The molarity (m) of an aqueous solution containing 22.5 g of glucose (c6h12o6) in 35.5 ml of solution is ________.

To find the grams of H3PO4 in 255 mL of a 4.50 M solution, we can use the formula Molarity (M) = moles of solute / liters of solution. By converting the volume from mL to L and using the molar mass of H3PO4, we can calculate that there are 112.25 grams of H3PO4 in the solution.

To find the grams of H3PO4 in 255 mL of a 4.50 M solution, we need to use the formula:

Molarity (M) = moles of solute / liters of solution

First, we need to convert the volume from mL to L:

255 mL = 0.255 L

Next, we can use the formula to find the moles of H3PO4:

4.50 M = moles / 0.255 L

moles = 4.50 M * 0.255 L = 1.1475 moles

Finally, we can use the molar mass of H3PO4 (97.99 g/mol) to find the grams of H3PO4:

grams = moles * molar mass = 1.1475 moles * 97.99 g/mol = 112.25 g

Therefore, there are 112.25 grams of H3PO4 in 255 mL of a 4.50 M solution.

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The molar solubility of Mg(OH)2 in 0.200 M NaOH can be calculated by considering the common ion effect and using the Ksp value provided. An ICE table is typically used to determine the concentrations at equilibrium, accounting for the initial NaOH concentration and the Ksp expression for Mg(OH)2's dissociation.

To determine the molar solubility of Mg(OH)2 in the presence of an additional source of OH− ions from NaOH, we must account for the common ion effect. The solubility product constant, Ksp, is given for magnesium hydroxide and can be used to establish a relationship between the concentrations of magnesium and hydroxide ions in a saturated solution.

The dissociation of Mg(OH)2 in water is represented by the equation:
Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH− (aq).

The expression for Ksp is:
Ksp = [Mg2+][OH−]².

However, the initial concentration of OH− ions from NaOH must be considered, and common ion effect will reduce the molar solubility of Mg(OH)2 compared to its solubility in pure water. The precise calculation would require solving the equilibrium expressions taking into account the initial NaOH concentration, the dissociation of Mg(OH)2, and the Ksp value provided. This involves setting up an ICE table (Initial, Change, Equilibrium) and solving for the equilibrium concentrations.

The reaction SO3 (g) + H2O (l) → H2SO4 (aq) is classified as an acid-base reaction, specifically demonstrating the behavior of sulfur trioxide (an acid) reacting with water (a base) to produce sulfuric acid. Also, this reaction can be considered a type of synthesis reaction.

The equation provided, SO3 (g) + H2O (l) → H2SO4 (aq), is an example of an acid-base or a synthesis reaction. In acid-base reactions, an acid (SO3 in this case) combines with a base (H2O here) to form water and an ionic compound (H2SO4 here, a strong acid). In a more general sense, this is also a synthesis reaction because two or more simple substances (SO3 and H2O) combine to create a more complex one (H2SO4).

Moreover, taking into consideration the

acid-base behavior

of substances and their reactions, an acid would donate a proton and a base would accept it, thus, in our reaction, H2O acts as a base accepting a proton from SO3.

Additionally, the process of synthesizing sulfuric acid from sulfur dioxide and water is an important industrial method in the production of this widely used chemical.

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The reaction [tex]SO_3 (g) + H_2O (l) \rightarrow H_2SO_4(aq)[/tex] is classified as a synthesis reaction. It involves the combination of sulfur trioxide and water to form sulfuric acid. Sulfuric acid in water acts as a dibasic acid, forming bisulfate and sulfate anions.

The classification for the reaction [tex]SO_3 (g) + H_2O (l) \rightarrow H_2SO_4(aq)[/tex] is a synthesis reaction. This type of reaction occurs when two or more simple substances (in this case, sulfur trioxide and water) combine to form a more complex substance (sulfuric acid). Sulfuric acid is a strong protic acid and in water, it acts as a dibasic acid, forming bisulfate ([tex]HSO_4^-[/tex]) and sulfate ([tex]SO_4^{2-[/tex]) anions.

First Ionization:

[tex]H_2SO_4 (aq) + H_2O (l) \rightarrow H_3O^+ (aq) + HSO_4^- (aq)[/tex]

Second Ionization:

[tex]HSO_4^- (aq) + H_2O (l) \rightarrow H_3O^+ (aq) + SO_4^{2- (aq)[/tex]

In conclusion, the reaction of [tex]SO_3[/tex] with [tex]H_2O[/tex] forming [tex]H_2SO_4[/tex] is a classic example of a synthesis reaction in chemistry.

The limiting reagent in a chemical reaction between copper sulfate and zinc or aluminum is usually the metal (zinc or aluminum).

The limiting reagent in a chemical reaction is the substance that is completely consumed when the chemical reaction is complete. In the reaction between copper sulfate and zinc or aluminum, the metal (zinc or aluminum) is typically the limiting reagent. This is because the copper sulfate is usually present in excess.

In this specific scenario, when zinc or aluminum reacts with copper sulfate, the zinc or aluminum displaces the copper. This displacement reaction leads to the formation of zinc sulfate or aluminum sulfate and copper metal.

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A flask contains 0.220 mole of liquid bromine, Br2. The number of bromine molecules present in the flask is 1.32506 x 10²³ moles.

Molecules are defined as the lowest identifiable unit into which a pure material can be divided while maintaining its chemical makeup and attributes is made up of two or more atoms. These four types of molecules are usually referred to as "molecules of life." Nucleic acids, lipids, proteins, and carbohydrates make up the four basic building blocks of life. Each and every one of the four groups is essential for every living thing on Earth.

The amount of liquid Br2 = 0.220 mole

We know that 1 mole of Br2 = 6.023 x 10²³

So, the amount of 0.22 mole

= 0.220 x 6.023 x 10²³

=  1.32506 x 10²³ moles

Thus, a flask contains 0.220 mole of liquid bromine, Br2. The number of bromine molecules present in the flask is 1.32506 x 10²³ moles.

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Final answer:

The number of bromine molecules in a flask containing 0.220 mol of bromine is 1.32484 x 10^23 molecules.

Explanation:

To determine the number of bromine molecules present in a flask that contains 0.220 mol of liquid bromine (Br2), we use Avogadro's constant, which is 6.022 × 10^23 molecules/mol. The calculation is as follows:

Number of molecules = moles × Avogadro's number

Number of molecules = 0.220 mol × 6.022 × 10^23 molecules/mol

Number of molecules = 1.32484 × 10^23 molecules of Br2

The molar mass of phenol is [tex]\boxed{94.113{\text{ g/mol}}}[/tex] .

Further Explanation:

Structure of phenol:

The phenol is a white crystalline aromatic compound which contains a hydroxyl group (-OH). Phenol is generally an active ingredient of phenol spray which helps in sore throat. It is also an active ingredient of some more oral analgesics which are used medically to reduce the pain.

The structure of phenol is composed of a benzene ring where one of the hydrogens is replaced by the –OH group. Since the molecular formula of benzene is [tex]{{\text{C}}_6}{{\text{H}}_6}[/tex] , therefore, after replacing of one hydrogen atom with –OH group, the molecular formula becomes,

[tex]\begin{aligned}{\text{Molecular formula}}\left( {{\text{Phenol}}} \right)&= {\text{Benzene}}-{\text{H}}\left({{\text{one}}}\right)+{\text{OH}}\\&={{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}} - {\text{H}}+{\text{OH}}\\&={{\text{C}}_{\text{6}}}{{\text{H}}_5}+{\text{OH}}\\&={{\text{C}}_{\text{6}}}{{\text{H}}_5}{\text{OH}}\\\end{aligned}[/tex]

Now calculate the molar mass of phenol as follows:

The formula to calculate the molar mass of phenol [tex]\left({{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}}\right)[/tex] is,

[tex]{\text{Molar mass}}=\left[\begin{gathered}\left({{\text{Total number of C}}}\right)\left({{\text{Atomic mass of C}}}\right)+\hfill\\\left({{\text{Total number of H}}}\right)\left({{\text{Atomic mass of H}}}\right)+\hfill\\\left({{\text{Total number of O}}}\right)\left({{\text{Atomic mass of O}}}\right)\hfill\\\end{gathered}\right][/tex]

The atomic mass of carbon atom is 12.011 g/mol.

The atomic mass of oxygen atom is 15.999 g/mol.

The atomic mass of hydrogen atom is 1.008 g/mol.

Substitute the respective values in the above formula,

Molar mass = [(Total number of C)(Atomic mass of C) + (Total number of H)(Atomic mass of H) + (Total number of O)(Atomic mass of O)]

[tex]\begin{aligned}{\text{Molar mass}}&=\left[\begin{gathered}\left( {\text{6}}\right)\left({{\text{12}}{\text{.011 g/mol}}} \right)+\left({\text{6}} \right)\left({{\text{1}}{\text{.008 g/mol}}}\right)+\hfill\\\left({\text{1}} \right)\left({{\text{15}{\text{.999gmol}}}\right)\hfill\\\end{gathered}\right]\\&=94.113{\text{ g/mol}}\\\end{gathered}[/tex]

Therefore, the molar mass of phenol is 94.113 g/mol.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: Phenol, molar mass, moles, atomic mass, structure of phenol, benzene ring, hydrogen atoms, carbon atoms, benzene ring, OH group, 94.113 g/mol.

The molar mass of phenol is 94.11 g/mol

Phenol is known mainly as chemical and a natural substance. It is said to be of no color but white solid at its pure state

It has a unique odor that is sweet and tarry and it does evaporates more slowly than water.

Phenol is known as the family of organic compounds characterized by a hydroxyl (―OH) group attached to a carbon atom.

The term phenol is the specific name for its simplest member, monohydroxybenzene (C6H5OH), called benzenol, or carbolic acid.

Conclusively, It is used in tiny quantity as a disinfectant in household cleaners, mouthwash etc.

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Answer : The volume of water added = 9.411 ml

Solution : Given,

Absorbance decreases by 32% upon dilution means,

Let,  initial absorbance = 100 [tex]Lmol^{-1}Cm^{-1}[/tex]

then  final absorbance = 100 - 32 = 68 [tex]Lmol^{-1}Cm^{-1}[/tex]

initial concentration = 1.2 M

initial volume = 20 ml = 0.02 L         ( 1 L = 1000 ml )

According to Beer-Lambert law, the absorbance is directly proportional to the concentration of an absorbing species.

A  ∝  C

[tex]\frac{A_{inital}}{A_{final}}=\frac{C_{initial}}{C_{final}}[/tex]      

Now put all the given values in this formula, we get

[tex]\frac{100}{68}=\frac{1.2}{C_{final}}[/tex]

[tex]C_{final}[/tex] = 0.816 M

Now, calculating the number of moles,

Moles = concentration × volume

Moles = 1.2 × 0.02 = 0.024 moles

Now, Calculating the final volume by this formula.

[tex]C_{final}=\frac{moles}{V_{final}}[/tex]

[tex]V_{final}[/tex] = [tex]\frac{0.024}{0.816}[/tex] = 0.0294 L = 29.411 ml

The inital volume is 20 ml and final volume is 29.411 ml.

Volume of water added = final volume - initial volume = 29.411 - 20 = 9.411 ml

Final answer:

Approximately 9.38 mL of water was added to the original 20.0 mL of a 1.20 M copper(II) ion solution to achieve a 32.0% reduction in absorbance.

Explanation:

When diluting a solution, the concentration is reduced, and this often leads to a decrease in the absorbance of the solution if the compound being measured absorbs light. Since the absorbance of the copper(II) ion solution decreases by 32.0% upon dilution, we can calculate the volume of water added using the formula C1V1 = C2V2, where C1 and V1 are the initial concentration and volume and C2 and V2 are the final concentration and volume. The initial absorbance does not need to be known since we have the percent decrease and initial volume.

Let's assume the initial absorbance, A1, correlates linearly with concentration, which drops to 68.0% of its original value. Consequently, the final concentration, C2, is also 68.0% of C1. Substitution yields:

Now applying the dilution formula:

Rearranging and solving for Vw:

Therefore, approximately 9.38 mL of water was added to the original 20.0 mL of a 1.20 M solution of Cu2+(aq) to achieve the 32.0% reduction in absorbance.

The element with half life of 1.3 billion years have the decay constant of 0.533 billion yrs⁻¹ will take 2.01 billion years to decay to 123 mg from the initial amount of 12 g. Hence, the age of the rock is 2.01 billion years.

Heavy unstable radioactive isotopes undergo nuclear decay by the emission of charged particles. Nuclear decay is a first order reaction.

Thus, the decay constant k  = 1/t ln W0/Wt

where t is the time of decay, W0 be the initial amount and Wt be the amount after time t.

The half life of the element =  1.3 billion years

decay constant, k = 0.693/1.3 billion yrs  = 0.53 billion yrs⁻¹.

The time taken to decay to 3 mg of its original amount is calculated as follows:

t = 1/k  ln Wo/(Wt )

= ln( 12 mg/3 mg)/  0.53 billion yrs⁻¹.

= 2.01 billion years

Therefore, the time before the element decay to 3 mg from 12 mg is 2.01 billion years. Hence, the age of the rock is 2.01 billion years.

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The compound formed between Na+ (sodium ion) and O2- (oxide ion) is Sodium Oxide, with the chemical formula Na2O.

The compound formed between Na+ (sodium ion) and O2- (oxide ion) is called Sodium Oxide. The formula of this compound is Na2O. In this compound, two sodium atoms each give their one electron to one oxygen atom which has two free spaces in its outer shell. This results in a stable compound with complete outer electron shells for each atom involved. The positive and negative charges are balanced in this compound as the ionic bond is formed.

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Answer:

Extrusive igneous rock

Explanation:

The igneous rocks that are crystallized on the earth's surface are commonly known as the extrusive igneous rock. These rocks are characterized by the presence of fine-grained crystals which is due to the rapid cooling of the magma, as it does not get enough time for the crystals to form. These are also sometimes comprised of volcanic glasses.which are formed on the surface due to the eruption of volcanoes.

When the hot intruding magma crystallizes below the earth's surface, then it is called intrusive igneous rocks. For example, granite, diorite and gabbro.

On the other hand, when the hot uprising magma crystallizes on or near the earth's surface, then it is known as extrusive igneous rocks. For example, basalt, rhyolite and andesite.

Final answer:

Buffers maintain a steady pH in a solution but not necessarily a neutral pH of 7. They can maintain acidic, neutral, or basic pH levels depending on their composition and are essential for processes requiring stable pH conditions, such as in human blood.

Explanation:

The question about whether a buffer is supposed to keep the pH of a solution at 7 (neutral) reflects a common misconception. Buffers do not necessarily maintain a pH of 7. They are solutions that consist of a weak acid and its conjugate base and function to moderate pH changes when an acid or base is added. This helps maintain a steady pH, but the specific pH maintained by a buffer depends on the properties of the acid-base pair it is made from. For example, the pH of a buffer can be acidic, neutral, or basic, depending on its composition. The crucial function of a buffer is its capacity to absorb excess H⁺ ions or OH⁻ ions, thus preventing significant shifts in pH.

It's important to understand that the pH scale ranges from 0 to 14, with 7 being neutral. Solutions with a pH lower than 7 are considered acidic, and those with a pH higher than 7 are basic. Buffers can be designed to maintain any pH within this range, depending on the needs of the environment or process they are used in. For instance, human blood has a buffer system that maintains its pH around 7.35 to 7.45, slightly alkaline, which is crucial for physiological functions.

There are only two possible structures for an octahedral molecule with a formula of ax4y2. One is when the two y’s are next to each other and the other one is when the two y’s are of the opposite side of the molecule. Different structures can be drawn but only two types of molecule can be formed.

An ion is an atom in which the total number of electrons is not the same with the total number of protons, giving it a net positive or negative electrical charge.

The electron configuration of chlorine is [Ne]3s23p5. If a chlorine atom adds an electron to form the chloride ion (Cl-), it will have an electron configuration of [Ne]3s23p6. While Argon has no charge.

The monatomic ion with a charge of 1- and the condensed electron configuration [ne]3s23p6 is the Chloride ion (Cl-).

The monatomic ion with a charge of 1- and the condensed electron configuration [ne]3s23p6 is the Chloride ion (Cl-).