Answer:
(+)-cholesterol = 71.88%
(-)-cholesterol = 28.12%
Explanation:
Asuming 1 gram of sample is dissolved in 1mL of water and the sample cell was 1dm long.
Enantiomeric excess is defined as the amount of pure enantiomer in a sample. The formula is:
ee = [α]mixture / [α]pure enantiomer.
Replacing:
ee = 14° / 32°×100 = 43.75%
As the sample is 14°, There is an excess of (+)-cholesterol and 56.25% is a 1:1 mixture of enantiomers.
That means percent composition of enantiomers is:
(+)-cholesterol = 43.75% + 56.25%/2 = 71.88%(-)-cholesterol = 56.25%/2 = 28.12%when dissolved in water, an acid or a base breaks down into a. a proton and an electron b. two negative ions c. a positive and a negative ion d. a positive ion and a proton
Answer:
C. A positive and a negative ion
Explanation:
Acids and bases are made up of charged particles known as ions. The ions present in acids are oppositely charged and are held together by strong electrostatic forces. When acids or bases are dissolved in water, the electrostatic forces holding their individual molecules together are weakened and these ions are free to move apart in a process known as dissociation. Dissociation occurs because of the attraction between the positive and negative ions in the acid and bases and the negative and positive polarity of water.
For example, when an acid like hydrochloric acid is dissolved in water it dissociates into positive and negative ions as follows:
HCl(aq) -----> H+ + Cl-
When a base like sodium hydroxide is dissolved in water, it dissociates into positive and negative ions as follows:
NaOH(aq) ----> Na+ + OH-
Answer:
yeah C is correct
Explanation:
A 3.00-g sample of an alloy (containing only Pb and Sn) was dissolved in nitric acid (HNO3). Sulfuric acid was added to this solution, which precipitated 2.93 g of PbSO4. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample? (molar mass of PbSO4 = 303.3 g/mol)
Answer:
33.3% of Sn in the sample
Explanation:
The addition of SO₄⁻ ions produce the selective precipitation of Pb²⁺ to produce PbSO₄.
Moles of PbSO₄ (molar mass 303.26g/mol) in 2.93g are:
2.93g ₓ (1mol / 303.26) = 9.66x10⁻³ moles PbSO₄ = Moles Pb²⁺.
As molar mass of Pb is 207.2g/mol, mass in 9.66x10⁻³ moles of Pb²⁺ is:
9.66x10⁻³ moles of Pb²⁺ ₓ (207.2g / mol) = 2.00g of Pb²⁺
As mass of the sample is 3.00g, mass of Sn²⁺ is 3.00g - 2.00g = 1.00g
And the percentage of Sn in the sample is:
1.00g / 3.00g ₓ 100 =
33.3% of Sn in the sampleA pentavalent cation atom has 20 and 15 neutrons as protons. Find the electron quantity and mass number respectively. (40 pts.) a) 20 and 15 b) 15 and 20 c) 15 and 35 d) 35 and 15 e) 10 and 20
Answer:
C.
Explanation:
Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.
Draw the structure 2 butylbutane
Answer:
please look at the picture below.
Explanation:
Which element has the largest atomic radius
Answer:
Francium
Explanation:
The atomic radius increases from top to bottom in a group, and decreases from left to right across a period.
so francium (Fr) is the largest atom or has highest radii.
Hope this helps & please mark as brainiest!
Answer:
Francium has the largest atomic radius.
The general trend for atomic radii is increasing from top to bottom and decreasing from left to right so the one with the largest atomic radius will be in the bottom left of the periodic table.
When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.430 g of a particular hydrocarbon was burned in air, 0.446 g of CO, 0.700 g of CO2, and 0.430 g of H2O were formed.
Required:
a. What is the empirical formula of the compound?
b. How many grams of O2 were used in the reaction?
c. How many grams would have been required for complete combustion?
Answer:
(a) The empirical formula of the compound is
m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).
(b) The grams of O2 that were used in the reaction is 1.146 g
(c) The amount of O2 that would have been required for complete combustion is 1.401 g.
Explanation:
a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)
(b) Using law of conservation of mass from above
m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)
m(O2) = 0.446 + 0.700 + 0.430 - 0.430
m(O2) = 1.146 g
The grams of O2 that were used in the reaction is 1.146 g
(c) for complete combustion, we need to oxidized CO to CO2
Then, 2CO +O2 = 2CO2
m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}
m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g
Note; Molar mass of O2 = 32, CO = 28
m(total)(O2) = m(O2) + m(add)(O2)
m(total)(O2) = 1.146 + 0.255 = 1.401 g
The amount of that grams would have been required for complete combustion is 1.401 g.
Note (add) and (total) were used subscript to "m"
Write the empirical formula
Answer:
[tex]1) NH_{4}IO_{3}\\2) Pb(IO_{3})_{4} \\3) NH_{4}(C_{2}H_{3}O_{2})\\4) Pb(C_{2}H_{3}O_{2})_{4}[/tex]
Explanation:
[tex]1) NH_{4}^{+}IO_{3}^{-} ---> NH_{4}IO_{3}\\2) Pb^{4+}(IO_{3}^{-})_{4} --->Pb(IO_{3})_{4} \\3) NH_{4}^{+}(C_{2}H_{3}O_{2})^{-} ---> NH_{4}(C_{2}H_{3}O_{2})\\4) Pb^{4+}(C_{2}H_{3}O_{2})^{-} _{4} --->Pb(C_{2}H_{3}O_{2})_{4}[/tex]
What is the mass of 3.45 moles
NO2?
(N = 14.01 g/mol, O = 16.00 g/mol)
Answer:
158.7 g
Its the right answer
How many oxygen molecules are needed to make 10 carbon dioxide molecules according to the following balanced chemical equation? 2 CO + O2 → 2 CO2
five oxygen molecules
step by step explanation.
according to the equation,one molecule of oxygen is enough to react with two carbon molecules thus 10 carbon molecules need 5oxygen molecules
Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [
Answer: pH=2.38
Explanation:
To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.
HCOOH ⇄ H⁺ + HCOO⁻
I 1.0M 0 0
C -x +x +x
E 1.0-x x x
For the steps below, refer to the ICE chart above.
1. Since we were given the initial of HCOOH, we can fill this into the chart.
2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.
3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.
4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].
5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.
[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]
6. Once we plug this into the quadratic equation, we get x=0.00415.
7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).
-log(0.00415)=2.38
Our pH for the weak acid solution is 2.38.
what is the molarity of a solution that contains 49.8 grams of nai and is dissolved in enough water to make 1.50 liters
Answer: The molarity of solution is 0.221 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in L
moles of [tex]NaI[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{49.8g}{149.89g/mol}=0.332mol[/tex]
Now put all the given values in the formula of molality, we get
[tex]Molarity=\frac{0.332mol}{1.50L}[/tex]
[tex]Molarity=0.221mol/L[/tex]
Therefore, the molarity of solution is 0.221 M
Write the empirical formula
Answer:
[tex]Pb(CO_{3})_{2} \\Pb(NO_{3})_{4} \\FeCO_{3}\\Fe(NO_{3})_{2}[/tex]
Explanation:
[tex]Pb^{4+}(CO_{3}^{2-})_{2} --->Pb(CO_{3})_{2} \\Pb^{4+} (NO_{3}^{-})_{4} --->Pb(NO_{3})_{4} \\Fe^{2+} CO_{3}^{2-} --->FeCO_{3}\\Fe^{2+} (NO_{3}^{-})_{2}--->Fe(NO_{3})_{2}[/tex]
Description (with words) of water just above melting temperature. What intermolecular forces do you expect to find in water in liquid state
Answer:
intermolecular dipole-dipole hydrogen bonds
Explanation:
Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.
Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and placed in 112.0 mL of water at 20.5 ∘C . What is the final temperature of the water when thermal equilibrium is reached?
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
The mathematics of combining quantum theory with wave motion of atomic particles is known as _____.
Combining quantum theory with wave motion of atomic particles is: Wave Mechanics
Which correctly lists three characteristics of minerals?
solid, crystal structure, definite chemical composition
organic, crystal structure, definite chemical composition
human-made, solid, organic
crystal structure, definite chemical composition, human-made
Answer:a
Explanation:
The three characteristics of minerals are that they are solid, have definite crystal structure and definite chemical composition.
What are minerals?Minerals are defined as a chemical compound which has a well -defined composition and possesses a specific crystal structure.It occurs naturally in the pure form.
If a compound occurs naturally in different crystal structure then each structure is considered as a different mineral.The chemical composition of a mineral varies depending on the presence of small impurities which are present in small quantities.
Some minerals can have variable proportions of two or more chemical elements which occupy equivalent position in the crystal structure.It may also have variable composition which is split into separate species.
Physical properties of minerals include color,streak, luster,specific gravity and cleavage.
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A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hydrazoic acid = 1.9 x 10-5.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M
Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M
GIven that :
Ka = [tex]1.9 x 10^{-5}[/tex]
Thus; it's pKa = 4.72
[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]
[tex]pH =4.72 + log(1.1376)[/tex]
[tex]pH =4.72 + 0.05598[/tex]
[tex]pH =4.77598[/tex]
pH ≅ 4.80
The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.
How we calculate the pH?pH of the given solution will be used by using the following equation:
pH = pKa + log[conjugate base] / [weak acid]
Given chemical reaction will be represented as:
HN₃ + OH⁻ → N₃⁻ + H₂O
Moles will be calculated as:
n = M×V, where
M = molarity
V = volume
Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375 mol
Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol
From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:
0.00375 - 0.001995 = 0.001755 mol
And 0.001995 mol of N₃⁻ is preduced by the reaction.
Total volume of the solution = 0.025 + 0.0133 = 0.0383 L
To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:
[N₃⁻] = 0.001995 mol / 0.0383 L = 0.0521 M
[HN₃] = 0.001755 mol / 0.0383 L = 0.0458 M
Ka for HN₃ = 1.9 × 10⁻⁵
pKa = -log( 1.9 × 10⁻⁵ ) = 4.72
On putting all these values on the above equation, we get
pH = 4.72 + log (0.0521) / (0.0458)
pH = 4.80
Hence, pH of the solution is 4.80.
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g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empirical formula of the compound?
Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O
Hypochlorous acid is formed in situ by reaction of aq. sodium hypochlorite solution with acetic acid. Draw balanced chemical equations to show the formation of hypochlorous acid and protonated hypochlorous acid.
Answer:
NaClO + CH₃COOH ----> HClO + CH3CO- + Na
Explanation:
This reaction occurs between the combination of a salt and an acid, that is, an oxide-reduction reaction
which factor is most responsible for the fact that water is a liquid rather than a gas at room temperature
The answer is Hydrogyn bonding. It keeps the water molocules bonded together and in a liquid state, without it it'd be in a gashious state.
Answer:Hydrogen bonds keep them together in room temperature, hope this helps!
Explanation:
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What is the specific heat of a 85.01 g piece of an unknown metal that exhibits a 45.2°C temperature change upon absorbing 1870 J of heat?
Answer:
[tex]0.48~\frac{J}{g~^{\circ}C}[/tex]
Explanation:
In this question, we have to remember the relationship between Q (heat) and the specific heat (Cp) the change in temperature (ΔT), and the mass (m).
[tex]Q=m*Cp*ΔT[/tex]
The next step is to identify what values we have:
[tex]Q~=~1870~J[/tex]
[tex]m~=~85.01~g[/tex]
[tex]ΔT~=~45.2~^{\circ}C[/tex]
[tex]Cp~=~X[/tex]
Now, we can plug the values and solve for "Cp":
[tex]1870~J=~85.01~g~*Cp*45.2~^{\circ}C[/tex]
[tex]Cp=\frac{1870~J}{85.01~g~*45.2~^{\circ}C}[/tex]
[tex]Cp=0.48~\frac{J}{g~^{\circ}C}[/tex]
The unknow metal it has a specific value of [tex]0.48~\frac{J}{g~^{\circ}C}[/tex]
I hope it helps!
What volume of 6.00 M hydrochloric acid is needed to prepare 500 mL of 0.100 M solution?
Answer:
8.33mL or .0083L
Explanation:
Use m1 * V1 = m2 * V2
6.00M(x) = 0.100M(500mL)
solve for x
x= (.1 * 500) / 6
x=8.333 mL
Cual es la diferencia entre agua pesada y agua ligera a) el agua pesada contiene mas minerales que el agua ligera b) el agua ligera es liquida mientras el agua pesada es solida c) el agua ligera es agua purificada y el agua pesada es agua contaminada d) el agua pesada contiene mas elementos estearato de sodio
Answer:
d) El agua pesada contiene mas elementos
Explanation:
La diferencia fundamental entre el agua pesada y el agua ligera es que la primera tiene una proporción mayor de deuterio que la segunda. El deuterio es un ión del hidrógeno que tiene un peso atómico mayor que el hidrógeno común y corriente. Por ende, la opción D ofrece la mejor aproximación.
Answer:
....................
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Explanation:ki
g Which statement is incorrect regarding oxidation? Oxidation is a "gain" of electrons. Oxidation is the combination with O atoms. Oxidation is an increase in oxidation state. Oxidation is always accompanied by reduction. none of these
Answer:
The incorrect statement from the options is OXIDATION IS A "GAIN" OF ELECTRONS
Explanation:
Oxidation in a redox reaction is the loss of electrons. It is also the increase in the oxidation states of an atom or ion or atoms in a molecule. A redox reaction is a type of chemical reaction in which there is a transfer of electrons from an atom or ion to another resulting in a change in oxidation states of the substances involved. The reducing agent in the reaction is undergoes oxidation by losing electrons while the oxidating agent is reduced that is it gains electrons at the end of the reaction. The atom or ion from which electron is lost is said to be oxidized while the other atom or ion involved in the reaction is reduced.
Oxidation is also the combination with O atoms and it is always accompanied by reduction because oxidation forms a half of the whole redox reaction. A substance cannot be oxidized except it has reduced another substance by losing electrons to it.
Pb(OH)Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe: PbO(s) NaCl(aq) H2O(l) --> Pb(OH)Cl(s) NaOH(aq) How many grams of PbO and how many grams of NaCl would be required to produce 10.0 g of Pb(OH)Cl
Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is
[tex]= \frac{Mass}{Molar\ mass}[/tex]
[tex]= \frac{10.0 g}{259.65g / mol}[/tex]
= 0.0385 mol
Mass of PbO needed is
[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]
After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is
[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]
After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.
(a) How many stereoisomers are possible for 4-methyl-1,2-cyclohexanediol? ___ (b) Name the stereoisomers formed by oxidation of (S)-4-methylcyclohexene with osmium tetroxide. If there is only one stereoisomer formed, leave the second space blank. Isomer #1: Isomer #2: (c) Is the product formed in step (b) optically active? _____
Answer:
See explanation
Explanation:
For the first part of the question, we have to check the chiral carbons in 4-methyl-1,2-cyclohexanediol. In this case carbons, 1 and 2 are chiral, if we have 2 chiral carbons we will have 4 isomers. We have to remember that formula 2^n in which "n" is the number of chiral carbons, so:
2^n = 2^2 = 4 isomers
And the isomers that we can have are:
1) (1R,2S)-4-methylcyclohexane-1,2-diol
2) (1S,2S)-4-methylcyclohexane-1,2-diol
3) (1S,2S)-4-methylcyclohexane-1,2-diol
4) (1S,2R)-4-methylcyclohexane-1,2-diol
See figure 1
For the second part of the question, we have to remember that the oxidation with [tex]OsO_4[/tex] is a syn addition. In other words, the "OHs" are added in the same plane. In this case, we have the methyl group with a wedge bond, so the "OH" groups will have a dashed bond due to the steric hindrance. Due to this we only can have 1 isomer ((1S,2R,4S)-4-methylcyclohexane-1,2-diol). Finally, on this molecule, we dont have any symmetry planes (this characteristic will cancel out the optical activity), so the product of this reaction has optical activity.
See figure 2
I hope it helps!
which process is used to produce gases from solutions of salts dissolved in water or another liquid?
A.Electrolysis
B.Metallic bonding
C.Ionic bonding
D. Polar covalent bonding
Answer:
A.Electrolysis
Explanation:
A.Electrolysis
For example, electrolysis of solution of NaCl in water gives H2 and O2.
Lewis structure of methyl metcaptain
Answer:
The lewis structure of the compounds can be drawn by making the skeleton of the molecule first. Then the different atoms are arranged and the electrons are arranged in their bonding pattern. The lone pair of the atoms, which are not involved in the bonding are represented by the dots.
So the lewis structures of both the compound methyl mercaptan has been in the attached image:
Spelling of methyl metcaptain is wrong, the correct spelling is methyl mercaptan.
Answer:
Methyl mercaptan is also known as Methanethiol with the chemical formula CH3SH and it is an organosulfur compound.
For lewis structure of methyl mercaptan (CH3SH), there are total 14 valence electrons. Four hydrogen atoms has one valence electron each, carbon has four valence electrons and sulfur has six valence electrons. Carbon form one bond with three hydrogen atoms by sharing one electron with each, carbon form one single bond with sulfur atom by sharing one electron with it and sulfur form one single bond with hydrogen. Sulfur left with four unpair electrons.