All organic molecules have what element in common?
A) carbon
B) nitrogen
C) phosphorus
D) sulfur

To calculate the mass of the excess reactant remaining, we need to find the limiting reactant and use stoichiometry. In this case, the mass of the excess reactant (F2) that is left is 78.86 g.

To determine the mass of the excess reactant left, we need to first calculate the amount of S that reacts completely with F2. We can do this by using stoichiometry and the balanced chemical equation. The balanced equation for the reaction between S and F2 is:

2S + 3F2 → 2SF6

From the equation, we can see that 2 moles of S react with 3 moles of F2. We can convert the given masses of S and F2 to moles using their molar masses:

Using these molar masses, we can calculate the number of moles of S and F2:

Since the reaction is 2:3, we can calculate the limiting reactant and the amount of excess reactant remaining. Since we have more F2 than needed, S is the limiting reactant and F2 is in excess.

To calculate the mass of the excess reactant remaining, we need to know how much F2 reacted completely with the S. We can do this by using the stoichiometric ratio:

Therefore, the mass of the excess reactant (F2) that is left is 78.86 g.

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Compounds can be exemplified in two main ways. These are Chemical Formula and Molecular Model.

Chemical formula includes:

Structural formula – this is a formula that shows the arrangement of atoms in the molecule of a compound.

Molecular formula – this is a formula giving the number of atoms of each elements to present in one molecule of a specific compound.

Empirical formula – this is a formula giving the proportions of elements present in a compound.

Molecular model:

Space-filling model - represents atoms fill the space between each other.

Ball and stick model - represents atoms as balls and chemical bonds as sticks.

Remember that a compound is a substance that is formed when two or more elements are bonded together.

Compounds can be represented in several ways including Molecular Formulas, Empirical Formulas, Structural Formulas, Ball-and-Stick Models, and Space-Filling Models. Each method provides different levels of detail and helps us understand the composition and structure of chemical compounds.

Compounds can be represented in several ways: Molecular Formulas, Empirical Formulas, Structural Formulas, Ball-and-Stick Models, and Space-Filling Models.

Molecular formulas use chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. For instance, the molecular formula for water is H2O, indicating there are two hydrogen atoms and one oxygen atom.

Empirical formulas give the simplest whole-number ratio of atoms in a compound. For water, the empirical formula would still be H2O.

Structural formulas indicate the bonding arrangement of the atoms in the molecule, giving more detailed information about the molecule's structure.

Ball-and-stick models show the geometric arrangement of the atoms in a molecule with atomic sizes not to scale. These models often use balls to represent atoms and sticks to represent chemical bonds.

Lastly, Space-filling models also show the geometric arrangement of atoms, but in these models, the relative sizes of the atoms are represented accurately and the model gives a more realistic view of the molecule's shape.

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Answer: 20.187 amu

Explanation:

20(0.905)+21(0.003)+22(0.092)

Wood is a heterogeneous mixture due to its composition of various easily distinguishable components such as cellulose fibers and lignin.

Wood is considered a heterogeneous mixture because it is composed of different components that can be easily distinguished. These components include cellulose fibers, lignin, and extracts such as resins and oils. The presence of knots, grain patterns, and different colors in wood further supports its classification as a heterogeneous mixture.

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To create a buffer with a pH of 9.50 using 2.45 L of 0.165 M [tex]NH_3[/tex], approximately 38.5 grams of ammonium chloride must be added, as calculated using the Henderson-Hasselbalch equation and considering the molar mass of [tex]NH_4Cl[/tex].

To calculate the mass of ammonium chloride (NH4Cl) needed for a buffer solution with a desired pH, we use the Henderson-Hasselbalch equation: pH = pKa + log([Base]/[Acid]). Ammonium chloride, when dissolved, provides the ammonium ion ([tex]NH_4^+[/tex]), which acts as the acid in this buffer system, whereas ammonia ([tex]NH_3[/tex]) acts as the base.

First, we need to find the pKa of [tex]NH_3[/tex], which is the negative logarithm of the dissociation constant (Ka) of its conjugate acid, [tex]NH_4^+[/tex]. If the pKa of [tex]NH_3[/tex] is 9.25, then using the desired pH of 9.50, we can set up the equation as follows:

[tex]pH = 9.25 + log(\frac{[NH_3]}{[NH_4^+]})[/tex]

[tex]9.50 = 9.25 + log(\frac{[0.165 M]}{[NH_4^+]})[/tex]

Solving for [[tex]NH_4^+[/tex]], we find that:

[tex]log(\frac{[NH_4^+]}{0.165}) = 9.50 - 9.25[/tex]

[tex]log(\frac{[NH4+]}{0.165}) = 0.25[/tex]

[tex][NH_4^+] = 10^{0.25} [NH_4^+] = 1.778 \frac{[NH4+]}{0.165} = 1.778[/tex]

[tex][NH_4^+] = 0.293 M[/tex]

Now, to find the mass of NH4Cl required, we use the formula:

mass = molarity volume molar mass

mass = (0.293 M) (2.45 L) (53.491 g/mol)

mass = 38.511 g (rounded to three significant figures)

Therefore, to prepare a buffer solution with a pH of 9.50 using a 2.45 L solution of 0.165 M [tex]NH_3[/tex], you must add approximately 38.5 grams of [tex]NH_4Cl[/tex].

Density is the amount of matter within a given amount of space. 

Water has a density of 1.0 gram per centimeter squared.

Mercury has a density of 13.6 grams per centimeter squared. 

Thus, within the same volume, mercury has more mass than water. Thus, mercury has a higher density. 

Density is an intensive property as it does not depend on the quantity of the substances Whereas mass and volume are extensive property. Mercury is more denser than water.

Density tells about the compactness of the substances, how much dense is the substances in other words. Object that is more denser than water they just sink in the water.

Mathematically,

Density = Mass of the substance ÷volume of substance

From above formula we can see that density is directly proportional to mass of substances at constant volume.

The  volume of  water  and mercury is same that is 1 liter. Since the mass of  mercury that is 13.6 kilograms is more than mass of water that is 1 kilogram. So, the density of mercury is more than the density of water which means mercury is denser than water.

Therefore mercury is more denser than water.

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Answer is: Identical particles were produced no matter which metal he used.

J.J. Thomson placed two oppositely charged electric plates around the cathode ray. He did experiments using different metals as electrode materials and found that the properties of the cathode ray remained constant no matter what cathode material he used.

Tomson concluded that atoms are divisible and that the corpuscles are their building blocks (atoms are made up of smaller particles).

J. J. Thomson discovered the electron in 1897.

His "plum pudding" model (1904) suggested: the electrons are embedded in the positive charge.

With this model, he abandoned his earlier hypothesis (the atom was composed of immaterial vortices).  

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

Answer:

0.2 moles O

Answer : The moles of oxygen present in the sample is, 0.2 moles

Explanation : Given,

Moles of carbon = 0.200 mole

Moles of hydrogen = 0.400 mole

Mass of unknown compound = 6.00 g

Molar mass of carbon = 12 g/mole

Molar mass of hydrogen = 1 g/mole

Molar mass of oxygen = 16 g/mole

First we have to calculate the mass of carbon and hydrogen.

[tex]\text{Mass of carbon}=\text{Moles of carbon}\times \text{Molar mass of carbon}=(0.200mole)\times (12g/mole)=2.4g[/tex]

[tex]\text{Mass of hydrogen}=\text{Moles of hydrogen}\times \text{Molar mass of hydrogen}=(0.400mole)\times (1g/mole)=0.4g[/tex]

Now we have to calculate the mass of oxygen.

Total mass of unknown compound = Mass of carbon + Mass of hydrogen + Mass of oxygen

6.00 = 2.4 + 0.4 + Mass of oxygen

Mass of oxygen = 3.2 grams

Now we have tom calculate the moles of oxygen.

[tex]\text{Moles of oxygen}=\frac{\text{Mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.2g}{16g/mole}=0.2moles[/tex]

Therefore, the moles of oxygen present in the sample is, 0.2 moles

The formula for calculating urms is given as:

urms = sqrt (3 R T / M)

where,

R is gas constant = 8.314 J / mol K

T is absolute temperature = ?

M is molar mass of H2 = 2 x 10^-3 kg/mol

Calculating for T:

1.8 * (1.84×10^3) = sqrt (3 * 8.314 * T / 2x10^-3)

T = 879.59 K      (ANSWER)

Answer:

The temperature will be 879.59K for 21.8 times higher than this value.

Explanation:

The Equation for calculating Root Mean Square Velocity([tex]\rm U_r_m_s[/tex])

[tex]\rm \mathbf{U_r_m_s}=\sqrt\frac{3RT}{M}[/tex]

Where,

[tex]\rm \mathbf{U_r_m_s}[/tex] is Root Mean Square Velocity in m/s.

R is gas constant which is [tex]\rm \texttt{8.314 J/mol K}[/tex]

T is the temperature at Kelvin, what do we need to calculate here?

M is the molar mass of  [tex]\rm H_2[/tex] which is [tex]\rm 2\times10^-^3\texttt{kg/mol}[/tex]

Now, put the value according to the equation above,

[tex]\rm 1.8 \times (1.84\times10^3)=\sqrt\frac{3 \times8.314 \times T}{2\times10^-^3} \\\rm \mathbf{T=879.59K}[/tex]

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Sodium nitrate and lead (II) acetate both dissolve in water and do not react with each other; thus, there is no reaction and no precipitate is formed.

The student is inquiring about the chemical reaction that occurs when solutions of sodium nitrate and lead (II) acetate are mixed. When mixing these two compounds, there is no reaction because both sodium nitrate and lead (II) acetate are soluble in water. As such, no precipitate forms, and the ions remain in solution. Therefore, the correct response to this question is that no reaction occurs.

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Answer:

number of gold atoms = [tex]1.528X10^{22}[/tex]

Explanation:

The mass of one atom of gold is given to be = [tex]3.271X10^{-22}[/tex]

The mass of gold is given to be = [tex]5.00g[/tex]

In order to determine the number of atoms present in the 5 g of gold, we will divide the mass of the gold with the mass of each atom of the gold.

the number of atoms = [tex]\frac{massofgold}{massofatom}=\frac{5}{3.271X10^{-22}} =1.528X10^{22}[/tex]

Thus number of gold atoms = [tex]1.528X10^{22}[/tex]

Substances are classified based on their chemical composition into elements (e.g., gold), compounds (e.g., salt, pure water, carbon dioxide), homogeneous mixtures (e.g., salt water, pure air), and heterogeneous mixtures (e.g., soil, bronze).

Substances can be classified into four main categories based on their chemical composition: elements, compounds, homogeneous mixtures, and heterogeneous mixtures. The classification for each of the given substances is as follows:

An element is a pure substance that cannot be broken down into simpler substances. A compound is a pure substance that consists of two or more elements chemically combined in a fixed ratio. A homogeneous mixture, or solution, has uniform composition throughout, whereas a heterogeneous mixture has a composition that varies from point to point.

Explanation:

An element is a substance that is made up of only one type of atoms. It cannot be broken down into simpler substances.

For example, sodium, aluminium, nickel etc are all elements.

Whereas when two or more different elements combine together then it results in the formation of a compound.

For example, [tex]MgCl_{2}[/tex] is a compound.

Therefore, we can conclude that a substance that cannot be broken down into simpler substances by chemical or physical means is called an element.

A substance that cannot be broken down into simpler substances by any chemical or physical means is called a: chemical element.

A chemical element can be defined as a pure substance that comprises atoms having the same atomic number (number of protons) in its nuclei and as such it is the primary constituent of matter.

Basically, a chemical element is a pure substance that can't be broken down, decomposed or transformed by into simpler substances chemical or physical means.

In Chemistry, some examples of a chemical element include the following:

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Answer:

Explanation:

We can fill maximum two electrons in one orbital. The s subshell has 1 orbital that can filled upto two electrons, the p subshell has 3 orbitals that can filled upto six electrons, the d subshell has 5 orbitals that can filled upto ten electrons, the f subshell has 7 orbitals that can filled upto fourteen electrons.

According to the energy level diagram, the 2p, 3p, 4p each can hold 6 electrons because they have 3 orbitals, and 3d, 4d each can hold 10 electrons because they have 5 orbitals, the 5f, 6f, each can hold 14 electrons because they have 7 orbitals. Therefore, the maximum number of electrons would be present in f subshell or 6f orbital.

This is an example of law of conservation of mass, which states that the total quantity of mass doesn't change means, in an isolated system mass is neither created nor destroyed by any chemical reactions or physical transformations.

According to law of mass conservation in a chemical reaction:

Mas of the Products = Mass of the Reactants