Where do scientists think the abundance of water on Earth came from?

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.

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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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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The mass in grams of 6.5×10²⁰ molecules of aspirin (C₉H₈O₄) is 0.194 g

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole of C₉H₈O₄

1 mole of C₉H₈O₄ = (12×9) + (1×8) + (16×3)

= 180 g

Therefore, we can say that:

6.02×10²³ molecules = 180 g of C₉H₈O₄

6.02×10²³ molecules = 180 g of C₉H₈O₄

Therefore,

6.5×10²⁰ molecules = [tex]\frac{180 * 6.5*10^{20} }{6.02*10^{23}}\\\\[/tex]

6.5×10²⁰ molecules = 0.194 g

Thus, the mass of 6.5×10²⁰ molecules of aspirin (C₉H₈O₄) is 0.194 g

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The mass in grams of the aspirin is 0.194 g

From the question,

We are to determine the mass in grams of  6.5×10²⁰ molecules of aspirin (C₉H₈O₄)

First, we will determine the number of moles of aspirin present

Using the formula

[tex]Number\ of\ moles = \frac{Number\ of\ molecules}{Avogadro's\ constant}[/tex]

Avogadro's constant = 6.022 × 10²³ mol⁻¹

From the given information

Number of molecules of aspirin = 6.5×10²⁰ molecules

∴ Number of moles of aspirin present = [tex]\frac{6.5 \times 10^{20} }{6.022 \times 10^{23} }[/tex]

Number of moles of aspirin present = 0.0010793756 moles

Number of moles of aspirin present = 1.0793756 ×10⁻³ mole

Now, to determine the mass of aspirin present

From the formula

Mass = Number of moles × Molar mass

Molar mass of aspirin = 180.158 g/mol

∴ Mass of aspirin = 0.0010793756 × 180.158

Mass of aspirin = 0.194458 g

Mass of aspirin = 0.194 g

Hence, the mass in grams of the aspirin is 0.194 g

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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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2. Stereoisomer for 1,2-difluoro-1,2-dimethylcyclopentane:https://brainly.com/question/8803065

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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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 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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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.

Final answer:

The molarity of the HNO3 solution is calculated by first determining the number of moles of borax used in the titration, then finding the equivalent moles of HNO3 based on the reaction stoichiometry, and finally dividing the moles of HNO3 by the volume of solution used in the titration.

Explanation:

Calculating the Molarity of HNO3:

To determine the molarity of the HNO3 solution, we need to follow a series of steps that involve the titration of borax with HNO3. First, we need to calculate the number of moles of borax using its molar mass. We know that the molar mass of borax is 381.372 g/mol and that 0.2619 g of borax was used in the titration. The number of moles of borax can be calculated as:

moles borax = (0.2619 g) / (381.372 g/mol)

Once we know the moles of borax, we can use the titration reaction to find the moles of HNO3 that react with the borax. Assuming a 1:1 reaction (which is typical for a neutralization reaction), the moles of HNO3 would be equal to the moles of borax. We then divide the number of moles of HNO3 by the volume of HNO3 used (in liters) to get the molarity:

molarity of HNO3 = moles of HNO3 / volume of HNO3 (in L)

Finally, we can fill in the known values and calculate the molarity of the HNO3 solution precisely.

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.

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-).

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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Explanation:

Molarity is the number of moles present in a liter of solution.

Mathematically,     Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]

And, no. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

Molar mass of [tex]C_{6}H_{12}O_{6}[/tex] is 180.15 g/mol. Therefore, number of moles present will be as follows.

   No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                         = [tex]\frac{22.5 g}{180.15 g/mol}[/tex]

                          = 0.124 mol

Hence, calculate the molarity as follows.

                Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]

                       = [tex]\frac{0.124 mol}{0.0355 L}[/tex]      (as 1 L = 1000 mL)

                       = 3.51 M

Thus, we can conclude that molarity of the solution is 3.51 M.

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.