How many grams are in 4.63 x 1024

molecules of CCl4?

Answer:

32.1 g

Explanation:

Step 1: Write the balanced combustion reaction

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 97.4 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

97.4 g × 1 mol/44.01 g = 2.21 mol

Step 3: Calculate the moles of butane that produced 2.21 moles of carbon dioxide

The molar ratio of C₄H₁₀ to CO₂ is 1:4. The moles of C₄H₁₀ required are 1/4 × 2.21 mol = 0.553 mol

Step 4: Calculate the mass corresponding to 0.553 moles of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

0.553 mol × 58.12 g/mol = 32.1 g

Answer:

208 grams

Explanation:

Half life is the time it takes for a sample to reduce to half it's initial concentration.

In this question, 3 half lives go by. This is given as;

Let original mass be represented by x

First Half life;

Starting mass = X

Remaining mass = X/2

Second Half life;

Starting mass = X/2

Remaining mass = (X/2) / 2 = x/4

Third Half life;

Starting mass = X/4

Remaining mass = (X/4) / 2 = X/8

Since 26g was left, it means;

X/8 = 26

X = 26 * 8  = 208 grams

Answer:

Mass = 14.0 g

Explanation:

Given data:

Mass of mercury nitrate = 22.17 g

Mass of mercury formed = ?

Solution:

Chemical equation:

Hg(NO₃)₂ + 2K    →    2KNO₃ + Hg

Number of moles of mercury nitrate:

Number of moles = mass/molar mass

Number of moles = 22.17 g / 324.6 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of Hg(NO₃)₂ and mercury.

                Hg(NO₃)₂       :        Hg

                     1                :          1

                  0.07             :       0.07

Mass of mercury:

Mass = number of moles × molar mass

Mass = 0.07 mol × 200.6 g/mol

Mass = 14.0 g

Approximately 13.70 grams of mercury (Hg) will be formed in the reaction.

Calculating the molar masses of the reactants and products in the balanced chemical equation and using stoichiometry will give us the mercury mass.

[tex]Hg(NO_{3})_2[/tex] molar mass:

Mercury (Hg) weighs 200.59 g/mol.

Nitrogen (N) weighs 14.01 g/mol.

O (oxygen) weighs 16.00 g/mol.

Since[tex]Hg(NO_{3})_{2}[/tex]  has two nitrate ions ([tex]NO_{3}[/tex]), we double the molar mass of [tex]NO_{3}[/tex] (62.00 g/mol) by 2.

The molar mass of [tex]Hg(NO_{3})_2[/tex]  is 324.61 g/mol: 200.59 g/mol + 2(14.01 g/mol + 3(16.00 g/mol)) = 200.59 g/mol + 2(48.00 g/mol) = 200.59 + 2(62.01 g/mol) = 200.59 + 124.02 = 324.61.

From the balanced equation, Hg(NO3)2 and Hg have a 1:1 molar ratio. Hg(NO3)2 creates one mole of Hg.

Calculate [tex]Hg(NO_{3})_2[/tex]moles:

22.17 g [tex]Hg(NO_{3})_2[/tex]

22.17 g/324.61 g/mol = 0.0682 mol.

Hg forms 0.0682 mol due to the 1:1 molar ratio.

Hg mass:

Hg's mass is 0.0682 mol × 200.59 g/mol = 13.70 g.

The process will produce 13.70 grammes of mercury (Hg).

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Answer:

0.448 atm

Explanation:

So, we have;

n = 2 moles

V = 126 L

T = 346 K

a = 6.20 atmL^2/mol^2

b = 0.0852 L/mol

R = 0.082 atmLK-1mol-1

P = nRT/V -nb - an^2/V^2

Substituting values;

P=2 * 0.082 * 346/126 -(2 * 0.0852) - 6.20 * 2^2/126^2

P=56.744/125.8296 -24.8/15876

P= 0.45 - 0.001562

P= 0.448 atm

Answer:

Valence electrons are the electrons in the outermost shell, or energy level, of an atom.

Answer: The outermost shell

Explanation:

Valence electrons are located in the outermost shell of the atom, or the energy level as it is also known.

Answer:

Explanation:

The answer is (4) 4.0 mol. This is a stoichiometry problem. You start with 2.0 mol of C2H6 and obtain the moles of C by multiplying 2.0 by the mole ratio, in this case 2. 2.0*2=4.0mol.

Answer:

d) Sugar and water​

Explanation:

The sugar and water solution will have a clear and transparent outlook.

When we look at a solution of a mixture of sugar and water, it a homogenous one.

A homogenous solution have their solution in one phase.

The solute is sugar and the solvent is water.

The solution obtained is colorless therefore, light will pass through without any obstruction.

The other solutions will be translucent or opaque

The distance of the earth to the sun in Mm = 1.5 x 10⁵

Given

The distance of the earth to the sun : 1.50 x 10⁸ km

Required

The distance in Mm

Solution

In converting units we must pay attention to the conversion factor.

the conversion factor :

1 kilometer(km) = 10⁻³ megameter(Mm)

So the distance conversion :

1.5 x 10⁸ x 10⁻³ = 1.5 x 10⁵ Mm

Answer: Oxygen

Explanation:

Producer organisms include plants and algae and refer to organisms that are able to produce their own food by synthesizing it from inorganic matter. This is what plants and algae are able to do through the process of photosynthesis.

During photosynthesis, these organisms produce food for themselves and in doing so produce oxygen as a by-product. This is what contributes mostly to the concentration of oxygen in the atmosphere.

Answer:

The products would not be formed or the reaction could not be carried out.

Explanation:

Sometimes in these situations it is chosen to place a chemical compound that accelerates the reaction and increases the kinetic energy of the reactants, without altering their chemical composition, and this is called a catalyst, which can be anything from a protein to a macromolecule.

Answer: my epic on fortnite is

Explanation:Bookboy24 ok add me and arena or creative

Answer: D

Explanation: Hope this helps!

Answer: i’m taking a test w this question rn and i think the best answer is D. When the water begins to boil, some of the water molecules will gain enough energy to change into a gas. i could def be wrong tho

Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).

Explanation:

This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7

Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4

Answer:

C

Explanation:

Answer:

See explanation

Explanation:

Looking at the structure of the starting material, we can see that the carbocation intermediate is a tertiary carbocation.

The tertary carbocation is very stable.  Indeed, it is the most stable carbocation intermediate. As such, none of this other product is obtained due to this intermediate.

The mechanism of this reaction is shown in the image attached to this answer.

The central charge is another name for the effective nuclear charge that an electron experiences.. It has the lowest effective nuclear charge of the four new elements.

An element is a type of atoms, including the pure material made entirely of that species, that have a specific number of protons in their nuclei.

Chemical elements, in contrast to chemical compounds, cannot be reduced by any chemical process into simpler molecules. The oxidation number of the atom can be used to calculate the strength of the nuclear charge.

The metals and non-metals are the other two basic categories into which the elements may be separated. Metals are frequently bright, malleable, and carry electricity.

All living and non-living things are made up of elements, which are fundamental building blocks. They serve as the universe's center. Elements make up every particle, no matter how little or large. Our body is composed of five main components are Calcium, Carbon, Hydrogen, Oxygen, and Nitrogen

Thus, option E is correct.

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Hi there!

[tex]\large\boxed{\text{B. Producer}}[/tex]

The only organisms in a food web that do not get energy from other organisms are producers. These are autotrophs (ex: plants or phytoplankton) that use photosynthesis to produce energy. This means that these organisms only intake water and sunlight rather than nutrients from other organisms (consumers).

Answer:

25.39

Explanation:

Given parameters:

Abundance of X-25  = 80.5%

Abundance of X - 27  = 19.5%

Unknown:

Average atomic mass of X  = ?

Solution:

The average atomic mass of X can be derived using the expression below:

Average atomic mass = (abundance x mass of X - 25) + (abundance x mass of X - 27)

Average atomic mass  =  (80.5%  x 25) +   (19.5% x 27)  = 25.39

Answer: write anything given below in the diagram except for the neutral substances

Explanation:

Answer:

Alkaline

1 drain cleaner

2 oven cleaner

3 baking soda

4 drain unblocker

Acid

1 drain cleaner

2 bowl cleaner

3 house bleach

4 diluted soap

Answer:

2

[tex]0.4167\ \text{M}^{-1}\text{min}^{-1}[/tex]

Explanation:

Half-life

[tex]{t_{1/2}}A=2\ \text{min}[/tex]

[tex]{t_{1/2}}B=4\ \text{min}[/tex]

Concentration

[tex]{[A]_0}_A=1.2\ \text{M}[/tex]

[tex]{[A]_0}_B=0.6\ \text{M}[/tex]

We have the relation

[tex]t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}[/tex]

So

[tex]\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}[/tex]

Comparing the exponents we get

[tex]1=n-1\\\Rightarrow n=2[/tex]

The order of the reaction is 2.

[tex]t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}[/tex]

The rate constant is [tex]0.4167\ \text{M}^{-1}\text{min}^{-1}[/tex]

Answer:

Matrix and Fibre.

Explanation:

The two main components composites are made up of are matrix and fibre. The matrix is the base material while the fiber is what reinforces the material.

Answer:

volcanoes and volcanic activity

Explanation:

I had the same Question and I got that.

Answer: Volcanoes and Volcanic Activity :)

Explanation: I got the same question :D

Answer:

8.2mL of the concentrated solution must be taken diluting to 1.30L with water

Explanation:

The concentrated HNO₃ comes in a presentation of 70% = 15.8M

As the concentrated solution is of 15.8M and we want to obtain a 0.100M, the dilution must be of:

15.8M / 0.100M = 158 times must be the concentrated solution diluted

As the total volume we want to prepare is 1.30L, the volume of the concentrated solution must be:

1.30L / 158 = 8.2x10⁻³L =

Answer:

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Contents Home   Bookshelves   Physical & Theoretical Chemistry   Supplemental Modules (Physical and Theoretical Chemistry)   Electronic Structure of Atoms and Molecules Expand/collapse global location

Predicting the Hybridization of Simple Molecules

Last updatedAug 16, 2020

Predicting the Bond-Order of Oxides based Acid Radicals

Prediction of Aromatic, Anti Aromatic and Non Aromatic Character of Heterocyclic Compounds along with their Omission Behavior- Innovative Mnemonics

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Prof. Linus Pauling (1931) first developed the Hybridization state theory in order to explain the structure of molecules such as methane (CH4).1This concept was developed for simple chemical systems but this one applied more widely later on and from today’s point of view it is considered an operative empirical for excusing the structures of organic and inorganic compounds along with their related problems. An innovative method proposed for the determination of hybridization state on time economic ground 2,3,4.

Prediction of sp, sp2, sp3 Hybridization state

We Know, hybridization is nothing but the mixing of orbital’s in different ratio to form some newly synthesized orbitals called hybrid orbitals. The mixing pattern is as follows:

s + p (1:1) - sp hybrid orbital; s + p (1:2) - sp2 hybrid orbital ; s + p (1:3) - sp3 hybrid orbital

Formula used for the determination of sp, sp2 and sp3 hybridization state:

Power on the Hybridization state of the central atom = (Total no of σ bonds around each central atom -1)

All single (-) bonds are σ bond, in double bond (=) there is one σ and 1π, in triple bond (≡) there is one σ and 2π. In addition to these each lone pair (LP) and Co-ordinate bond can be treated as one σ bond subsequently.

Eg.:

a. In NH3: central atom N is surrounded by three N-H single bonds i.e. three sigma (σ) bonds and one lone pair (LP) i.e. one additional σ bond. So, in NH3 there is a total of four σ bonds [3 bond pairs (BPs) + 1 lone pair (LP)] around central atom N. Therefore, in this case power of the hybridization state of N = 4-1 = 3 i.e. hybridization state = sp3.

b. In H2O: central atom O is surrounded by two O-H single bonds i.e. two sigma (σ) bonds and two lone pairs i.e. two additional σ bonds. So, altogether in H2O there are four σ bonds (2 bond pairs + 2 lone pairs) around central atom O, So, in this case power of the hybridization state of O = 4-1 =3 i.e. hybridization state of O in H2O = sp3.

c. In H3BO3:- B has 3σ bonds (3BPs but no LPs) and oxygen has 4σ bonds (2BPs & 2LPs) so, in this case power of the hybridization state of B = 3-1 = 2 i.e. B is sp2 hybridized in H3BO3. On the other hand, power of the hybridization state of O = 4-1= 3 i.e. hybridization state of O in H3BO3 is sp3.

d. In I-Cl: I and Cl both have 4σ bonds and 3LPs, so, in this case power of the hybridization state of both I and Cl = 4 - 1 = 3 i.e. hybridization state of I and Cl both are sp3.

e. In CH2=CH2: each carbon is attached with 2 C-H single bonds (2 σ bonds) and one C=C bond (1σ bond), so, altogether there are 3 sigma bonds. So, in this case, power of the hybridization state of both C = 3-1 = 2 i.e. hybridization state of both C’s are sp2.

Prediction of sp3d, sp3d2, and sp3d3 Hybridization States

Answer:

Option D: Cations

Explanation:

Do you need only one answer .

Answer:

Explanation:

It changes from a liquid to a solid. This change occurred due to the change in temperature. As the mixture cooled, the molecules condensed and the mixture formed a solid shape.Which formed the jelly.

Answer:

Above and below the plane of a square

Explanation:

The number of electron groups often determine the shape of a molecule. When we have a coordination number of six, one of the possible arrangement of groups is the hexagonal planar geometry. In this arrangement, six electron groups are arranged in the same plane.

If two among the groups are replaced by lone pairs, the substituents will arrange themselves at the corners of a square while the lone pairs will be found above and below the plane of the square. This is the square planar geometry.

The two lone pairs will be located at the top and at the bottom of the plane.

The VSEPR theory is a model that helps to predict the shape of a molecule based on the repulsions associated with the electron groups on the central atom.

Given that a hexagonal planar electron geometry possibly has 6 electron groups. The description of how the electron groups are arranged can be seen in the first image below.

If two of the six groups are lone pairs, we know that lone pairs repel each other. As such, in this arrangement, the lone pairs will occpy the opposite direction which is the top and the bottom of the plane in order to have minimum repulsion as possible.

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