipole consists of two opposite charges separated by a distance, and is seen in natural structures like water molecules. A permanent dipole is due to molecular structure, while an induced dipole occurs due to an external electric field. The dipole moment represents the strength and direction of this separation of charges.
Explanation:
An electric dipole consists of two equal but opposite charges separated by a distance. This model is crucial in understanding atomic and molecular interactions. A common example of a natural dipole is the water molecule. The unequal distribution of electron density throughout a molecule can lead to a positive end and a negative end, resulting in a dipole moment.
Natural vs. Induced Dipole
A permanent dipole is inherent within a molecule and results from an unequal distribution of electron density due to its molecular structure. In contrast, an induced dipole occurs when an external electric field influences a neutral atom or molecule, causing the displacement of charges and creating the dipole moment. The induced dipole moment will be aligned with the external electric field.
The strength and direction of an electric dipole are expressed by the dipole moment, a vector quantity that represents the size of the charge separation and the distance between the charges. The physical significance of the dipole moment lies in its alignment parallel to an external electric field and its role in decreasing the total electric field within the dipole region, which has applications in areas like capacitors.
Asprin is a compond with the molecular formula, C9H8O4. What is the percent composition?
Answer:
Oxygen 35.6 percent
Carbon 60 percent
Hydrogen 4.4 percent
Explanation:
Molecular mass of aspirin[tex](C_9H_8O_4)=180 gm/mol.[/tex]
Mass of carbon in 180 gm of aspirine[tex]=12\times 9=108\ gm.[/tex]
Therefore, percentage of carbon in aspirin is [tex]=\dfrac{108}{180}\times 100=60\ percent.[/tex]
Similarly, mass of hydrogen in 8 gm of aspirine[tex]=1\times 8=8\ gm[/tex].
Hydrogen's percentage in aspirin[tex]=\dfrac{8}{180}\times 100=4.4\ percent.[/tex]
Also, mass of oxygen in aspirin is [tex]=16\times 4=64\ gm.[/tex]
Oxygen's percentage in aspirin[tex]=\dfrac{64}{180}\times 100=35.6\ percent.[/tex]
Hence, this is the required solution.
Final answer:
The percent composition of aspirin (C9H8O4) is calculated based on its molecular mass of 180.15 amu. It is 60.03% carbon, 4.48% hydrogen, and 35.49% oxygen.
Explanation:
To determine the percent composition of aspirin, which has the molecular formula C9H8O4, we need to calculate the percentage of each element in the compound based on its molecular mass. The molecular mass of aspirin is 180.15 amu. This is calculated by adding the masses of nine carbon (C) atoms, eight hydrogen (H) atoms, and four oxygen (O) atoms together.
To find the percent composition, we first calculate the total mass of each type of atom in one molecule of aspirin and then divide it by the molecular mass of aspirin. To express it as a percentage, we multiply by 100%. The atomic masses are approximately 12.01 amu for carbon, 1.008 amu for hydrogen, and 16.00 amu for oxygen.
The calculations for percent composition are as follows:
Carbon: (9 atoms × 12.01 amu/atom) / 180.15 amu × 100% = 60.03%Hydrogen: (8 atoms × 1.008 amu/atom) / 180.15 amu × 100% = 4.48%Oxygen: (4 atoms × 16.00 amu/atom) / 180.15 amu × 100% = 35.49%Therefore, the percent composition of aspirin (C9H8O4) is 60.03% carbon, 4.48% hydrogen, and 35.49% oxygen.
A 0.1014 g sample of a purified CHO compound was burned in a combustion apparatus and produced 0.1486 g CO2 and 0.0609 g of H2O. Mass spectrometry analysis revealed that the CHO compound had a molar mass of 180 g/mol. What is the molecular formula for this compound?
Answer:
[tex]C_6H_{12}O_6[/tex]
Explanation:
Mass of water obtained = 0.0609 g
Molar mass of water = 18 g/mol
Moles of [tex]H_2O[/tex] = 0.0609 g /18 g/mol = 0.00338 moles
2 moles of hydrogen atoms are present in 1 mole of water. So,
Moles of H = 2 x 0.00338 = 0.00676 moles
Molar mass of H atom = 1.008 g/mol
Mass of H in molecule = 0.00676 x 1.008 = 0.00681 g
Mass of carbon dioxide obtained = 0.1486 g
Molar mass of carbon dioxide = 44.01 g/mol
Moles of [tex]CO_2[/tex] = 0.1486 g /44.01 g/mol = 0.00337 moles
1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,
Moles of C = 0.00337 moles
Molar mass of C atom = 12.0107 g/mol
Mass of C in molecule = 0.00337 x 12.0107 = 0.04047 g
Given that the compound only contains hydrogen, oxygen and carbon. So,
Mass of O in the sample = Total mass - Mass of C - Mass of H
Mass of the sample = 0.1014 g
Mass of O in sample = 0.1014 - 0.04047 - 0.00681 = 0.05412 g
Molar mass of O = 15.999 g/mol
Moles of O = 0.05412 / 15.999 = 0.00338 moles
Taking the simplest ratio for H, O and C as:
0.00676 : 0.00338 : 0.00337
= 2 : 1 : 1
The empirical formula is = [tex]CH_2O[/tex]
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 1×12 + 2×1 + 16= 30 g/mol
Molar mass = 180 g/mol
So,
Molecular mass = n × Empirical mass
180 = n × 30
⇒ n = 6
The formula of compound = [tex]C_6H_{12}O_6[/tex]
Final answer:
The molecular formula of the compound is (CHO)6.
Explanation:
To determine the molecular formula of the compound, we need to find the empirical formula first. We can do this by finding the moles of carbon and hydrogen in the sample using the masses of CO2 and H2O produced.
Mass of CO2 = 0.1486 g
Mass of H2O = 0.0609 g
Now, we need to convert the masses of CO2 and H2O to moles and divide by the smallest value to find the ratio of moles of carbon to hydrogen.
After finding the ratio, we multiply it by a common factor to get whole numbers. In this case, the ratio is 1:1, so the empirical formula is CHO.
The empirical formula has a molar mass of 30 g/mol.
To find the molecular formula, we divide the given molar mass of 180 g/mol by the empirical formula molar mass. The result is 6, which means the molecular formula is (CHO)6.
Assume that the complete combustion of one mole of fructose, a monosaccharide, to carbon dioxide and water liberates 2810 kJ (ΔG°\' = –2810 kJ/mol). If the energy generated by the combustion of fructose is entirely converted to the synthesis of a hypothetical compound X, calculate the number of moles of the compound that could theoretically be generated. Use the value ΔG°\'compound X = − 64.1 kJ/mol kJ/mol. Round your answer to two significant figures.
Explanation:
The reaction equation will be as follows.
[tex]C_{6}H)_{12}O_{6}(s) + 6O_{2}(g) \rightarrow 6CO_{2}(g) + 6H_{2}O(l)[/tex]
It is given that the total energy liberated is -2810 kJ/mol. As the sign is negative this means that energy is being released. Also, it is given that the energy required to synthesis is -64.1 kJ/mol.
Therefore, calculate the number of moles of compound as follows.
No. of moles = [tex]\frac{\text{total energy}}{\text{energy necessary to synthesise 1 mole of compound X}}[/tex]
= [tex]\frac{-2810 kJ}{-64.1 kJ/mol}[/tex]
= 43.83 mol
= 44 mol (approx)
Thus, we can conclude that the number of moles of compound is 44 mol.
The number of moles of the compound that could theoretically be generated resulting in approximately 44 moles when rounded to two significant figures.
To calculate the number of moles of hypothetical compound X that can be generated from the combustion of one mole of fructose, we use the provided Gibbs free energy change (ΔG°') of fructose and compound X. The energy liberated from the combustion of fructose is given as -2810 kJ/mol, and the energy required to synthesize one mole of compound X is -64.1 kJ/mol.
By dividing the total energy released by combustion by the energy required to synthesize one mole of compound X, we can find out the number of moles of compound X that can be theoretically produced.
Here’s the calculation:
Energy released by combustion of fructose: -2810 kJEnergy required for synthesis of one mole of compound X: -64.1 kJ/molNumber of moles of compound X = Total energy released / Energy required per mole of compound XNumber of moles of compound X = 2810 kJ / 64.1 kJ/molNumber of moles of compound X = 43.84 molRounded to two significant figures, the number of moles of compound X that can be theoretically generated is 44 moles.
Question 25 An aqueous solution at has a concentration of . Calculate the concentration. Be sure your answer has the correct number of significant digits.
The given question is lacking some details, the complete question is following
Question:
An aqueous solution at 25 °C has a OH⁻ concentration of 2.5 x 10⁻⁴ M . Calculate the H₃O⁺ concentration. Be sure your answer has the correct number of significant digits
Answer:
Concentration of H₃O⁺ is:
[tex][H_{3}O^{+}]=4.0X10^{-9} M[/tex]
Explanation:
In aqueous solutions the product of the concentration of hydronium ions H₃O⁺ and hydroxide ions OH⁻ is 1.0 x 10⁻¹⁴. This value is the dissociation or ionization constant of water at 25 °C. Its formula is given as:
[tex]Kw = [H_{3} O^{+}][OH^{-} ][/tex]
[tex]1.0 X 10^{-14}= [H_{3}O^{+}](2.5 X 10^{-4})[/tex]
[tex][H_{3}O^{+}]= \frac{1.0X10^{-14}}{2.5 X 10^{-4}}[/tex]
[tex][H_{3}O^{+}]=4.0X10^{-9} M[/tex]
P.S: As the smallest number of significant figure in the ratio was two, so the answer contains two significant figures.
To calculate the hydronium ion concentration from the hydroxide ion concentration of 0.001 M at 25 °C, use the water ion-product constant, Kw (1.0 × 10^-14), and the inverse relationship between [H3O+] and [OH-] to find [H3O+] = 1.0 × 10^-11 M.
Explanation:To calculate the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C, we use the ion-product constant for water (Kw), which is 1.0 × 10-14 at 25 °C. The concentration of hydronium ions [H3O+] and hydroxide ions [OH-] are inversely proportional, which means as the concentration of one goes up, the other goes down. Therefore, the calculation for the hydronium ion concentration can be done using the equation:
Kw = [H3O+] × [OH-]
Substituting in the values we have:
1.0 × 10-14 = [H3O+] × 0.001
Therefore, the concentration of hydronium ions [H3O+] is:
[H3O+] = ⅖{1.0 × 10-14}{0.001}
[H3O+] = 1.0 × 10-11 M
Make sure that your final answer has the correct number of significant figures, which is guided by the number of significant figures in the given hydroxide ion concentration (0.001 M has one significant figure).
A mixture of He and Ne need to be separated. What will be the ratio of the rates of effusion of He and Ne?
Answer:
Ratio of rates of effusion of He to Ne is 2.245
Explanation:
According to Graham's law rate of effusion is inversely proportional to square root of molar mass of a gas
So, [tex]\frac{r_{He}}{r_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{He}}}[/tex]
where, [tex]r_{He}[/tex] and [tex]r_{Ne}[/tex] are rate of effusion of He and Ne respectively. [tex]M_{He}[/tex] and [tex]M_{Ne}[/tex] are molar mass of He and Ne respectively.
Molar mass of He = 4.003 g/mol
Molar mass of Ne = 20.18 g/mol
So, [tex]\frac{r_{He}}{r_{Ne}}=\sqrt{\frac{20.18}{4.003}}=2.245[/tex]
So, ratio of rates of effusion of He to Ne is 2.245
A chemist mixes 71.1 g of carbon tetrachloride with 65.4 g of isopropenylbenzene and 4.02 g of 2-ethyltoluene. Calculate the percent by mass of each component of this solution. Be sure each of your answer entries has the correct number of significant digits mass percentー1 □ component 0% carbon tetrachloride isopropenylbenzene 2-ethyltoluene
Answer:
∴ wt% CCl4 = 50.597 %
∴ wt% C9H10 = 46.541 %
∴ wt% C9H12 = 2.861 %
Explanation:
wt% = ((mass solute)/(mass solution))×100⇒ mass sln = 71.1 g CCl4 + 65.4 C9H10 + 4.02 g C9H12
⇒ mass sln = 140.52 g
∴ CCl4:
⇒ wt% CCl4 = ((71.1 g CCl4)/(140.82 g sln))×100
⇒ wt% CCl4 = 50.597 %
∴ C9H10:
⇒ wt% C9H10 = ((65.4 g C9H10)/(140.52 g sln))×100
⇒ wt% C9H10 = 46.541 %
∴ C9H12:
⇒ wt% C9H12 = ((4.02 g C9H12)/(140.52 g sln))×100
⇒ wt% C9H12 = 2.861 %
The percent by mass of each component in the solution is calculated by dividing each component's mass by the total mass and multiplying by 100. The results are 50.62% for carbon tetrachloride, 46.53% for isopropenylbenzene, and 2.86% for 2-ethyltoluene.
Explanation:In order to calculate the percent by mass of each component of the solution, the total mass of the solution needs to be determined first. The total mass would be obtained by summing the individual masses of the carbon tetrachloride, isopropenylbenzene, and 2-ethyltoluene. That is, 71.1 g + 65.4 g + 4.02 g = 140.52 g.
Then, the mass of each component is divided by the total mass and multiplied by 100 to obtain the percentage.
For carbon tetrachloride, the calculation would be (71.1 / 140.52) x 100 = 50.62%. For isopropenylbenzene, the calculation would be (65.4 / 140.52) x 100 = 46.53%. For 2-ethyltoluene, the calculation would be (4.02 / 140.52) x 100 = 2.86%.
Our three significant figures in each of the percentages are due to the fact that our least precise measurement, 4.02 g of 2-ethyltoluene, has four significant figures.
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The percent yield of a reaction between elemental zinc and an aqueous solution of 0.50 M hydro-chloric acid is known to be 78.0%. We need to produce 35.5 g of zinc chloride, what is the minimum amount in mL of hydrochloric acid that are required, given that zinc is in excess
Answer:
1.3 × 10³ mL
Explanation:
Let's consider the following reaction.
Zn + 2 HCl → ZnCl₂ + H₂
The percent yield is 78.0%. The real yield (R) of zinc chloride is 35.5 g. The theoretical yield (T) of zinc chloride is:
35.5 g (R) × (100 g T/ 78.0 g R) = 45.5 g T
The molar mass of zinc chloride is 136.29 g/mol. The moles corresponding to 45.5 g of zinc chloride is:
45.5 g × (1 mol/ 136.29 g) = 0.334 mol
The molar ratio of HCl to ZnCl₂ is 2:1. The moles of HCl that react with 0.334 moles of ZnCl₂ are 2 × 0.334 mol = 0.668 mol.
We need 0.668 moles of a 0.50 M HCl solution. The volume required is:
0.668 mol × (1000 mL/0.50 mol) = 1.3 × 10³ mL
Halogenated compounds are particularly easy to identify by their mass spectra because chlorine and bromine occur naturally as mixtures of two abundant isotopes. Chlorine occurs as 35Cl (75.8%) and 37Cl (24.2%); bromine occurs as 79Br (50.7%) and 81Br (49.3%). For the compound Chlorocyclohexane, C6H11Cl: At what masses do the molecular ions occur? (List in order of increasing mass separated by commas, e.g. 120,122.) What are the percentages of each molecular ion?
Answer:
The increasing order of masses of molecule ions:
118 g/mol(75.8%) , 120 g/mol(24.2%)
Explanation:
Chlorine occurs as 35-Cl (75.8%) and 37-Cl (24.2%).
Atomic mass of 35-Cl = 35 g/mol
Atomic mass of 37-Cl = 37 g/mol
Mass of Chlorocyclohexane in which 35-cl is present as a chlorine atom: [tex]C_6H_{11}Cl[/tex]
[tex]=6\times 12 g/mol+11\times 1 g/mol+1\times 35 g/mol=118 g/mol[/tex]
Mass of Chlorocyclohexane in which 37-Cl is present as a chlorine atom: [tex]C_6H_{11}Cl[/tex]
[tex]=6\times 12 g/mol+11\times 1 g/mol+1\times 37 g/mol=120 g/mol[/tex]
The increasing order of masses of molecule ions:
118 g/mol(75.8%) < 120 g/mol(24.2%)
A noble gas atom in the liquid phase is on average surrounded by 10 other atoms. Each pair of atoms contribute g
Answer: noble gases are in reactive.
Explanation: noble gases are present in the right most corner of the periodic table in the 8th group. So their outermost shells are complete. Their boiling point, mass increases down the group. the have strong forces of interaction. their ionization energy decreases down the group
You need to prepare an acetate buffer of pH 5.31 5.31 from a 0.784 M 0.784 M acetic acid solution and a 2.85 M KOH 2.85 M KOH solution. If you have 930 mL 930 mL of the acetic acid solution, how many milliliters of the KOH KOH solution do you need to add to make a buffer of pH 5.31 5.31 ? The p K a pKa of acetic acid is 4.76. 4.76.
Answer:
214 milliliters of KOH needs to be added in 1 litre of 0.784 M acetic acid to make a acetate buffer of 5.31
Explanation:
To solve the problem, let us first use the Henderson-Hasselbalch equation to determine the amount of acetate needed to make a buffer of pH 5.31.
Henderson-Hasselbalch equation:
[tex]pH=pKa + log(\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]})[/tex]
Here, let us consider the moles of both species instead of the molar concentration, as the volume for both is the same. Also, acetate will be formed by the neutralization of acetic acid, hence the final moles of acetic acid will be the difference of initial moles of acetic acid and the moles of acetate formed. Now the equation becomes as follows:
[tex]pH=pKa + log(\frac{n_{ CH_{3}COO^{-}}}{n_{iCH_{3}COOH}-n_{ CH_{3}COO^{-}}}})[/tex]
From given data
pH = 5.31
pKa = 4.76
n(CH₃COO⁻) = ?
ni(CH₃COOH) = 0.784 mol (initial moles of acetic acid)
Placing the data in the equation, we get:
[tex]5.31=4.76 + log(\frac{n_{ CH_{3}COO^{-}}}{0.784-n_{ CH_{3}COO^{-}}}})\\ \\ n_{ CH_{3}COO^{-}}=10^{5.31-4.76}(0.784-(n_{ CH_{3}COO^{-}}))\\ \\ n_{ CH_{3}COO^{-}}= 2.78 mol-3.55(n_{ CH_{3}COO^{-}})\\ \\ n_{ CH_{3}COO^{-}}= 0.61mol[/tex]
The molar ratio of KOH and CH₃COOH is 1:1, i.e 1 mol of KOH will react with CH₃COOH and give 1 mol of acetate (CH₃COO⁻). Hence, 0.61 mol of KOH will give 0.61 mol of KOH. Now to determine the volume of 2.85 M KOH that contains 0.61 moles:
[tex]M_{KOH} =\frac{n_{KOH} }{V_{KOH} (L)}[/tex]
[tex]2.85=\frac{0.61}{V_{KOH} (L)}\\ \\ V_{KOH} (L)=\frac{0.61}{2.85}\\ \\ V_{KOH}=0.214 litre[/tex]
Finally convert liter into milliliter dividing by 1000 (mL/L)
Volume of KOH required = 214 milliliters
What challenges would you encounter with the titration if you had used apple cider vinegar or balsamic vinegar as the analyte instead of white vinegar?
Explanation:
Apple cider vinegar or balsamic vinegar might contain various synthetics which might interact with or nullified the findings, whereas white vinegar comprises of acidic acid only. In fact, apple cider vinegar and balsamic vinegar are deep in appearance this would make it very difficult to determine the color. This is why it is preferable to use white vinegar in the Titration process.
The challenge that would be encountered with the titration if you had used balsamic vinegar as an analyte is the inability to determine the endpoint of the titration with your naked eye.
An analyte, also known as a titrand is a solution whose concentration is needed to be determined. During titration, it will be difficult and impossible to determine the endpoint of titration because it will be difficult to catch the color changes with visible eyes.
To eliminate this difficulty, balsamic vinegar needs to be diluted with enough water to detect the color change, or the use of white vinegar is used.
Similarly, apple cider, as well as balsamic vinegar, comprises additional acid than acetic acid, this will increase the level of acidity during the titration, unlike white vinegar that contains only acetic acid.
Thus, the acid in apple cider vinegar or balsamic vinegar will be more and higher during titration compared to that of white vinegar.
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A 8-kg plastic tank that has a volume of 0.18 m3 is filled with liquid water. Assuming the density of water is 1000 kg/m3, determine the weight of the combined system.
Answer:
1.84 × 10³ N
Explanation:
To determine the weight of the system, we will determine its mass which is equal to the sum of the masses of the plastic tank (8 kg) and the water.
We have 0.18 m³ of water with a density of 1000 kg/m³. Its mass is:
0.18 m³ × 1000 kg/m³ = 180 kg
The mass of the system is 180 kg + 8 kg = 188 kg.
We can find the weight (w) of the system using Newton's second law of motion.
w = m × g = 188 kg × 9.81 m/s² = 1.84 × 10³ N
where
g is the gravity
The weight of the combined system, including the plastic tank and the water, is 180 kg.
Explanation:The weight of an object depends on its mass and the gravitational force acting on it. To find the weight of the combined system, we need to determine the mass of the plastic tank and the water it contains. The 8-kg plastic tank does not contribute to the weight of the system, but the weight of the water can be calculated using its density.
The volume of the water is given as 0.18 m³, and the density of water is 1000 kg/m³. Multiplying these values together will give us the mass of the water:
Weight of the water = density × volume = 1000 kg/m³ × 0.18 m³ = 180 kg
Therefore, the weight of the combined system, including the plastic tank and the water, is 180 kg.
Nitrogen monoxide and hydrogen react to form nitrogen and water, like this: 2NO+2 H2(g)-N29)+2H2O(9) Write the pressure equilibrium constant expression for this reaction
Answer:
Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)² = Kp
Explanation:
The reaction is:
2NO + 2H₂ → N₂ + 2H₂O
The expression for Kp (pressure equilibrium constant) would be:
Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)²
There is another expression for Kp, where you work with Kc (equilibrium constant)
Kp = Kc (R.T)^Δn
where R is the Ideal Gases constant
T° is absolute temperature
Δn = moles of gases formed - moles of gases, I had initially
The pressure equilibrium constant expression for the reaction 2NO + 2H2 -> N2 + 2H2O is Kp = (p(N2))^1 * (p(H2O))^2 / (p(NO))^2 * (p(H2))^2.
Explanation:The pressure equilibrium constant expression is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients.
For the reaction 2NO + 2H2 -> N2 + 2H2O, the pressure equilibrium constant expression can be written as:
Kp = (p(N2))^1 * (p(H2O))^2 / (p(NO))^2 * (p(H2))^2
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Give the balanced equation for the following.
The combustion of liquid butanol (C4H9OH) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
Answer:
The balanced chemical reaction is given as:
[tex]C_4H_9OH(l)+6O_2(g)\rightarrow 4CO_2(g)+5H_2O(g)[/tex]
Explanation:
Combustion is defined as chemical reaction which an organic compounds reacts with oxygen gas to to give water and carbon dioxide as a products along with releases of heat energy.
The combustion of liquid butanol gives water vapor and carbon dioxide as a product and this reaction is given as:
[tex]C_4H_9OH(l)+6O_2(g)\rightarrow 4CO_2(g)+5H_2O(g)[/tex]
According to reaction, 1 mole of butanol reacts with 6 moles of oxygen gas to gives 4 moles of carbon dioxide gas and 5 moles of water vapors.
The combustion of liquid butanol, in the presence of oxygen, forms carbon dioxide and water vapor. The balanced chemical equation is: C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(g).
Explanation:The combustion of liquid butanol (C4H9OH) in the presence of oxygen (O2) produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation for the combustion of liquid butanol can be written as:
C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(g)
Here, C4H9OH(l) represents liquid butanol, O2(g) represents oxygen gas, CO2(g) represents carbon dioxide gas and H2O(g) represents water vapor. This equation states that one mole of butanol reacts with six moles of oxygen to form four moles of carbon dioxide and five moles of water vapor.
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A chemist adds of a zinc nitrate solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.
Answer:
5.3 × 10⁻³ kg
Explanation:
There is some info missing. I think this is the original question.
A chemist adds 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
We have 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution. The moles of zinc nitrate are:
0.1350 L × 0.21 mol/L = 2.8 × 10⁻² mol
The molar mass of zinc nitrate is 189.36 g/mol. The mass corresponding to 2.8 × 10⁻² moles is:
2.8 × 10⁻² mol × 189.36 g/mol = 5.3 g
1 kilogram is equal to 1000 grams. Then,
5.3 g × (1 kg/1000 g) = 5.3 × 10⁻³ kg
A solid substance has a vapor pressure of 100 mm Hg at the melting point (100 °C).
Assuming an atmospheric pressure of 760 mm Hg, describe the behavior of this solid as the temperature is raised from room temperature to its melting point.
Answer:
The melting point would be 250. That is the answer I think
Explanation:
The solid substance will turn into vapor when the temperature is increased from room temperature into meeting point temperature.
The given parameters;
Vapor pressure of the solid, [tex]P_{vap}[/tex] = 100 mmHgMelting point of the solid, T = 100 °C.Atmospheric pressure, [tex]P_{atm}[/tex] = 760 mmHgRoom temperature, [tex]T_0[/tex] = 20 °CA substance boils when the atmospheric pressure is equal to the vapor pressure.
Since the vapor pressure of the solid substance occurs at the meeting temperature of the substance, once the temperature of the solid substance is increased to the meeting point temperature, the solid substance will turn into vapor (gas).
Thus, we can conclude that the solid substance will turn into vapor when the temperature is increased from room temperature into meeting point temperature.
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A chemist prepares a solution of mercury(II) iodide by weighing out of mercury(II) iodide into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.
The question is incomplete, here is the complete question:
A chemist prepares a solution of mercury(II) iodide [tex](HgI_2)[/tex] by measuring out 0.0122 µmol of mercury(II) iodide into a 400 mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.
Answer: The molarity of chemist's mercury (II) iodide solution is [tex]3.05\times 10^{-8}mol/L[/tex]
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
We are given:
Moles of mercury (II) iodide = [tex]0.0122\mu mol=0.0122\times 10^{-6}mol[/tex] (Conversion factor: [tex]1mol=10^6\mu mol[/tex] )
Volume of solution = 400. mL
Putting values in above equation, we get:
[tex]\text{Molarity of }HgI_2=\frac{0.0122\times 10^{-6}\times 1000}{400.}\\\\\text{Molarity of }HgI_2=3.05\times 10^{-8}mol/L[/tex]
Hence, the molarity of chemist's mercury (II) iodide solution is [tex]3.05\times 10^{-8}mol/L[/tex]
A covalent compound is composed of a nitrogen atom, a hydrogen atom, and a carbon atom. How many electrons should the Lewis structure of the covalent compound contain?
Answer:
The answer is 10 electrons
Explanation:
From periodic table its possible to find
Nitrogen atom has 5 VE (Valence)
Hydrogen atom has 1 VE (Valence)
Carbon atom has 4 VE ( Valence)
5+1+4=10 Electrons
The Lewis structure of a covalent compound containing one nitrogen atom, one hydrogen atom, and one carbon atom would contain 10 electrons. This is calculated by summing up the valence electrons of each atom: nitrogen (5), hydrogen (1), and carbon (4).
Explanation:The Lewis structure of a covalent compound represents the arrangement of atoms and the bonding electrons. In this case, we have one nitrogen atom, one hydrogen atom, and one carbon atom. Nitrogen has 5 valence electrons, hydrogen has 1, and carbon has 4. Altogether, the Lewis structure of this covalent compound should contain 10 electrons.
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luminum and oxygen react according to the following equation: 4Al(s) +3O2(g) --> 2Al2O3(s) What mass of Al2O3, in grams, can be made by reacting 4.6 g Al with excess oxygen?
Answer: 8.7 grams
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.6g}{27g/mol}=0.17moles[/tex]
[tex]4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)[/tex]
As oxygen is in excess, Aluminium is the limiting reagent and limits the formation of products.
According to stoichiometry:
4 moles of aluminium give = 2 moles of [tex]Al_2O_3(s)[/tex]
Thus 0.17 moles of aluminium give=[tex]\frac{2}{4}\times 0.17=0.085mol[/tex]
Mass of [tex]Al_2O_3=moles\times {\text {molar mass}}=0.085\times 102g/mol=8.7g[/tex]
Thus the mass of [tex]Al_2O_3(s)[/tex] is 8.7 grams
The results from an iodine-deficient diet, which causes a hyperplasia or overgrowth of the thyroid gland ______
Answer:
Goiter.
Explanation:
The thyroid gland is responsible for the secretion of the triidotyronine and thyroxine hormone in the body. The thyroid hormone regulates the metabolism and amount of particular ions in the body.
The iodine acts as the precursor for the maintenance of the thyroxine. The deficiency of the thyroid hormone can cause goiter. The thyroid gland gets swollen and cause problem in the breathing. The production of mucus and cough is common in goiter.
Thus, the answer is goiter.
Ne ( g ) effuses at a rate that is ______ times that of Cl 2 ( g ) under the same conditions.
Answer: 1.88times as that of Cl2
Explanation:
According to Graham law of effusion , the rate of effusion is inversely proportional to the square root of the molar mass
Rate= 1/√M
R1/R2 =√M2/M1
Let the rate of diffusion of Ne= R1
And rate of diffusion of Cl2 = R2
M1 ,molar mass of Ne= 20g/mol
M2,molar mass of Cl2 =71g/mol
R1/R2 = √ (71/20)
R1/R2 = 1.88
R1= 1.88R2
Therefore the Ne effuses at rate that is 1.88times than that of Cl2 at the same condition.
Using Graham's law of effusion, Ne gas is found to effuse approximately 1.88 times faster than Cl₂ gas under the same conditions.
This is because the rate of effusion is inversely proportional to the square root of molar mass.
To determine the rate at which Neon (Ne) effuses compared to Chlorine (Cl₂) under the same conditions, we can use Graham's law of effusion. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
The molar mass of Ne is approximately 20 g/mol, and the molar mass of Cl₂ is approximately 71 g/mol. Applying Graham's law:
(Rate of Ne) / (Rate of Cl₂) = [tex]\sqrt{sqrt(Molar mass of Cl_{2} / Molar mass of Ne)}[/tex]
This means Ne gas effuses approximately 1.88 times faster than Cl₂ gas under the same conditions.
Concentrated hydrochloric acid is an aqueous solution that is 34.70 % HCl. The density of the solution is 1.19 g/mL. What mass of HCl is contained in 0.356 L of solution?
The mass of HCl that is contained in the solution is 147 g HCl
Why?
To find the mass of HCl we have to apply what is called a conversion factor. In a conversion factor we put the units we don't want at the bottom, and the ones we want at the top.
For this question, we want to go from liters of solution to mass of HCl, and the conversion factor is laid out as follows:
[tex]0.356Lsolution*\frac{1000mL}{1L}*\frac{1.19 g solution}{1 mL solution}*\frac{34.70 g HCl}{100 g solution}=147 g HCl[/tex]
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Under equilibrium condition, what is the probability of an electron state being occupied if it is located at the Fermi level
Answer:
Hi
The Fermi Level is a term used to describe the set of electron energy levels at a temperature of absolute zero. Fermi's energy concept is important for the understanding of the electrical and thermal properties of solids. Both electrical and thermal processes involve energy values of a small fraction of an electron-volt. In thermal equilibrium, the net current of both electrons and holes is zero.
Explanation:
Under equilibrium conditions, the probability of an electron state being occupied if it is located at the Fermi level is 50 percent probability.
What is Fermi Level?Fermi Level is the term that is used to define the highest or most optimal energy level that an electron can occupy when the temperature is at absolute zero.
The Fermi level is equidistant between the valence band and conduction band. This is because when the temperature is at absolute zero, the electrons are all in the lowest state of energy.
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For each of the following reactions, give the balanced equation for the reaction and state the meaning of the equation in terms of the numbers of individual molecules and in terms of moles of molecules.
a. MnO2(s) + Al(s)-Mn(s) + Al2O3(s)
b. B203(s) + CaF2(s)-BF3(g) + CaO(s)
c. NO2(g) + H2O)> HNO;(aq) + NO(g)
d. C6H6(g) + H2(g)â C6H12(g) 12)
Answer:
a. 3MnO2 + 4Al —> 3Mn+ 2Al2O3
3 moles of MnO2 required 4 moles of Al to produce 3 moles of Mn and 2moles of 2Al2O3
b. B203 + 3CaF2 —> 2BF3 + 3CaO
1mole of B203 requires 3 moles of CaF2 to produce 2moles of BF3 and 3 moles of CaO
c. 3NO2 + H2O —> 2HNO3 + NO
3 moles of NO2 requires 1mole of H2O to produce 2moles of HNO3 and 1mole of NO
d. C6H6 + 3H2 —> C6H12
1mole of C6H6 requires 3 moles of H2 to produce 1mole of C6H12
Final answer:
To balance the equations, ensure there are equal numbers of each atom. The balanced equations and their meanings in terms of molecules and moles are...
Explanation:
In order to balance the equations, you need to ensure that there are equal numbers of each type of atom on both sides of the equation. Here are the balanced equations for each reaction:
a. MnO2(s) + 2Al(s) -> Mn(s) + Al2O3(s)
b. B2O3(s) + 3CaF2(s) -> 2BF3(g) + 3CaO(s)
c. 2NO2(g) + H2O(l) -> HNO2(aq) + NO(g)
d. C6H6(g) + 3H2(g) -> C6H12(g)
In terms of the number of individual molecules, the balanced equation shows the ratio in which the reactants combine to form the products. In terms of moles of molecules, the balanced equation allows you to calculate the amount of each substance involved in the reaction using the mole ratio.
If the pH of a solution is 2.78 with an absolute uncertainty of ±0.05, what is the LaTeX: \left[H^+\right][ H + ] and its uncertainty?
Answer:
[H⁺] = 16.6 (±1.9) x10⁻⁴ M
Explanation:
The pH is defined as:
pH = -log[H⁺]So we can calculate [H⁺]:
2.78 = -log[H⁺]-2.78 = log[H⁺][H⁺] = [tex]10^{-2.78}[/tex][H⁺] = 1.66x10⁻³ M
The relative uncertainty in [H⁺] is
uR/R = 2.303 x AbsoluteUncertaintyuR/R = 2.303 * 0.05 = 0.115Thus the uncertainty in the concentration is:
1.66x10⁻³ M * 0.115 = 1.91x10⁻⁴ M
Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = 6.304 x 10^-3 in.) can be produced from the copper in 6.95 lb of covellite, an ore of copper is 66.0% copper by mass? (Hint: Treat the wire as a cylinder: V for cylinder = (pi)(r^2)(h); d of copper = 8.95 g/cm^3)
Answer:
11,547.67 meters of copper can be drawn.
Explanation:
Mass of mineral = 6.95 lb = 6.95 × 453.592 = 3,152.46 g
1 lbs = 453.592 g
An ore of copper is 66.0% copper by mass, So mas of copper in 3,152.46 grams of ore= m
[tex]m = \frac{66.0}{100}\times 3,152.46 g=2080.63 g[/tex]
Volume of copper = V
Density of copper = d = [tex]8.95 /cm^3[/tex]
[tex]V=\frac{m}{d}=\frac{2080.63 g}{8.95 /cm^3}=232.47 cm^3[/tex]
Diameter of the wire drawn from the [tex]232.47 cm^3[/tex] of copper= d
d = [tex]6.304\times 10^{-3} inch=6.304\times 10^{-3}\times 2.54 cm=0.01601 cm[/tex]
(1 inch= 2.54 cm)
Radius of the wire= r = 0.5 × d =0.5 × 0.01601 =0.008005 cm
Length of the wire = h
Volume of the cylindrical wire = [tex]\pi r^2 h[/tex]
[tex]V=\pi r^2 h[/tex]
[tex]232.47 cm^3=3.14\times (0.008005 cm)^2\times h[/tex]
Solving for h :
h =1,154,767.015 cm
1 cm = 0.01 m
h = 1,154,767.015 cm = 1,154,767.015× 0.01 m = 11,547.67 m
11,547.67 meters of copper can be drawn.
Formation of which type of product--byproduct, side product, both, or neither--always coincides with a loss in percent yield of the desired product?
Answer:
Side products coincides with a loss in percentage yield of the product
Explanation:
- The product yield is defined as
product yield= actual yield/theoretical yield
- Byproducts are undesired products that come from the same reaction pathway that our products → even if the reaction procedes ideally , we will always have byproducts → byproducts do not alter the Yield by themselves
- Side products , however , come from undesired products from side reactions and consume reactants due to these alternative pathways without producing our product → side products do alter the yield of the product