For the reaction 2hbr(g)⇌h2(g)+br2(g), k= 2.00×10−19 at 298 k what can be said about this reaction at this temperature? hints for the reaction , at 298 what can be said about this reaction at this temperature? the equilibrium lies far to the right. the reaction will proceed very slowly. the reaction contains significant amounts of products and reactants at equilibrium. the equilibrium lies far to the left.

In a methyl tert-butyl ether sample that contains 69.0 moles of hydrogen, there are 5.75 moles of oxygen.

Methyl tert-butyl ether is an organic compound that can be represented through the semi condensed formula (CH₃)₃COCH₃ or through the condensed formula C₅H₁₂O. As we can see, in methyl tert-butyl ether the molar ratio of H to O is 12:1. The number of moles of oxygen in a sample that contains 69.0 moles of H are:

[tex]69.0 mol H \times \frac{1mol O}{12 mol H} = 5.75 mol O[/tex]

In a methyl tert-butyl ether sample that contains 69.0 moles of hydrogen, there are 5.75 moles of oxygen.

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Silver (Ag) is the substance most likely to be malleable among the listed materials because it is a metal with characteristics of ductility and malleability, unlike the brittle ionic compound sodium chloride or the gaseous nitrogen and propane. Hence, correct option B.

Among sodium chloride (NaCl), silver (Ag), nitrogen, and propane, the substance that is likely malleable is silver (Ag). Malleability is a property characteristic of metals that allows them to be hammered or rolled into thin sheets without breaking. Sodium chloride, being an ionic compound, is hard, brittle, and not malleable. Nitrogen is a diatomic gas at room temperature, and propane is a hydrocarbon gas, neither of which has malleable properties. Silver, however, is a metal known for its excellent malleability and ductility. It is soft enough to be worked into various shapes and maintains its structural integrity when manipulated.

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.

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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.

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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

Answer:

c.  inertia.

Explanation:

According to Newton’s First Law of Motion or the Law of Inertia a body at rest continues to be at rest unless acted upon by some force. A body in uniform motion continues to move in uniform motion unless acted upon by some external force. The external motion may change the uniform motion of the object.

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

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.

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-).

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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.

I believe that the answer is A!!!

Hope this helps

-Abigail.

The answer is D- Vegetable soup

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.

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]

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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.

they have a higher mass

The answer is A. Na (apex answer)

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.

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.

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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

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The stability of an excited-state ion, in this case, the H⁻₂ ion (diatomic hydrogen anion), depends on several factors, including the energy of the excited state, the electronic configuration, and the propensity for the ion to release energy and return to a lower energy state.

In general, when an electron is excited to a higher-energy molecular orbital, the resulting ion is not stable in the long term. Excited states are typically higher in energy and have higher potential energy compared to the ground state. As a result, the excited-state ion is often in an energetically unfavorable condition.

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The excited-state H₂²- ion is unlikely to be stable because its ground state already has a bond order of zero, indicating instability, and excitation to an antibonding orbital further destabilizes it.

The stability of the H₂²- ion in an excited state can be assessed using a molecular orbital energy-level diagram. When an electron in this ion is excited to a higher energy level, such as from a bonding to an antibonding orbital due to the absorption of light, the bond order decreases, affecting stability. In the case of H₂²-, the bond order in the ground state is already zero, indicating that the ion is not stable. Exciting an electron to an antibonding orbital would not change the bond order but would further destabilize any remnant bonding interactions. Thus, it would be unlikely for the excited-state H₂²- ion to be stable.

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.

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:

Using 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

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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:

Using 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%

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.

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.