Answer:
a. [tex]Ag^0[/tex]
b. [tex]Ag^{+}[/tex]
c. [tex]Ca^{2+}F_2^-[/tex]
d. [tex]H_2^+S^{2-}[/tex]
e. [tex](C^{4+}O_3^{2-})^{-}[/tex]
f. [tex](Cr^{6+}O_4^{2-})^{2-}[/tex]
g. [tex]Cl_2^0[/tex]
h. [tex]Fe^0[/tex]
i. [tex]Cu^{2+}Cl_2^-[/tex]
j. [tex]C^{4-}H_4^+[/tex]
Explanation:
Hello there!
In this case, according to the concept of charge balance, which tell us that the overall charge is zero for any compound, except ions, it turns out possible to proceed as follows:
a. [tex]Ag^0[/tex]
b. [tex]Ag^{+}[/tex]
c. [tex]Ca^{2+}F_2^-[/tex]
d. [tex]H_2^+S^{2-}[/tex]
e. [tex](C^{4+}O_3^{2-})^{-}[/tex]
f. [tex](Cr^{6+}O_4^{2-})^{2-}[/tex]
g. [tex]Cl_2^0[/tex]
h. [tex]Fe^0[/tex]
i. [tex]Cu^{2+}Cl_2^-[/tex]
j. [tex]C^{4-}H_4^+[/tex]
Keep in mind lonely elements have 0 as their oxidation state.
Regards!
Write the number of sig. fig. in four numbers given in the sentence below. An (one) octopus has 8 legs. 13 octopi have 104 legs.
Give four answers.
A. Infinity, Infinity, Infinity, Infinity
B. 1, 1, 2, 3
C. Infinity, Infinity, 2, 3
D. No answer text provided.
Answer:
1, 1, 2, 3
Explanation:
The numbers 1 and 8 both have 1 sig. fig.
The number 13 has 2 sig. figs.
The number 104 has 3 sig. figs.
A scientific hypothesis is
ANSWER:
predictive.
testable.
explanatory.
all of the above.
Answer:
All of the above.
Explanation:
For a scientific hypothesis to be considered a hypothesis, it has to be testable. When conducting a lab experiment, it also allows the tester to predict what might occur during and after the experimentation. They are also explanatory. For example, theories are hypotheses that have been verified and can explain why something in nature takes place.
An atom has 20 electrons. Find out
i. It’s atomic numbers and total number of p-electrons
ii. The value of azimuthal quantum number (l) and magnetic quantum number (m) of the 19th electron of the atom.
iii. It’s group position in the periodic table.
Answer:
it's atomic number is 5 and total number is 10
The atom has an atomic number of 20 and has a total of 12 p electrons.
The azimuthal quantum number (l) of the 19th electron is 0 and the magnetic quantum number (m) of the 19th electron is 0.
It is an element of group 2
The number of electrons in the neutral atom is equal to the number of protons and is also the atomic number of an atom.
An atom is known to be electrically neutral. This is because the number of electrons in the atom is equal to the number of protons in the neutral atom.
The number of protons in the neutral atom is called the atomic number of the atom.
For an element that has 20 electrons, its electronic configuration is;
1s2 2s2 2p6 3s2 3p6 4s2.
The 19th electron is in the 4s orbital hence both the azimuthal and magnetic quantum numbers are zero.
The element has outermost electron configuration ns2 so it mus belong to group 2 of the periodic table.
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Suppose you ran this reaction without triethylamine and simply used an excess of reactant 1. At the end of the reaction, your methylene chloride solution would contain mostly reactant 1 and the product. What would you do to remove reactant 1 from the solution
ummm is that chemistry?
Answer:
is this chem
Explanation:
True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.
Answer:
True
Explanation:
The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.
The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.
Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.
Which equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
Given 0.60 mol CO2, 0.30 mol CO, and 0.10 mol H20, what is the partial pressure of the CO if the total pressure of the mixture was 0.80 atm?
Answer:
Explanation:
/ means divided by
* means multiply
1. formula is
partial pressure = no of moles(gas 1)/ no of moles(total)
0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->
.3/(.6+.3+.1) =
.3/1 =
.3 =
partial pressure of CO
2.
.3 * .8 atm = .24
khanacademy
quizlet
The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.
Dalton's Law of Partial pressureDalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.
Dalton's Law of partial pressure using mole fraction of gas
Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure
Now, we have to find the first mole fraction of CO
Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]
= [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]
= [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]
= [tex]\frac{0.30}{1}[/tex]
= 0.3
Now, put the value in above equation, we get that
Partial pressure of carbon monoxide (CO)
= Mole fraction of carbon monoxide (CO) × Total pressure
= 0.3 × 0.8
= 0.24 atm
Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.
Learn more about the Dalton's Law of partial Pressure here: https://brainly.com/question/14119417
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If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?
Answer:
1.88 A
Explanation:
Let's consider the reduction of copper in an electrolytic cell.
Cu²⁺ + 2 e⁻ ⇒ Cu
We can calculate the charge used to deposit 12.3 g of Cu using the following relations.
The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).The charge used is:
[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]
We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.
5.50 h × 3600 s/1 h = 1.98 × 10⁴ s
The current used is:
I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A
What would be the specific mathematical effect on the reaction rate if you carried out the sodium iodide-in-acetone reactions on the alkyl halides using an iodide solution half as concentrated? ("Slower" or "faster" is not specific enough.)
Answer:
Slower
Explanation:
The reaction between alkyl halides and sodium iodide-in-acetone is an SN2 reaction. The rate of reaction depends on the concentration of the alkyl halide as well as the concentration of the sodium iodide. It is a bimolecular reaction.
This means that if the concentration of any of the reactants is halved, the rate of reaction decreases accordingly.
Therefore, if the iodide solution is half as concentrated, the reaction is observed to be slower in accordance with the rate law;
Rate = k[alkyl halide] [iodide]
You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4. What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this
Answer:
e. HCOOH and NaCHOO
Explanation:
For a buffer solution, both an acid and its conjugate base are required.
With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.
Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:
H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO
State two conditions necessary for an esterification reaction to take place
Explanation:
Esterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.
Answer:
The Esterification ProcessThe Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.
The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.Once the -OH has been removed, the hydrogen on the alcohol can be removed and that oxygen can be connected to the carbon. Because the oxygen was already connected to a carbon, it is now connected to a carbon on both sides, and an ester is formed.
The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.Once the -OH has been removed, the hydrogen on the alcohol can be removed and that oxygen can be connected to the carbon. Because the oxygen was already connected to a carbon, it is now connected to a carbon on both sides, and an ester is formed.The methyl acetate that was formed is an ester. In this image, the green circle represents what was the carboxylic acid (in this case acetic acid), and the red circle represents what was the alcohol (in this case methanol):
This reaction lost an -OH from the carboxylic acid and a hydrogen from the alcohol. These two also combine to form water. So any esterification reaction will also form water as a side product.
bio-chemisty of protain
Answer:
Protein biochemistry is the study of proteins. Protein biochemistry is a scientific field dedicated to the study of proteins, complex chains of amino acids which make up the building blocks of all living organisms.
Explanation:
I hope that helped
Copy and Pasted!
Answer:
Listen to what guy said on top.
Explanation:
polypeptide structures consisting of one or more long chains of amino acids residue.....
or my answer
refer to pic plssss
Answer:
fgufyifyifyiyduhyufyiddjyfjyf86yif
Que es la actividad física y en qué mejora
what is the machine used to check melting point called?
Answer:
Melting-point apparatus
Based upon the intermolecular forces present, rank the following substances according to the expected boiling point for the substance.
a. HCl
b. NaCl
c. N2
d. H2O
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
9.36 g
Explanation:
The equation of the reaction is;
C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)
Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles
1 mole of octane yields 9 moles of water
0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water
Number of moles of oxygen = 23g/32g/mol = 0.72 moles
12.5 moles of oxygen yields 9 moles of water
0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water
Hence oxygen is the limiting reactant;
Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g
PLEASE HELP!!
How does temperature, agitation, and particle size affect solubility?
Answer:
At higher temperatures, particles move faster and collide more, increasing solubility rates.
Agitation increases solubility rates as well, by bringing fresh solvent into contact with the undissolved solute
The smaller the particle size, the higher (faster) solubility rate. Vice versa, the bigger the particle size, the lower (slower) solubility rate.
Explanation:
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Answer:
molar heat of combustion = -5156 *10³ kJ/mol
Explanation:
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Step 1: Data given
Mass of naphthalene = 1.435 grams
Initial temperature of water = 20.28 °C
Final temperature of water = 25.95 °C
heat capacity of the bomb plus water was 10.17 kJ/°C
Molar mass naphtalene = 128.2 g/mol
Step 2:
Qcal = Ccal * ΔT
⇒with Qcal =the heat of combustion
⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C
⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C
Qcal = 10.17 kJ/°C * 5.67 °C
Qcal = 57.7 kJ
Step 3: Calculate moles
Moles naphthalene = 1.435 grams / 128.2 g/mol
Moles naphthalene = 0.01119 moles
Step 4: Calculate the molar heat of combustion
molar heat of combustion = Qcal/ moles
molar heat of combustion = -57.7 kJ/ 0.01119 moles
molar heat of combustion = -5156 *10³ kJ/mol
Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.
2CO(g) + O2(g) ⇌ 2CO2
Answer:
Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.
[tex]2CO(g) + O2(g) <=> 2CO2[/tex]
Explanation:
When the reactants concentration increases, then the equilibrium will shift towards products and when the concentration of products increases, then equilibrium will shift towards reactants.
So, increases in concentration of carbon monoxide (CO) shifts the equilibrium to favor the formation of carbondioxide.
Similarly increase in concentration of oxygen also favor the formation of product carbon dioxide.
Increase in concentration of CO2 favors the formation of CO and O2.
Decrease in product concentration also favors the formation of product.
Decrease in reactant concentration favors the formation of reactants only.
4.005 X 74 X 0.007 = 2.10049
Answer:
2.07459
Explanation:
this is the correct answer.
A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of energy in joules when the system releases 1.5 KJ of heat?
Answer:
the change in the internal energy of the system is 3,752.67 J
Explanation:
Given;
initial volume of the gas, V₁ = 30.6 L
final volume of the gas, V₂ = 1.8 L
constant pressure of the gas, P = 1.8 atm
Energy released by the system, Q = 1.5 kJ = 1,500 J
Apply pressure-volume work equation, to determine the work done on the gas;
w = -PΔV
w = -P(V₂ - V₁)
w = - 1.8 atm(1.8 L - 30.6 L)
w = 51.84 L.atm
w = 51.84 L.atm x 101.325 J/L.atm
w = 5,252.67 J
The change in the internal energy of the system is calculated as;
ΔU = Q + w
Since the heat is given out, Q = - 1,500 J
ΔU = -1,500 J + 5,252.67 J
ΔU = 3,752.67 J
Therefore, the change in the internal energy of the system is 3,752.67 J
En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción
Answer:
88.9%
Explanation:
Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:
5.97 g ÷ 180 g/mol = 0.0332 molDespués calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:
0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:
0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 molFinalmente calculamos el rendimiento porcentual:
0.0590 mol / 0.0664 mol * 100% = 88.9%discuss the benefits of observing good safety measures in relation to increase in productivity within a pharmaceutical laboratory?
Answer:
Pharmaceutical laboratory helps in devloping and conducting research, vaccines. Various kinds of drugs and chemical substances used and are produced at a Pharmaceutical laboratory.
The pharmaceutical laboratories performs with various hazardous substances that results in exposure to various chemicals, biological substances and radiation. To avoid any injury or infection labs need to maintain all safety measures.
Spillage and relaseing chemical substances can be lethal during transportaions by safety measures for heling in for manufacturing of such therapeutic agents spillage and avoid wastage.
Maintaining good safety standards in the pharmaceuticals laboratory will help promote the health of technicians and workers which in turn will increase productivity and attain positive outcomes.
1.rain pours from the sky
2.leaves of the plant dried
3.fluffy clouds form in the sky
4.bathing suit dries after swim
5.water puddles disappear
A.Evaporation
B.Condensation
C.Precipitation
D.Transpiration
Yan po pag pipilian
Answer:
1.Precipitation
2.Transpiration
3.Condensation
4.Evaporation
5.Evaporation
3.Condensation
Explanation:
Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.
4.106
Calculate the moles and the mass of solute in each of the following solutions.
(a) 150.0 mL of 0.245 M CaCl2
molarity = moles of solute / volume of solution
moles of solute = molarity × volume of solution
moles of solute = 0.245 mol/L × 0.1500 L
moles of solute = 0.03675 mol
moles of solute = 0.0368 mol
-----------------------------------------------------------
Solution: (mass of solute)Step 1: Calculate the molar mass of solute.
molar mass of solute = (40.08 g/mol × 1) + (35.45 g/mol × 2)
molar mass of solute = 110.98 g/mol
Step 2: Calculate the mass of solute.
mass of solute = moles of solute × molar mass of solute
mass of solute = 0.03675 mol × 110.98 g/mol
mass of solute = 4.08 g
Note: The volume of solution must be expressed in liters (L).
Answer:
[tex]\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}[/tex]
[tex]\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}[/tex]
Explanation:
1. Moles of SoluteMolarity is a measure of concentration in moles per liter.
[tex]molarity= \frac {moles \ of \ solute}{liters \ of \ solution}[/tex]
In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.
First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.
[tex]{150 \ mL * \frac{1 \ L}{1000 \ mL}= \frac{150}{1000} \ L = 0.150 \ L[/tex]Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.
[tex]0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}[/tex]
We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.
[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L[/tex]
[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L=x[/tex]
The units of liters cancel.
[tex]0.150 *0.245 \ mol \ CaCl_2 =x[/tex]
[tex]0.03675 \ mol \ CaCl_2[/tex]
The original measurements have 3 significant figures, so our answer must have the same.
We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.
[tex]\bold {0.0368 \ mol \ CaCl_2}[/tex]
2. Mass of the SoluteWe can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.
The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.
Ca: 40.08 g/mol Cl: 35.45 g/molNotice that chlorine has a subscript of 2. We must multiply the molar mass by 2.
Cl₂: 35.45 *2= 70.9 g/molAdd calcium's molar mass.
CaCl₂: 40.08 + 70.9 =110.98 g/molUse the molar mass as a ratio.
[tex]\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]
Multiply the moles of calcium chloride we calculated above.
[tex]0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]
The units of moles of calcium chloride cancel.
[tex]0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }[/tex]
[tex]4.084064 \ g\ CaCl_2[/tex]
Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.
[tex]\bold {4.08 \ g \ CaCl_2}[/tex]
If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?
Answer:
0.74 M
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 5.90 M
Volume of stock solution (V₁) = 0.250 L
Volume of diluted solution (V₂) = 2 L
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.90 × 0.250 = M₂ × 2
1.475 = M₂ × 2
Divide both side by 2
M₂ = 1.475 / 2
M₂ = 0.74 M
Thus, the molarity of the diluted solution is 0.74 M
What are the uses of Sulphuric acid?
Answer:
The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
What is the difference between conjugate acid-base pair?
a. a H atom. c. a mole water
b. a H+ ion d. a OH– ion
Answer:
b. a H+ ion
Explanation:
The concept of conjugate acid-base pair is related to Bronsted-Lowry acid-base theory and according to this theory, acid is a proton acceptor.
In short,
conjugate base is formed when an acid donates a proton.
conjugate acid is formed when a base accepts a proton.
2- A 0.60 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24 grams of carbon, 0.040 grams of hydrogen, with the rest being oxygen. If the molecular weight of the substance is 90 grams/n, what is the molecular formula
Answer:
C₃H₆O₃
Explanation:
To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).
With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:
Moles C -Molar mass:12.0g/mol-
0.24g * (1mol/12.0g) = 0.020 moles C
Moles H = Mass H because molar mass = 1g/mol:
0.040 moles H
Moles O -Molar mass: 16g/mol-
Mass O: 0.60g - 0.24g - 0.040g = 0.32g O
0.32g O * (1mol/16g) = 0.020 moles O
Ratio of atoms (Dividing in moles of C: Lower number of moles):
C = 0.020 moles C / 0.020 moles C = 1
H = 0.040 moles H / 0.020 moles C = 2
O = 0.020 moles O / 0.020 moles C = 1
Empirical formula:
CH₂O.
Molar mass CH2O:
12g/mol + 2*1g/mol + 16g/mol = 30g/mol
As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:
C₃H₆O₃The molecular formula of the organic acid would be C3H6O3
Molecular formulaMolecular formula = [empirical formula]n
Where n = molar mass/mass of empirical formula
Empirical formula
C = 0.24/12 = 0.02
H = 0.040/1 = 0.04
O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02
Divide by the smallest
C = 1
H = 2
O = 1
Empirical formula = CH2O
Empirical formula mass = 12 + 2 + 16 = 30
n = 90/30 = 3
Molecular formula = [CH2O]3
= C3H6O3
More on molecular formula can be found here: https://brainly.com/question/1247523
