To find the percent dissociation of a 0.015M solution of hydrofluoric acid with a Ka of 6.3x10⁻⁴, set up an ICE table, use the equilibrium expression for Ka, solve for x (the concentration of dissociated ions), and then calculate percent dissociation.
Explanation:To calculate the percent dissociation of a 0.015M solution of hydrofluoric acid (HF) with a Ka of 6.3x10⁻⁴, we can set up an ICE table for the dissociation and use the given Ka value to solve for the concentrations of the products.
The dissociation reaction for HF is as follows: HF(aq) ⇌ H⁺(aq) + F⁻(aq)
Starting with an initial concentration of HF (0.015M) and assuming x is the amount dissociated, we have:
Initial: [HF] = 0.015, [H⁺] = 0, [F⁻] = 0Change: [HF] = -x, [H⁺] = +x, [F⁻] = +xEquilibrium: [HF] = 0.015 - x, [H⁺] = x, [F⁻] = xUsing the equilibrium expression for Ka:
Ka = [H⁺][F⁻] / [HF]
We plug in the values and solve for x:
6.3x10⁻⁴ = (x)(x) / (0.015 - x) => x² / (0.015 - x) = 6.3x10⁻⁴
Assuming x is much smaller than 0.015, which is reasonable given the small Ka value, this simplifies to:
x² / 0.015 ≈ 6.3x10⁻⁴
Solving for x gives us the concentration of the dissociated ions. Then the percent dissociation is:
Percent dissociation = (concentration of dissociated HF / initial concentration of HF) * 100
Learn more about Percent Dissociation here:https://brainly.com/question/32005232
#SPJ12
The percent dissociation of a 0.015M solution of hydrofluoric acid (HF) with a Ka of 6.3 x 10⁻⁴ can be calculated as approximately 20.47% by using an ICE table and simplifying the equilibrium expression. The dissociation reaction for HF is HF ⇌ H3O⁺ + F⁻. Percent dissociation is found to be x/0.015, where x is the concentration of dissociated ions.
Dissociation of Hydrofluoric Acid (HF):
To find the percent dissociation of a 0.015M solution of hydrofluoric acid (HF) with a given acid dissociation constant (Ka) of 6.3 x 10⁻⁴, follow these steps:
Dissociation Reaction:
The dissociation reaction for hydrofluoric acid can be written as:
HF (aq) ⇌ H₃O⁺ (aq) + F⁻ (aq)
Here, HF is the weak acid, H₃O⁺ is the hydronium ion (conjugate acid), and F⁻ is the fluoride ion (conjugate base).
Calculation Using ICE Table:
To calculate the percent dissociation, we can use an ICE table:
Initial: [HF] = 0.015M, [H₃O⁺] = 0, [F⁻] = 0Change: [HF] = -x, [H₃O⁺] = +x, [F⁻] = +xEquilibrium: [HF] = 0.015 - x, [H₃O⁺] = x, [F⁻] = xUsing the equilibrium expression for Ka:
Ka = [H₃O⁺][F⁻] / [HF]
6.3 x 10⁻⁴ = (x)(x) / (0.015 - x)
Assuming x is small compared to 0.015M, we approximate:
6.3 x 10⁻⁴ ≈ x2 / 0.015
x² ≈ 9.45 x 10⁻⁶
x ≈ √9.45 x 10⁻⁶
x ≈ 3.07 x 10⁻³ M
Percent dissociation can be calculated as:
% dissociation = (x / 0.015) * 100 ≈ (3.07 x 10⁻³ / 0.015) * 100 ≈ 20.47%
During an investigation, similar glow sticks were placed in two beakers containing water at different temperatures. A record of the investigation is shown below: Investigation RecordGlow Stick Beaker Water Temperature Duration of Glow Stick in the Beaker A 50 °F 2 minutes B 140 °F 2 minutes When the glow sticks were bent, chemicals inside reacted to produce a glow. Which statement is most likely correct?
I believe that the answer is A!!!
Hope this helps
-Abigail.
The substance gray tin is found to crystallize in a cubic lattice, with an edge length of 646.0 pm. if the density of solid gray tin is 5.850 g/cm3, how many sn atoms are there per unit cell?
First find the volume of the cubic lattice given edge length of 646 pm or 6.46x10^-8 cm.
volume = (6.46x10^-8 cm)^3
volume = 2.7x10^-22 cm^3
The total mass of the lattice is:
mass = (5.850 g/cm^3) * 2.7x10^-22 cm^3
mass = 1.577x10^-21 grams
The molar mass of tin is 118.71 g/mol and the Avogadros number is 6.022 x 10^23 atoms/mol, hence:
Sn atoms = [1.577x10^-21 g / (118.71 g/mol)] * 6.022 x 10^23 atoms/mol
Sn atoms = 8 atoms
The number of Sn atoms in the unit cell of grey tin depends on the type of cubic cell structure (simple, body-centered, or face-centered). The question does not provide sufficient information to determine this. With full information, one can use the density and molar mass of Sn to calculate the number of atoms per unit cell.
Explanation:To determine the number of Sn atoms per unit cell, we must first determine the type of cubic cell structure. Metals typically crystallize in one of three types of cubic unit cells: simple cubic (sc), face-centered cubic (fcc), or body-centered cubic (bcc). Unfortunately, information is not provided regarding the specific type of gray tin, so we can't definitively answer your question.
However, here's a brief explanation of how you would calculate this if the necessary details were given:
For a simple cubic lattice, there is one atom per unit cell. For a body-centered cubic lattice, there are two atoms per unit cell. For a face-centered cubic lattice, there are four atoms per unit cell.
The density, given that it's in g/cm³, can be converted to atoms per cm³ using the molar mass of Sn (118.7 g/mol), from which we can calculate atoms per unit cell, providing the structure type of the unit cell is known.
Learn more about Unit Cell Structure here:https://brainly.com/question/29140332
#SPJ11
A firecracker exploding would be an example of a (n) ___? A. Physical change B. Exothermic reaction C. Formation of a precipitate D. Endothermic reaction
what happens when an object speeds up,slows down, or changes direction A) velocity B) time C) deceleration D) acceleration
To determine the atomic mass of an element, the masses of all the naturally occurring isotopes of that element are averaged together. Why are some isotopes more heavily weighted than others? A. They occur more frequently in nature. B. They have a higher mass. C. They have a higher number of orbitals. D. They have more protons.
they have a higher mass
Classify the substances according to the strongest solute-solvent interaction that will occur between the given substances and water during dissolution. please match. 1. ion-dipole forces 2. dipole-dipole forces 3. hydrogen bonding 4. london dispersion forces
A. AlCl3
B. FeBr3
C. NH3
D. C2H5OH
Answer: [tex]NH_3[/tex] and [tex]C_2H_5OH[/tex] are hydrogen boding, [tex]AlCl_3[/tex] and [tex]FeBr_3[/tex] are ion-dipole forces.
Explanation: If the bond is formed between a metal and a non metal then it is known as ionic bond and ionic compounds have positive and negative ions and so they have ion-dipole forces. Aluminium chloride and Iron(III)bromide both are ionic as they have a bond between a metal and non metal and so both of these have ion-dipole forces.
Ammonia and ethanol both are polar molecules and we know that polar molecules have dipole-dipole forces.
A hydrogen bond could form if hydrogen is bonded with more electron negative atom(N, O or F).
In ammonia, H is bonded to N and in ethanol, H is bonded to O, so both of these molecules must have hydrogen bonding. Since hydrogen bond is stronger as compared to dipole - dipole forces, we will say that both ammonia and ethanol have hydrogen bonding.
Answer: [tex]AlCl_3[/tex] and [tex]FeBr_3[/tex] have ion-dipole interactions.
[tex]NH_3[/tex] and [tex]C_2H_5OH[/tex] have hydrogen bonding.
Explanation:
[tex]AlCl_3[/tex] and [tex]FeBr_3[/tex] have ion-dipole interactions.[tex]NH_3[/tex] and [tex]C_2H_5OH[/tex] have dipole-dipole forces. ‘H’ atom is bonded to ‘N’ in ammonia and ‘H’ atom is bonded to ‘O’ in ethanol, therefore; both ammonia and ethanol must have hydrogen bonding. Hydrogen bond is stronger with compared to dipole-dipole forces, hence; hydrogen bond gets the priority than dipole-dipole forces.
Further explanation:
Inter molecular forces are the physical forces between molecules.
London dispersion forces are occurring between two non-polar molecules and these are the weakest inter molecular forces of the inter molecular forces.
Dipole-dipole forces are occurs between two polar molecules. Many molecules are polar and hence this is a common inter molecular force.
Ion-dipole forces occur when an ion encounters a polar molecule. A cation will attract to the negative part of the molecule and an anion will attract to the positive part of the molecule.
Hydrogen bonds are the attraction of molecules which are already in other chemical bonds and these types of bonds are primary electrostatic force.
Ammonia [tex](NH_3)[/tex] is a polar molecule. Ethanol [tex](C_2H_5OH)[/tex] is a very polar molecule. [tex]AlCl_3[/tex] and [tex]FeBr_3[/tex] are ionic compounds and dissociate to give ions.
Hence [tex]AlCl_3[/tex] and [tex]FeBr_3[/tex] have ion-dipole interactions.
[tex]NH_3[/tex] and [tex]C_2H_5OH[/tex] have dipole-dipole forces. ‘H’ atom is bonded to ‘N’ in ammonia and ‘H’ atom is bonded to ‘O’ in ethanol, therefore; both ammonia and ethanol must have hydrogen bonding. Hydrogen bond is stronger with compared to dipole-dipole forces, hence; hydrogen bond gets the priority than dipole-dipole forces.
Learn more:
1. Intermolecular forces : https://brainly.com/question/13159906 (answer by lucic)
2. London dispersion forces: https://brainly.com/question/13188977 ( answer by Tirana)
3. Hydrogen bonding : https://brainly.com/question/2161098 (answer by Jamuuj)
Keywords:
Intermolecular forces, London dispersion forces, Dipole-dipole forces, Ion-dipole forces, Hydrogen bonds
A photon has 4.75x10^-19 J energy. What is the frequency of this photon?
How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?
The complete balanced chemical reaction is:
2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S
First let us calculate the number of moles of AgNO3.
moles AgNO3 = 0.315 M * 0.035 L
moles AgNO3 = 0.011025 mol
From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:
moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)
moles Na2S required = 5.5125 x 10^-3 mol
Therefore volume required is:
volume Na2S = 5.5125 x 10^-3 mol / 0.260 M
volume Na2S = 0.0212 L = 21.2 mL
21.20 mL of 0.260 M [tex]\(\text{Na}_2\text{S}\)[/tex] is needed to react with 35.00 mL of 0.315 M [tex]\(\text{AgNO}_3\).[/tex]
To determine the volume of 0.260 M [tex]\(\text{Na}_2\text{S}\)[/tex] needed to react with 35.00 mL of 0.315 M [tex]\(\text{AgNO}_3\)[/tex], we need to use the stoichiometry of the reaction between [tex]\(\text{Na}_2\text{S}\) and \(\text{AgNO}_3\).[/tex]
The balanced chemical equation for the reaction is:
[tex]\[ \text{Na}_2\text{S} + 2\text{AgNO}_3 \rightarrow \text{Ag}_2\text{S} + 2\text{NaNO}_3 \][/tex]
From the balanced equation, we see that 1 mole of [tex]\(\text{Na}_2\text{S}\)[/tex] reacts with 2 moles of [tex]\(\text{AgNO}_3\).[/tex]
Step 1: Calculate moles of [tex]\(\text{AgNO}_3\)[/tex]
First, we calculate the moles of [tex]\(\text{AgNO}_3\)[/tex] in the 35.00 mL solution.
[tex]\[ \text{Moles of AgNO}_3 = \text{Molarity} \times \text{Volume (in L)} \][/tex]
[tex]\[ \text{Moles of AgNO}_3 = 0.315 \, \text{M} \times 0.03500 \, \text{L} \][/tex]
[tex]\[ \text{Moles of AgNO}_3 = 0.011025 \, \text{moles} \][/tex]
Step 2: Determine moles of [tex]\(\text{Na}_2\text{S}\)[/tex] needed
From the balanced equation, 1 mole of [tex]\(\text{Na}_2\text{S}\)[/tex] reacts with 2 moles of [tex]\(\text{AgNO}_3\).[/tex]
[tex]\[ \text{Moles of Na}_2\text{S} = \frac{\text{Moles of AgNO}_3}{2} \][/tex]
[tex]\[ \text{Moles of Na}_2\text{S} = \frac{0.011025 \, \text{moles}}{2} \][/tex]
[tex]\[ \text{Moles of Na}_2\text{S} = 0.0055125 \, \text{moles} \][/tex]
Step 3: Calculate the volume of 0.260 M [tex]\(\text{Na}_2\text{S}\)[/tex] solution needed
Now, we need to find the volume of 0.260 M [tex]\(\text{Na}_2\text{S}\)[/tex] that contains 0.0055125 moles of [tex]\(\text{Na}_2\text{S}\).[/tex]
[tex]\[ \text{Volume (in L)} = \frac{\text{Moles of Na}_2\text{S}}{\text{Molarity}} \][/tex]
[tex]\[ \text{Volume (in L)} = \frac{0.0055125 \, \text{moles}}{0.260 \, \text{M}} \][/tex]
[tex]\[ \text{Volume (in L)} = 0.0212019 \, \text{L} \][/tex]
Convert the volume from liters to milliliters:
[tex]\[ \text{Volume (in mL)} = 0.0212019 \, \text{L} \times 1000 \, \text{mL/L} \][/tex]
[tex]\[ \text{Volume (in mL)} = 21.20 \, \text{mL} \][/tex]
Therefore, 21.20 mL of 0.260 M [tex]\(\text{Na}_2\text{S}\)[/tex] is needed to react with 35.00 mL of 0.315 M [tex]\(\text{AgNO}_3\).[/tex]
What mass of caco3 (in grams) is needed to make 10.5 l of a 1.50 m caco3 solution?
the use of high-pressure chambers to control disease processes is known as
Which statement is true?
A) Unicellular organisms cannot obtain energy or reproduce, but they can maintain structure (homeostasis).
B) Unicellular organisms don't have to do anything - they are only just one cell!
C) Unicellular organisms have to meet the 3 main challenges of life. They have to obtain energy, reproduce, and maintain structure (homeostasis).
D) Unicellular organisms are not living.
Final answer:
The true statement about unicellular organisms is that they must obtain energy, reproduce, and maintain homeostasis, despite being single-celled.
Explanation:
The statement that is true about unicellular organisms is: C) Unicellular organisms have to meet the 3 main challenges of life. They have to obtain energy, reproduce, and maintain structure (homeostasis).
All living organisms, including unicellular ones, need to carry out several basic functions to sustain life. These functions include obtaining energy to power their biological processes, growing and reproducing to ensure the survival of their species, and maintaining a stable internal environment through the process of homeostasis. Even though they are made of only one cell, unicellular organisms are complex and very much alive. They are capable of responding to their environment and have complex chemistry. Additionally, these organisms must exchange matter with their surroundings in order to grow, reproduce, and maintain their organization.
The scientific method helps scientists avoid injecting which of the following into their experiments? a. information b. conclusions c. hypotheses d. bias
The oxidation of magnesium to form magnesium oxide is shown by which balanced chemical equation?
Answer : The balanced chemical reaction will be,
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO[/tex]
Explanation :
Balanced chemical reaction : It is the reaction in which the number of atoms of an individual elements present on reactant side must be equal to the product side.
As per question, when 2 moles of magnesium react with 1 mole of oxygen the it reacts to give 2 moles of magnesium oxide as a product.
The balanced chemical reaction will be,
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO[/tex]
Pure acetic acid (hc2h3o2) is a liquid and is known as glacial acetic acid. calculate the molarity of a solution prepared by dissolving 50.00 ml of glacial acetic acid at 25 °c in sufficient water to give 500.0 ml of solution. the density of glacial acetic acid at 25 °c is 1.05 g/ml
The molarity of a solution prepared is 1.735 M
From the question,
We are to calculate the molarity of a solution prepared by dissolving 50.00 ml of glacial acetic acid in sufficient water to give 500.0 ml of solution.
First, we will determine the mass of glacial acetic acid present in the 50.00 mL solution
From the question,
The density of glacial acetic acid at 25 °c is 1.05 g/ml
This means, there are 1.05 g of glacial acetic acid in 1 mL of the solution
If, 1mL of the solution contains 1.05 g of glacial acetic acid
Then, 50 mL of the solution will contain 50 × 1.05 g of glacial acetic acid
50 × 1.05 g = 52.5 g
∴ The mass of glacial acetic acid in the 50 mL solution is 52.5 g
Now, we will determine the number of moles of the glacial acetic acid
From the formula
[tex]Number \ of\ moles =\frac{Mass }{Molar\ mass}[/tex]
Mass of glacial acetic acid = 52.5 g
Molar mass of glacial acetic acid = 60.052 g/mol
∴ Number of moles of glacial acetic acid = [tex]\frac{52.5}{60.052 }[/tex]
Number of moles of glacial acetic acid = 0.86748 moles
Now, for the molarity (that is, concentration) of the solution
From the formula
Number of moles = Concentration × Volume
Then,
[tex]Concentration = \frac{Number \ of \ moles}{Volume}[/tex]
Number of moles of glacial acetic acid = 0.86748 moles
Volume of the final solution = 500.0 mL = 0.5 L
∴ Concentration of the solution = [tex]\frac{0.86748}{0.5}[/tex]
Concentration of the solution = 1.735 M
Hence, the molarity of a solution prepared is 1.735 M
Learn more here: https://brainly.com/question/21017309
The emission spectrum of gold shows a line of wavelength 2.76x10^-7m. What is the corresponding frequency of this light? How much energy is emitted in production of this radiation?
The formula relating wavelength and frequency is:
f = c / ʎ
where f is frequency, c is speed of light = 3 x 10^8 m/s, and ʎ is wavelength of light = 2.76 x 10^-7 m
f = (3 x 10^8 m/s) / 2.76 x 10^-7 m
f = 1.09 x 10^15 s-
The energy can be calculated using the formula:
E = h f
where h is Planck’s constant = 6.626 x 10^-34 J s
E = (6.626 x 10^-34 J s) * 1.09 x 10^15 s-
E = 7.2 x 10^-19 J
The frequency corresponding to the emission spectrum line of gold with wavelength 2.76 x 10⁻⁷m is approximately 1.0869565 x 10¹⁵ Hz. The energy emitted per photon for this radiation is about 7.202 x 10⁻¹⁹ joules.
The emission spectrum of gold shows a line of wavelength 2.76 x 10⁻⁷m. To find the corresponding frequency of this light, we can use the equation c = λf, where c is the speed of light in a vacuum (approximately 3 x 10⁸ m/s), λ is the wavelength, and f is the frequency.
Therefore, the frequency (f) is given by:
f = c / λ
f = (3 x 10⁸ m/s) / (2.76 x 10⁻⁷ m) = 1.0869565 x 10¹⁵ Hz
The energy emitted in the production of this radiation can be calculated using the equation E = hf, where h is Planck's constant (6.626 x 10⁻³⁴ J s) and f is the frequency we just calculated.
E = (6.626 x 10⁻³⁴ J s) (1.0869565 x 10¹⁵ Hz) = 7.202 x 10⁻¹⁹ J per photon
Classify these alcohols as primary (1°), secondary (2°), or tertiary (3°). g
The alcohols are classified as primary, secondary, or tertiary based on the number of carbon atoms directly bonded to the hydroxyl-bearing carbon. Examples include CH3-CH3CHOH (1°), OH-CH3CCH3-CH3 (2°), and OH-CH2CHCH3-CH3 (3°), indicating their respective structural arrangements.
The classification of alcohols as primary (1°), secondary (2°), or tertiary (3°) is based on the number of carbon atoms directly bonded to the carbon atom bearing the hydroxyl group. Here's the classification for the given alcohols:
a. CH3-CH3CHOH
- This alcohol has one carbon directly bonded to the carbon with the hydroxyl group. Therefore, it is a primary alcohol (1°).
b. OH-CH3CCH3-CH3
- This alcohol has two carbons directly bonded to the carbon with the hydroxyl group. Therefore, it is a secondary alcohol (2°).
c. OH-CH2CHCH3-CH3
- This alcohol has three carbons directly bonded to the carbon with the hydroxyl group. Therefore, it is a tertiary alcohol (3°).
Learn more about alcohols here:
https://brainly.com/question/13786374
#SPJ6
The probable question may be:
Classify these alcohols as primary (1°), secondary (2°), or tertiary (3°).
a. CH3-CH3CHOH b. OH-CH3CCH3-CH3 c. OH-CH2CHCH3-CH3
A certain amount of hydrogen peroxide was dissolved in 100. ml of water and then titrated with 1.68 m kmno4. what mass of h2o2 was dissolved if the titration required 13.3 ml of the kmno4 solution?
The solution for the problem is:
First, use the concentration of the volume of the thing you know to compute for the moles of that substance. Then, use the coefficient in the balanced equation to relay moles of that to moles of anything else in the chemical equation. Lastly, translate moles into mass by means of its molar mass, or into a concentration using the volume.
Applying what I have said earlier:
0.0133 L X 1.68 mol/L = 0.0223 mol KMnO4 X (1 mol H2O2 / 2 mol KMnO4) = 0.0112 mol H2O2
Mass H2O2 = 0.0112 mol H2O2 X 34.0 g/mol = 0.380 grams H2O2
A titration of hydrogen peroxide with potassium permanganate is conducted. Using the volume and molarity of KMnO4 we find the moles of KMnO4 used. From the balanced redox equation, we calculate the corresponding moles of H2O2 and finally its mass.
Explanation:This question is about a titration process where a certain amount of hydrogen peroxide (H2O2) is dissolved in water and then titrated with a solution of potassium permanganate (KMnO4). To solve this, we need to find the moles of KMnO4 used, which we can calculate by multiplying the volume (in Liters) of KMnO4 by its molarity. That gives us 0.022324 mol. The balanced redox equation between KMnO4 and H2O2 is 2MnO4^- (aq) + 5H2O2 (aq) + 6H^+ (aq) -> 5O2 (g) + 2Mn^2+ (aq) + 8H2O(l). From this, we can see that the mole ratio between KMnO4 and H2O2 is 2:5. So, we multiply the moles of KMnO4 by 5/2 to get the moles of H2O2, giving us 0.027905 mol. Lastly, to find the mass of H2O2, we calculate by multiplying the number of moles of H2O2 by its molar mass, giving approximately 0.94 g.
Learn more about Titration here:https://brainly.com/question/31271061
#SPJ11
How reactive is an atom of Sodium(Na) and why?
Question 1 options:
Sodium (Na) is very reactive because it does not have a full valence shell.
Sodium (Na) is very reactive because it does not have enough protons in the nucleus.
Sodium (Na) is not very reactive because it does not have a full valence shell.
Sodium (Na) is not very reactive because it can only bond with Chlorine (Cl) to become salt.
PLz plz plz plz help
How reactive is an atom of Sodium(Na) and why?
Peter is 5 ft, 11 inches tall. Paul is 176 cm tall. Who is taller?
Final answer:
To compare the heights of Peter and Paul, we convert their heights to the same unit. Peter is taller with a height of 71 inches compared to Paul's height of approximately 69.29 inches.
Explanation:
In order to compare the heights of Peter and Paul, we need to convert their heights to the same unit. Peter's height is given in feet and inches, so we need to convert it to inches. Since there are 12 inches in a foot, Peter's height is 5 ft * 12 inches/ft + 11 inches = 60 inches + 11 inches = 71 inches.
Now we can compare Peter's height of 71 inches with Paul's height of 176 cm. In order to convert centimeters to inches, we need to know that 1 inch is approximately equal to 2.54 centimeters. So, Paul's height is 176 cm * 1 inch/2.54 cm = 69.29 inches (approximately).
Comparing the heights, we can see that Peter is taller with a height of 71 inches compared to Paul's height of approximately 69.29 inches.
What is the number of carbon atoms in the ring portion of the haworth structure of glucose?
The number of carbon atoms in the ring portion of the Haworth structure of glucose is six.
In the Haworth structure of glucose, the ring portion is a six-membered ring formed by the reaction of the aldehyde group with the hydroxyl group on the fifth carbon atom. This forms a hemiacetal linkage and results in the formation of a six-membered ring known as a pyranose ring. In the Haworth projection, the carbon atoms are depicted as vertices of the ring, with oxygen and hydrogen atoms shown explicitly.
In the case of glucose, the ring portion consists of six carbon atoms. Each carbon atom is bonded to either a hydrogen atom or a hydroxyl group. The chemical formula for glucose is C6H12O6, which indicates that there are six carbon atoms in the molecule. In the ring structure of glucose, all six carbon atoms are part of the ring, forming a hexagonal shape.
Therefore, the number of carbon atoms in the ring portion of the Haworth structure of glucose is six. These carbon atoms play a crucial role in forming the stable cyclic structure of glucose and are involved in various biochemical processes within living organisms.
A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 30.00 gram sample of the alcohol produced 57.30 grams of CO2 and 35.22 grams of H2O. What is the empirical formula of the alcohol?
first we calculate the mass percent composition
of three elements carbon hydrogen and oxygen.
Mass of CO₂ = 44.0096 g/mol
Mass of H₂O = 8.0153 g/mol
percent composition of carbon = (57.30 g CO2) / (44.0096 g CO2/mol) x
(1/1) x (12.0108 g C/mol) / (30.00 g) = 0.521264 = 52.1264% C
Percent composition of hydrogen = (35.22 g H2O) / (18.0153 g H2O/mol) x (2/1) x
(1.0079 g H/mol) / (30.00 g) = 0.131363 = 13.1363% H
percent composition of oxygen = 100% - 52.1264% C 13.1363% H = 34.7373% O
Now calculate the number of moles in the compound;
(52.1264 g C) / (12.0108 g C/mol) = 4.33996 mol C
(13.1363 g H) / (1.0079 g H/mol) = 13.0333 mol H
(34.7373 g O) / (15.9994 g O/mol) = 2.17116 mol O
Now divide by the smallest number of moles:
(4.33996 mol C) / (2.17116 mol) = 1.999 = 2
(13.0333 mol H) / (2.17116 mol) = 6.003 = 6
(2.17116 mol O)/ (2.17116 mol) = 1.000 = 1
So, the empirical formula is :
C₂H₆O
The empirical formula of the alcohol cannot be determined with the given information.
Explanation:The empirical formula of the alcohol can be determined by analyzing the masses of carbon dioxide (CO₂) and water (H₂O) produced during combustion. From the given information, 57.30 grams of CO₂ and 35.22 grams of H₂O were produced. To find the empirical formula, we need to calculate the moles of carbon, hydrogen, and oxygen in the sample.
First, calculate the moles of CO₂ and H₂O produced:
CO₂: 57.30 g * (1 mol CO₂ / 44.01 g) = 1.30 mol CO₂H₂O: 35.22 g * (1 mol H₂O / 18.02 g) = 1.95 mol H₂ONext, calculate the moles of carbon and hydrogen:
Carbon: 1.30 mol CO₂ * (1 mol C / 1 mol CO₂) = 1.30 mol CHydrogen: 1.95 mol H₂O * (2 mol H / 1 mol H₂O) = 3.90 mol HFinally, divide the moles of each element by the smallest number of moles to get the mole ratio:
Carbon: 1.30 mol C / 1.30 mol C = 1Hydrogen: 3.90 mol H / 1.30 mol C = 3Oxygen: Since we don't have the mass of oxygen, we can assume it's the leftover mass from the sample. Oxygen = Total mass - (mass of carbon + mass of hydrogen) = 30.00 g - (57.30 g + 35.22 g) = -62.52 g (which is not physically possible)Based on these calculations, it is not possible to determine the empirical formula of the alcohol using the given information. Further information is needed to calculate the empirical formula.
A 50.0-ml volume of 0.15 m hbr is titrated with 0.25 m koh. calculate the ph after the addition of 11.0 ml of koh. express your answer numerically.
Final answer:
To calculate the pH after the addition of KOH to HBr, determine moles of each reactant, find the limiting reactant, and then calculate remaining moles of HBr. The remaining concentration of HBr gives the H+ concentration, allowing for pH calculation using the negative log of the H+ concentration. The pH is approximately 1.109.
Explanation:
To calculate the pH after the addition of 11.0 mL of 0.25 M KOH to a 50.0 mL volume of 0.15 M HBr, we will first need to find out whether the reaction goes to completion and determine the number of moles of the remaining reactant. Hydrobromic acid (HBr) is a strong acid and reacts completely with potassium hydroxide (KOH), a strong base, to form KBr and water:
HBr + KOH → KBr + H₂O
The initial moles of HBr are given by the product of its concentration and volume in liters:
n(HBr) = 0.15 mol/L × 0.050 L = 0.0075 mol
The moles of KOH added can be calculated as follows:
n(KOH) = 0.25 mol/L × 0.011 L = 0.00275 mol
Since KOH is the limiting reactant, it will be used up completely, leaving:
n(HBr remaining) = 0.0075 mol - 0.00275 mol = 0.00475 mol
The remaining HBr will dissociate completely as it is a strong acid. The concentration of HBr after the reaction is:
C(HBr remaining) = n(HBr remaining) / Total volume
Total volume = initial volume of HBr + volume of KOH added = 0.050 L + 0.011 L = 0.061 L
C(HBr remaining) = 0.00475 mol / 0.061 L ≈ 0.0779 M
The pH of the solution is calculated by taking the negative log of the H+ concentration, which is equal to the concentration of HBr after the KOH has been added:
pH = -log[0.0779 M] = -log[7.79 × 10-2] ≈ 1.109
Thus, the pH of the solution after the addition of 11.0 ml of 0.25 M KOH is approximately 1.109.
An element with a single electron in its highest
energy level is probably
An element with a single electron in its highest energy level is probably an
alkali metal.
Alkali metals are found in group 1 of the periodic table. They have
characteristic shiny and soft and are highly reactive due to the presence of
just one valence electrons in their outermost shell.
Example of alkali metals include potassium, sodium, caesium etc. They have
a single electron in its highest energy level which validates it being an alkali
metal.
Read more on https://brainly.com/question/19109836
Determine the [h3o+] of a 0.250 M solution of Formic acid
The [H3O+] of a 0.250 M solution of Formic acid is 0.0081 M, as calculated by substituting values into the percent ionization equation.
Explanation:The [H3O+] of a 0.250 M solution of Formic acid can be determined by using the given equation [H3O+] = 10^-2.09 = 0.0081 M. Further, if we substitute this value and the provided initial acid concentration into the percent ionization equation, it gives 8.1 × 10^-3 in percentage value. Please note, this is based on the assumption that Formic acid behaves as a weak acid, slightly ionizing in water to form hydronium ions (H3O+) and formate ions (HCOO-).
Learn more about Hydronium here:https://brainly.com/question/29286376
#SPJ3
Complete the balanced overall ionic equation for sodium iodide dissolving in water.
Answer: NaI (aq) → Na⁺ (aq) + I⁻(aq)
Explanation:
1) The formula of sodium iodide is NaI
2) Due to the great electronegativities of sodium (Na) and iodine (I), sodium iodide is a ionic compound.
3) Ionic compounds dissociate in water to produce the corresponding ions. In this case, since I has the greatest electronegativity, the ions are Na⁺ and I⁻ .
4) When you write the ionic equation, you have to take into account, the number of ions per unit formula and the phases.
So, in this case, the species are in water solutions, this is aqueous solutions. Then you use the symbol aq to indicate the aqueous phase.
The result is the ionic equation: NaI (aq) → Na⁺ (aq) + I⁻(aq)
The overall balanced ionic equation for sodium iodide dissolved in water is
[tex]\boxed{{{\text{H}}_2}{\text{O}}\left( l \right)\to{{\text{H}}^ + }\left({aq}\right) + {\text{O}}{{\text{H}}^ - }\left({aq}\right)}[/tex]
Further Explanation:
The three types of equations that are used to represent the chemical reaction are as follows:
1. Molecular equation
2. Total ionic equation
3. Net ionic equation
The reactants and products remain in undissociated form in molecular equation. In the case of total ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of net or overall ionic equation only the useful ions that participate in the reaction are represented.
The steps to write the overall ionic reaction are as follows:
Step 1. Write the molecular equation for the reaction with the phases in the bracket.
In the reaction, NaI reacts with [tex]{{\text{H}}_2}{\text{O}}[/tex] to form NaOH and HI. The balanced molecular equation of the reaction is as follows:
[tex]{\text{NaI}}\left({aq}\right)+{{\text{H}}_2}{\text{O}}\left( l \right)\to{\text{NaOH}}\left( {aq} \right)+{\text{HI}}\left( {aq}\right)[/tex]
Step2. Dissociate all the compounds with the aqueous phase to write the total ionic equation. The compounds with solid and liquid phase remain same. The total ionic equation is as follows:
[tex]{\text{N}}{{\text{a}}^ + }\left({aq} \right) + {{\text{I}}^ - }\left({aq} \right) + {{\text{H}}_2}{\text{O}}\left( l \right) \to{\text{N}}{{\text{a}}^ + }\left({aq} \right)+{\text{O}}{{\text{H}}^ - }\left( {aq}\right)+{{\text{H}}^ + }\left( {aq} \right)+{{\text{I}}^ - }\left( {aq} \right)[/tex]
Step3. The common ions on both the sides of the reaction get cancelled out to get the overall ionic equation.
[tex]\boxed{{\text{N}}{{\text{a}}^ + }\left( {aq} \right)}+\boxed{{{\text{I}}^ - }\left( {aq}\right)} + {{\text{H}}_2}{\text{O}}\left( l \right) \to\boxed{{\text{N}}{{\text{a}}^ + }\left( {aq}\right)}+{\text{O}}{{\text{H}}^ - }\left( {aq} \right)+{{\text{H}}^ + }\left( {aq}\right) + \boxed{{{\text{I}}^ - }\left( {aq}\right)}[/tex]
Therefore, the overall ionic equation obtained is as follows:
[tex]{{\text{H}}_2}{\text{O}}\left( l \right) \to{{\text{H}}^ + }\left({aq} \right)+{\text{O}}{{\text{H}}^ - }\left({aq} \right)[/tex]
Learn more:
1. Balanced chemical equation: https://brainly.com/question/1405182
2. Oxidation and reduction reaction: https://brainly.com/question/2973661
Answer details:
Grade: High School
Subject: Chemistry
Chapter: Chemical reaction and equation
Keywords: overall ionic equation, NaI, NaOH, H2O, HI, H+, I-, aqueous phase, dissociate, molecular equation, reaction, sodium iodide, water.
Is nitrogen triiodide an ionic compound or covalent compound?
Answer: The given compound is a covalent compound.
Explanation:
Covalent compound is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound. These are usually formed when two non-metals react.
An ionic compound is defined as the compound which is formed when electron gets transferred from one atom to another atom. These are usually formed when a metal reacts with a non-metal.
We are given:
A chemical compound having chemical name as nitrogen triiodide.
This compound is formed by the combination of nitrogen and iodine atoms. Both these elements are non-metals and thus form covalent compound.
The chemical formula for the given compound is [tex]NI_3[/tex]
Hence, the given compound is a covalent compound.
When 225 mg of anthracene was burned in a bomb calorimeter the temperature rose?
What process is primarily responsible for the formation of precipitation in warm clouds?
Which if the following items is an example of heterogeneous mixture? A- milk B- apple juice C- stainless steel D- vegetable soup
The answer is D- Vegetable soup