Drying of organic solvent
1) which labs its done
2) its use + definition
3) process

The statement that is not correct about carboxylic acids is (b) "carboxylic acids do not form intermolecular h-bonding with water." Carboxylic acids do form intermolecular H-bonding with water.

This is due to the polar nature of carboxylic acids, which results in the formation of H-bonds between the oxygen atoms of carboxylic acid molecules and the hydrogen atoms of water molecules. These H-bonds increase the solubility of carboxylic acids in water.

The presence of the carboxyl group (-COOH) in carboxylic acids also facilitates the formation of H-bonds between adjacent carboxylic acid molecules. This results in the formation of dimers and other aggregates in the solid state. Overall, carboxylic acids are polar molecules that are capable of forming intermolecular H-bonds with both water and other carboxylic acid molecules.

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The oxygen atom is sp3 hybridized in dialkyl ethers. In dialkyl ethers, the oxygen atom forms two sigma bonds with the two alkyl groups that are linked to it as well as two lone electron pairs.

As a result of hybridizing oxygen's one s orbital and three p orbitals, these four electron pairs now occupy four equivalent hybrid orbitals.

Resulting sp3 hybrid orbitals have bond angles of approximately 109.5° and are orientated in a tetrahedral configuration. Due to this hybridization, the oxygen atom may effectively establish covalent bonds with the two alkyl groups, resulting in a molecular structure that is stable.

Stability and reactivity of these compounds are greatly influenced by sp3 hybridization of the oxygen atom in dialkyl ethers.

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The pH of a saturated solution of calcium hydroxide at 25°C is (B) 12.14.

The acronym "pH" stands for "potential of Hydrogen." It is a gauge for a solution's acidity or basicity. The neutral pH value is 7, and the pH scale goes from 0 to 14. Acidic solutions are those with a pH below 7, whereas basic or alkaline solutions are those with a pH above 7. The quantity of hydrogen ions (H+) in the solution determines the pH of the solution.

The balanced chemical equation for the dissolution of calcium hydroxide in water is:

Ca(OH)₂ (s) ⇌ Ca²⁺ (aq) + 2OH⁻ (aq)

The Ksp expression for calcium hydroxide is:

Ksp = [Ca²⁺][OH⁻]₂

At equilibrium, the concentration of Ca²⁺ is equal to the concentration of OH-, so we can simplify the expression to:

Ksp = [Ca²⁺][OH⁻]² = x(x)² = x³

where x is the molar solubility of calcium hydroxide.

We can solve for x by substituting the Ksp value and solving for x:

Ksp = 1.3 × 10⁻⁶  = x³

x = (1.3 × [tex]10^{-6}^( \frac{1}{3} )[/tex] = 0.00522 M

The concentration of OH- in the saturated solution is twice the solubility concentration, so:

[OH⁻] = 2x = 0.0104 M

Now, we can use the relationship between pH and [OH⁻] to calculate the pH of the solution:

pH = 14 - log[OH⁻] = 14 - log(0.0104) = 12.14

Therefore, the pH of a saturated solution of calcium hydroxide at 25°C is (B) 12.14.

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The correct order of increasing vapor pressure of the three liquids at STP is as follows:

- Lowest: Liquid Z
- Middle: Liquid Y
- Highest: Liquid X
Vapor pressure is inversely related to boiling point.

A lower boiling point indicates higher vapor pressure, and a higher boiling point indicates lower vapor pressure.

Given that liquid X has the lowest boiling point and liquid Z has the highest boiling point, the order of increasing vapor pressure would be Z, Y, X.

Summary: The order of increasing vapor pressure at STP for the three liquids is Z (lowest), Y (middle), and X (highest).

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Quick clay can move like a liquid when vibration or shaking separates the water-coated particles.

Quick clay, also known as Leda clay, is a type of clay that has a unique behavior. When the clay is disturbed or shaken, the water-coated particles lose their bonding strength and the clay liquefies. This can cause landslides and other types of soil failures, which can be dangerous to people and property.

Quick clay is found in areas of Canada, Norway, Sweden, and other regions with a history of glaciation. It is formed from the deposition of clay particles by glacial meltwater, which creates a highly sensitive deposit. Quick clay is a type of clay and not sand, and its behavior is different from that of other types of clay due to its unique physical properties.

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The correct answer is answer is small , highly charged metal cations (e.g. Al³⁺ , Fe²⁺ etc)

According to Lewis acid base theory , electron rich species are called base while electron deficient species are called acids. According to Bronsted Lowry theory , an acid is a species which can donate H⁺ ion is solution while a base is a species which can accept H⁺ in solution. Cations are Lewis acid while anions are Lewis base. Lewis Acids are the chemical species which have empty orbitals and are able to accept electron pairs from Lewis bases.

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Complete question-

Which of the following types of substances are classified as acids only under the Lewis definition but not the Brønsted-Lowry definition?

Salts that contain the conjugate acid of a weak base (e.g., NH4Cl).

Molecules with atoms such as N or O that have electron pairs available to donate to another atom.

Weak acids (e.g., HCN).

Small, highly charged metal cations (e.g. Al3+, Fe2+, etc.).

The IUPAC name for the compound 13264a2,2-diethyl-22-hexanone is (2E)-2,2-diethylhex-2-en-5-one.

To break it down:
- The "2,2-diethyl" part refers to two ethyl groups (CH3CH2) attached to the second carbon of the main chain.
- The "22" indicates the location of the ketone functional group (C=O) on the 2nd carbon from the end of the main chain.
- The "hexanone" part indicates that there are six carbons in the main chain and a ketone functional group.
- The "2E" before the name refers to the stereochemistry of the double bond in the main chain (in this case, it is a trans double bond).

Overall, the IUPAC name for this compound is quite long due to the specific details included in the naming convention.

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if you combined 0.015 moles of salicylic acid with 0.051 moles of acetic anhydride, the theoretical yield (in grams) of acetylsalicylic acid that can be produced is 2.703 grams.

The balanced equation for the reaction of salicylic acid (SA) with acetic anhydride (AA) to produce acetylsalicylic acid (ASA) and acetic acid (AAc) is:

SA + AA → ASA + AAc

The molar mass of salicylic acid is 138.12 g/mol and the molar mass of acetic anhydride is 102.09 g/mol. The molar mass of acetylsalicylic acid is 180.16 g/mol.

To find the limiting reagent and theoretical yield of acetylsalicylic acid:

Calculate the number of moles of each reagent:

0.015 moles SA

0.051 moles AA

Determine the limiting reagent by comparing the mole ratio between SA and AA in the balanced equation (1:1). Since both reactants are in a 1:1 ratio, AA is not limiting.

Calculate the moles of acetylsalicylic acid that can be produced from the limiting reagent (SA):

0.015 moles SA × (1 mole ASA / 1 mole SA) = 0.015 moles ASA

Calculate the theoretical yield of acetylsalicylic acid in grams:

0.015 moles ASA × 180.16 g/mol = 2.703 g ASA

Therefore, the theoretical yield of acetylsalicylic acid that can be produced is 2.703 grams.

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The reaction which is half-cell reactions occurring in a Daniell cell the one which is oxidation is:   Zn(s)→Zn²⁺(aq)+2e⁻.

The Daniell cell is a type of electrochemical device developed in 1836 by British chemist and meteorologist John Frederic Daniell. It consists of a copper pot containing a copper (II) sulphate solution, a zinc electrode, and an unglazed earthenware container containing sulfuric acid. Using a second electrolyte to absorb the hydrogen created by the first, he developed a solution to the hydrogen bubble issue that was discovered in the voltaic pile. Sulfuric acid can be replaced by zinc sulphate. The Daniell cell represented a significant advancement over the earlier battery development technologies.

The International System of Units' volt, the unit of electromotive force, has its modern meaning based on the historical definition of the Daniell cell. The electromotive force of the Daniell cell would be around 1.0 volts according to the definitions of electrical units that were put forward at the 1881 International Conference of Electricians. The standard potential of the Daniell cell at 25 °C is really 1.10 V according to modern specifications.

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An aqueous solution at 25°C, the hydronium ion concentration of the solution is 2.8x10⁻¹¹ M

The quantity of a material, like salt, that is present in a specific volume of tissue or liquid, like blood, according to science. When there is less water present, the substance becomes more concentrated. For instance, when a person doesn't drink enough water, the concentration of salt in their urine may increase.

Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume. Mass concentration, molar concentration, number concentration, and volume concentration are four different categories of mathematical description. Any type of chemical mixture can be referred to by the term "concentration," but solutes and solvents in solutions are most frequently mentioned.

There are many types of molar (quantity) concentration, including normal concentration and osmotic concentration. By adding a solvent to a solution, for example, dilution is the lowering of concentration. The opposite of dilution is concentration increase, which is the meaning of the word concentrate.

You must know that [H₃O⁺] x [OH⁻] = Kw = 1x10⁻¹⁴

Substituting the know [OH-], we can solve for the [H3O+]

[H₃O⁺][3.6x10⁻⁴] = 1x10⁻¹⁴

[H₃O⁺] = 1x10⁻¹⁴ / 3.6x10⁻⁴

[H₃O⁺] = 2.8x10⁻¹¹ M.

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Phosphorus would be a good n-type dopant for the semiconductor silicon.

N-type doping involves adding impurities with extra electrons to the semiconductor material, creating excess negative charge carriers or electrons.

Phosphorus has five valence electrons and, when added to silicon, forms a donor level that provides one extra electron for the conduction band of the semiconductor, making it a good n-type dopant.

Summary: Phosphorus is a suitable n-type dopant for silicon because it provides an extra electron for the conduction band, making it a donor impurity.

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If the car now accelerates from 0 m/s to 30. 0 m/s in 5. 00 s. If the wheels have a radius of 24. 1 cm, then their angular acceleration is 24.9 rad/s².

To calculate the angular acceleration of the car's wheels, we need to first find their angular velocity. We know that the car accelerates from 0 m/s to 30.0 m/s in 5.00 s, so we can calculate the average acceleration:

a = (vf - vi)/t = (30.0 m/s - 0 m/s)/5.00 s = 6.00 m/s²

Next, we can use the formula for linear acceleration in terms of angular acceleration and radius:

a = αr

where α is the angular acceleration and r is the radius of the wheels. Solving for α, we get:

α = a/r = 6.00 m/s² / 0.241 m = 24.9 rad/s²

Therefore, the angular acceleration of the car's wheels is 24.9 rad/s².

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1 mole of phosphorus reacts with 3 moles of hydrogen gas to produce 1 mole of phosphine.
Therefore, 27.9 g of phosphorus and 77.6 L of hydrogen gas will produce 20.1 g of phosphine at STP.

To answer this question, we need to first write the balanced chemical equation for the reaction between phosphorus and hydrogen gas to produce phosphine. The balanced equation is P4 + 6H2 → 4PH3. This means that 1 mole of phosphorus reacts with 3 moles of hydrogen gas to produce 1 mole of phosphine.

To calculate the number of moles of phosphine formed, we need to convert the given quantities of phosphorus and hydrogen gas to moles. The molar mass of phosphorus is 30.97 g/mol, and therefore 27.9 g of phosphorus is equal to 0.901 mol. The volume of hydrogen gas at STP is equal to 22.4 L/mol, and therefore 77.6 L of hydrogen gas is equal to 3.47 mol.

From the balanced equation, we can see that the number of moles of phosphine formed is equal to the number of moles of phosphorus (0.901 mol). Therefore, the mass of phosphine formed is equal to the molar mass of phosphine (33.998 g/mol) multiplied by the number of moles of phosphine (0.901 mol), which is equal to 20.1 g.

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The percent ionization for the 0.10 M HOCl is 0.057 %.

The HOCl is the weak monoprotic acid that is dissociated is as :

HOCl ⇄ H⁺  +  OCl⁻

The ionization constant, Ka = 3.2 × 10⁻⁸

The concentration of the HOCl = 0.10 M.

The expression for ionization constant is as :

Ka = [ H⁺ ] [ OCl⁻ ] / [ HOCl ]

Ka = x² / 0.10

3.2 × 10⁻⁸ = x² / 0.10

x = [H⁺] = 5.7 × 10⁻⁵ M

The x is the hydrogen ion concentration = 5.7 × 10⁻⁵ M

The percent ionization is as :

The percent ionization = ( 5.7 × 10⁻⁵ / 0.10 ) × 100 %

The percent ionization = 0.057 %

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The pH of a 0.188 M NH3 solution at 25°C, with a Kb of 1.76 x 10^-5, is approximately 11.26.

determine the pH of a 0.188 M NH3 solution at 25°C. The Kb of NH3 is 1.76 x 10^-5.

Write the reaction for NH3 in water:
NH3 + H2O ⇌ NH4+ + OH-

Set up an ICE table (Initial, Change, Equilibrium):
NH3      + H2O    ⇌  NH4+    +  OH-
Initial: 0.188 M          0 M       0 M
Change: -x                  +x         +x
Equilibrium: 0.188-x M    x M      x M

Write the Kb expression and substitute the equilibrium concentrations:
Kb = [NH4+][OH-] / [NH3]
1.76 x 10^-5 = (x)(x) / (0.188-x)

Make the approximation that x is much smaller than 0.188, so the equation becomes:
1.76 x 10^-5 ≈ x^2 / 0.188

Solve for x, which represents [OH-]:
x = sqrt(1.76 x 10^-5 * 0.188)
x = 0.00183

Calculate the pOH:
pOH = -log10(0.00183)
pOH = 2.74

Determine the pH by subtracting the pOH from 14:
pH = 14 - 2.74
pH = 11.26

The pH of a 0.188 M NH3 solution at 25°C, with a Kb of 1.76 x 10^-5, is approximately 11.26.

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When the authorities units a tax, it ought to determine whether or not to levy the tax at the manufacturers or the purchasers. This is known as prison tax occurrence.

The maximum famous taxes are ones levied at the consumer, inclusive of Government Sales Tax (GST) and Provincial Sales Tax (PST). The authorities additionally units taxes on manufacturers, inclusive of the fueloline tax, which cuts into their profits. The prison occurrence of the tax is surely inappropriate while figuring out who's impacted through the tax. When the authorities levies a fueloline tax, the manufacturers will byskip a number of those fees on as an improved fee. Likewise, a tax on purchasers will in the end lower amount demanded and decrease manufacturer surplus. This is due to the fact the monetary tax occurrence, or who surely will pay withinside the new equilibrium for the occurrence of the tax, is primarily based totally on how the marketplace responds to the fee change – now no longer on prison occurrence.

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In a buffer solution made of acetic acid and sodium acetate, any acid that is added will react with hydroxide.

Acid is a substance that has a pH level of less than 7, meaning it is more acidic than pure water. Acids are found in nature and are also produced through chemical reactions. Acids are able to dissolve certain substances, such as metals, and can be used to break down proteins, fats, and carbohydrates. Acids are often used in industrial processes, such as in the production of fertilizers and in electroplating. Acids can be classified as either mineral acids, such as hydrochloric acid, or organic acids, such as acetic acid. Acids can be beneficial, such as when used in food processing and preservation, or hazardous, such as when exposed to skin or eyes. Acids can also be used in cleaning products, including to remove rust and calcium deposits.

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The standard enthalpy of formation of SO₃(g) is -49.5 kJ/mol.

The heat change caused by a chemical reaction at constant volume or constant pressure is referred to as enthalpy change. The enthalpy change indicates how much heat was absorbed or evolved during the reaction. It is represented by the letter ΔH.

We can use Hess's Law to calculate the standard enthalpy of formation (ΔfH°) of SO₃(g) using the given information:

ΔfH°[2SO₂(g)] + ΔfH°[O₂(g)] → ΔfH°[2SO₃(g)]

Since ΔrH° = -198 kJ for the reaction 2SO₂(g) + O₂(g) → 2SO₃(g), we know that:

ΔfH°[2SO₂(g)] + ΔfH°[O₂(g)] = -198 kJ/mol

Using the given values, we can find the enthalpies of formation of SO₂(g) and O₂(g):

ΔfH°[SO₂(g)] = -297 kJ/mol

ΔfH°[O₂(g)] = 0 kJ/mol (by definition)

Substituting these values into the equation above, we can solve for ΔfH°[2SO₃(g)]:

ΔfH°[2SO₃(g)] = ΔfH°[2SO₂(g)] + ΔfH°[O₂(g)] - ΔrH°

ΔfH°[2SO₃(g)] = (-297 kJ/mol) + (0 kJ/mol) - (-198 kJ/mol)

ΔfH°[2SO₃(g)] = -99 kJ/mol

Finally, we can divide by 2 to get the standard enthalpy of formation of SO₃(g):

ΔfH°[SO₃(g)] = ΔfH°[2SO₃(g)] / 2

ΔfH°[SO₃(g)] = (-99 kJ/mol) / 2

ΔfH°[SO₃(g)] = -49.5 kJ/mol

Therefore, the standard enthalpy of formation of SO₃(g) is -49.5 kJ/mol.

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Therefore, the molar solubility of Ba(IO3)2 in a solution of 0.010 M NaIO3 is 5.31 x 10^-4 M

The balanced equation for the dissolution of Ba(IO3)2 in water is:

Ba(IO3)2(s) → Ba2+(aq) + 2 IO3-(aq)

The solubility product expression for Ba(IO3)2 is:

Ksp = [Ba2+][IO3-]^2

Let x be the molar solubility of Ba(IO3)2 in the presence of NaIO3. The reaction between Ba(IO3)2 and NaIO3 will produce Ba(IO3)2 and NaIO3 in solution. Assuming that the NaIO3 does not affect the solubility of Ba(IO3)2, we can write the following equation for the dissolution of Ba(IO3)2:

Ba(IO3)2(s) ⇌ Ba2+(aq) + 2 IO3-(aq)

Initially, there is no Ba2+ or IO3- in solution, so we can write:

[Ba2+] = x

[IO3-] = 2x

The concentration of IO3- is not affected by the presence of NaIO3 because NaIO3 does not contain any IO3- ions.

The concentration of Ba2+ is related to the solubility of Ba(IO3)2 by the equation:

[Ba2+] = 2x + [NaIO3]

The [NaIO3] is 0.010 M.

Substituting these expressions into the solubility product expression:

Ksp = (2x + 0.010)(2x)^2

Solving for x using the given Ksp value of 2.2 x 10^-9:

2.2 x 10^-9 = (2x + 0.010)(2x)^2

x = 5.31 x 10^-4 M

Therefore, the molar solubility of Ba(IO3)2 in a solution of 0.010 M NaIO3 is 5.31 x 10^-4 M, and the correct answer is (C)

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An aromatic substitution reaction, the presence of iron in the reaction mixture enhances the electrophilicity of the bromine atom enough that it will react with benzene.

An electrophile substitutes an atom that is affixed to an aromatic ring in electrophilic aromatic substitution reactions, which are organic processes. Typically, in these reactions, an electrophile takes the place of a hydrogen atom from a benzene ring.

An electrophilic aromatic substitution process maintains the aromaticity of the aromatic system. The stability of the aromatic ring is retained, for instance, when bromobenzene is produced from the reaction of benzene and bromine.

A substitution reaction is when an atom or group of atoms in an organic molecule are directly replaced by another atom or group of atoms without causing any changes to the remaining components of the molecule. Electrophiles can start substitution reactions, which are referred to as electrophilic substitution reactions. Nucleophilic substitution reactions are those involving substitution that start with a nucleophile attack.

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The correct answer is d. 11.3 kJ released, which is the value obtained by rounding the actual heat of hydrofluoric acid released to two significant figures.

Based on the given thermochemical equation, the reaction releases 910.9 kJ/mol of SiO2 reacted. To determine the amount of heat released or absorbed in the reaction of 10.0 grams of SiO2 with excess hydrofluoric acid, we need to first convert the mass of SiO2 to moles.

Molar mass of SiO2 = 60.08 g/mol
10.0 g SiO2 ÷ 60.08 g/mol = 0.166 mol SiO2

Since the reaction ratio between SiO2 and heat is 1 mol SiO2 : 910.9 kJ, we can calculate the amount of heat released or absorbed by multiplying the number of moles of SiO2 by the heat released per mole:

0.166 mol SiO2 x 910.9 kJ/mol = 151.2 kJ

The heat released in the reaction of 10.0 grams of SiO2 with excess hydrofluoric acid is 151.2 kJ. However, the answer choices do not match this value exactly. The closest answer choice is e. 6.56 kJ released. This answer is incorrect, as it is much lower than the actual heat released in the reaction.

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The element with the lowest IE is K (Krypton) due to its large atomic radius. This means that the outermost electrons are held less tightly by the nucleus, and are therefore easier to remove.

Electrons are subatomic particles with a negative electric charge. They are found in atoms, and are responsible for electric currents and chemical reactions. Electrons are the smallest known particles and are a major part of all matter. In a normal atom, the number of electrons is equal to the number of protons. Electrons are important components of the electromagnetic force, which holds atoms together. They are also responsible for the electrical and magnetic properties of materials. In addition, electrons can be used to create energy in the form of electricity.

The next lowest IE is Ca (Calcium), which has a relatively small atomic radius and is just one electron away from a filled shell, making it slightly more difficult to remove an electron. Rb (Rubidium) is the next lowest, with a slightly larger atomic radius than Ca, making it easier to remove an electron. Finally, Br (Bromine) has the highest IE of the four elements, as its outermost electron is held more tightly than those of the other elements due to its smaller atomic radius.

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Concentration the moles of ClF₃ formed in the first 6s reaction will be 0.1176 moles . If the concentration of F₂(g) dropped

                              Cl₂ + 3 F₂  → 2ClF₃

Rate of reaction  = [tex]\frac{-1}{3} \frac{dF_{2} }{dt}[/tex]  = [tex]\frac{+ 1}{2} \frac{dClF_{3} }{dt}[/tex]

Initial concentration of F₂ = 0.901 M

Concentration after 19 s = 0.554 M

Change in concentration = concentration after 19 s --- Initial concentration  of F₂

                     = 0.554 - 0.901

                       =  - 0.347 M

Time taken = 19 s

                                       [tex]\frac{-1}{3} \frac{dF_{2} }{dt} = \frac{+1}{2}\frac{dClF_{3} }{dt}[/tex]

                                   0.347 × 2× 6 / 3 × 19 = [tex]\frac{dClF_{3} }{1}[/tex]

d[ClF₃] = 0.07305

concentration after the 6s - initial concentration = 0.07305

concentration of ClF₃ After 6 s = 0.07305 M

volume of reaction vessel = 1.61 L

                       Molarity = moles ÷ volume

                       Molarity × volume = moles of ClF₃

                         0.07305 × 1.61 = moles of ClF₃

Hence , the moles of ClF₃ formed in the first 6s will be 0.1176 .

The speed at which a chemical reaction occurs is referred to as the reaction rate or rate of reaction. It is proportional to the increase in product concentration per unit time and the decrease in reactant concentration per unit time. In chemistry, the term "reaction rate" refers to the rate of a chemical reaction.

It is frequently expressed in terms of either the concentration of a reactant that is consumed in a unit of time or the concentration of a product that is formed in a unit of time (amount per unit volume).

Incomplete question :

Consider the reaction: Cl2(g) + 3 F2(g) + 2 ClF3(g) In the first 19 s of this reaction, the concentration of F2(g) dropped from 0.901 M to 0.554 M. If the volume of the reaction vessel was 1.61 L, what amount of CIF3(g) (in moles) was formed during the first 6 s of the reaction? moles number (rtol=0.03, atol=1e-08)

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The number of moles of the product that is obtained is  137.5 moles

Chemical reactions classified as first-order reactions have rates of reaction that are inversely correlated with the concentration of a single component. In other words, the reaction rate is inversely proportional to the reactant concentration, expressed as a power of 1.

We know that;

-1/3 [dF2]/dt = 1/2[dClF3]/dt

-1/3 ( 0.554 - 0.901 )/19 - 0 = 1/2 [dClF3]/7 - 0

6.1 * 2 * 7 = [dClF3]

85.4 M

Hence the number of moles = 85.4 M *  1.61  L

= 137.5 moles

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Missing parts;

Consider the reaction: Cl2(g) + 3 F2(g) + 2 ClF3(g) In the first 19 s of this reaction, the concentration of F2(g) dropped from 0.901 M to 0.554 M. If the volume of the reaction vessel was 1.61 L, what amount of CIF3(g) (in moles) was formed during the first 7 s of the reaction?

The mass of magnesium sulfide produced when 12 grams of magnesium reacts with 18 grams of sulfur is 27.8 grams.

To determine the mass of magnesium sulfide (MgS) produced when a 12-gram sample of magnesium reacts with an 18-gram sample of sulfur, we need to use the law of conservation of mass. This law states that in a chemical reaction, the mass of the products must be equal to the mass of the reactants.
The balanced chemical equation for this reaction is:
Mg + S → MgS
From this equation, we can see that 1 mole of magnesium reacts with 1 mole of sulfur to produce 1 mole of magnesium sulfide. The molar mass of magnesium is 24.31 g/mol, and the molar mass of sulfur is 32.06 g/mol. Therefore, the number of moles of magnesium and sulfur are:
moles of Mg = 12 g / 24.31 g/mol = 0.494 moles
moles of S = 18 g / 32.06 g/mol = 0.561 moles
The limiting reactant in this reaction is magnesium, since it is the reactant that will be completely consumed in the reaction. Therefore, the number of moles of magnesium sulfide produced is also 0.494 moles.
The molar mass of magnesium sulfide is 56.30 g/mol. To find the mass of magnesium sulfide produced, we can use the following equation:
mass of MgS = moles of MgS x molar mass of MgS
mass of MgS = 0.494 mol x 56.30 g/mol
mass of MgS = 27.8 g
Therefore, the mass of magnesium sulfide most likely produced is 27.8 g.  a27.8 grams.

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Exchange of gases and metabolites between the blood and tissues occurs in the capillaries.

Capillaries are small and thin-walled blood vessels that connect arteries and veins. They are the smallest of the blood vessels, with diameters ranging from 5 to 10 micrometers, and they are so numerous that virtually every cell in the body is located within a few micrometers of a capillary.

Capillaries have several important functions in the body, but their most important role is to facilitate the exchange of gases, nutrients, and waste products between the blood and tissues. Oxygen and nutrients diffuse from the capillaries into the tissues, while carbon dioxide and other waste products diffuse from the tissues into the capillaries to be carried away and eliminated from the body.

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The correct balanced molecular equation for the complete neutralization of H2SO4 by KOH in aqueous solution is option D, which is H2SO4 (aq) + 2KOH (aq) → 2H2O (l) + K2SO4 (aq). This equation shows that for every one mole of sulfuric acid (H2SO4) that reacts with two moles of potassium hydroxide (KOH), two moles of water (H2O) and one mole of potassium sulfate (K2SO4) are formed.

This is a neutralization reaction, which means that the acid and base react to form a salt and water. The reaction proceeds through the transfer of protons (H+) from the acid to the hydroxide ions (OH-) in the base.

The result is the formation of water and a salt. The balanced equation shows the stoichiometry of the reaction, which is essential for accurately predicting the amounts of reactants and products that will be formed.

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The mixture always be called a solution when the mixture is homogeneous, option D.

A mixture is a concoction of two or more unrelated chemicals that are not linked chemically. The two categories of mixtures are homogeneous mixtures and heterogeneous mixtures. A mixture is considered homogenous if its composition is constant throughout. It is the kind of combination in which the components are evenly distributed or whose composition is continuous throughout.

A mixture that is homogenous has the same proportions of its constituent parts throughout a particular sample, whether it be a solid, liquid, or gaseous combination. Its makeup is constant throughout. In a homogenous combination, just one phase of matter is visible.

A solution is, for instance, salt or sugar dissolved in water. The size of the particles is less than 2 x 10⁻⁹ m. They are so tiny that it is impossible to tell the difference between the solute, which is being dissolved, and the solvent, which dissolves the solute.

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Complete question:

In which situation can a mixture always be called a solution?

when one of its components is a gas

when there is no solvent in the mixture

when its components are made up of different types of particles

when the mixture is homogeneous

There are 0.00826 moles of HCl present in a 35.67 mL sample of a 0.232 M solution. To calculate the number of moles of HCl present in a 35.67 mL sample of a 0.232 M solution, we need to use the formula:

moles = concentration (M) x volume (L)

Firstly, we need to convert the volume from mL to L by dividing it by 1000:

35.67 mL ÷ 1000 = 0.03567 L

Then we can plug in the values we have:

moles = 0.232 M x 0.03567 L

moles = 0.00826 mol

It's important to note that the concentration of a solution is the amount of solute (in this case, HCl) dissolved in a given amount of solvent (usually water) and is measured in moles per liter (M). This calculation is useful in chemistry when we want to know the amount of a substance present in a solution.

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To make a 1% agarose gel, weigh 1 g of agarose powder and dissolve it in 100 ml of electrophoresis buffer by heating and stirring.

Agarose gel electrophoresis is a widely used technique in molecular biology for separating DNA fragments. To prepare a 1% agarose gel, first, weigh 1 g of agarose powder and add it to 100 ml of electrophoresis buffer (such as TAE or TBE). Heat the mixture while stirring until the agarose powder dissolves completely. Allow the solution to cool until it reaches approximately 60°C.

Then, pour the solution into a casting tray with a comb inserted. Allow the gel to solidify at room temperature for about 30 minutes. After the gel is solidified, carefully remove the comb and place the gel into the electrophoresis chamber. Fill the chamber with electrophoresis buffer and carefully load the DNA samples into the wells. Run the gel at a suitable voltage until the desired separation is achieved.

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The nucleophilic species in the reaction of 1-butene with bromine in aqueous solution is water.

Water acts as the nucleophile attacking the carbocation intermediate formed during the addition of bromine to 1-butene. In the first step of the reaction, bromine is added to the double bond of 1-butene, forming a cyclic bromonium ion intermediate.

The bromonium ion is then attacked by a water molecule (nucleophile), which opens the ring and results in the formation of a carbocation intermediate. The carbocation intermediate is attacked by another water molecule to form 1-bromo-2-butanol as the main product Overall, the reaction proceeds via an electrophilic addition mechanism, where bromine acts as the electrophile, and water acts as the nucleophile.

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