How many molecules are in 15.0 g of silicon tetrafluoride?

According to the stoichiometry of the chemical reaction between magnesium and copper nitrate  42.72 g of precipitate is formed  if we start with 7 g of magnesium metal and 0.5 moles of copper(II) nitrate.

It is the determination of proportions of elements or compounds in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.

Stoichiometry is used in quantitative analysis for measuring concentrations of substances present in the sample.In the given reaction , 24.3 g magnesium gives 148.3 g magnesium nitrate thus, 7 g magnesium will give , 7×148.3/24.30=42.72 g

Thus, 42.72 g of precipitate is formed  if we start with 7 g of magnesium metal and 0.5 moles of copper(II) nitrate.

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Answer:

Calcium, potassium, aluminum, chlorine, bromine

Explanation:

I do not know all of them, but here are some.

Answer:

What homework? i dont see a picture, can u tell me the question?

Explanation:

\(Ba(OH)_2\rightarrow \bold{\green{Barium\: hydroxide}}\)

\( Ca(OH)_2\rightarrow \bold{\red{Calcium\: hydroxide}} \)

\(RbOH \rightarrow \bold{\blue{Rubidium \:hydroxide}}\)

Option A, In the polysaccharide cellulose, large numbers of monosaccharides are linked together through glycosidic linkages.

The most prevalent carbohydrates in nature are polysaccharides, which have several uses including energy storage and inclusion in plant cell walls. Large polymers known as polysaccharides are created by joining tens to thousands of monosaccharides together via glycosidic connections. Starch, glycogen, and cellulose are the three polysaccharides that are most prevalent. Because each of these three produces only one kind of monosaccharide (glucose) after full hydrolysis, they are collectively known as homopolymers. In addition to monosaccharides, heteropolymers may also contain sugar acids, amino sugars, or noncarbohydrate compounds. Non-reducing carbohydrates known as polysaccharides lack the sweetness and do not mutate.

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Answer:

A. its reactivity

Explanation:

It's reactivity because copper was exposed to air and if it is reactivity it must be exposed to air

Answer:

A.  reactivity

Explanation:

Boyle's Law, which states that at constant temperature, a gas's pressure and volume are inversely related, can be used to address this issue.

Boyle's Law says that P1V1 = P2V2.

Where: P1 = The gas's initial pressure (in torr)

V1 is the container's initial volume (in millilitres).

P2 stands for the gas's post-volume change pressure (unknown).

V2 is the revised volume of the container (in millilitres).

Let's solve for P2 by entering the provided values into the equation:

P₁ = 757.2 torr

V2 = 132.5 mL V1 = 81.4 mL

P2 = ambiguous

P₁V₁ = P₂V₂

P2 equals 757.2 torr x 81.4 mL x 132.5 mL.

P2 times 132.5 mL = 61702.08 torr * mL

61702.08 torr = P₂ * 132.5

P2 = 132.5 mL / 61702.08 torr

P₂ ≈ 465.69 torr

as a result, the gas's pressure

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The answer is (D) 3.20L; the gas will occupy a volume of 3.20 L at 45.0°C and 1.00 atm.

The combined gas law is a gas law that combines Boyle's law, Charles's law, and Gay-Lussac's law. It relates the pressure, volume, and temperature of a fixed amount of gas. The law states that the product of the pressure and volume of a gas is directly proportional to the product of its temperature and the number of moles of gas present.

To solve this problem, we can use the combined gas law,

(P1V1)/T1 = (P2V2)/T2

Where:

P1 = initial pressure

V1 = initial volume

T1 = initial temperature

P2 = final pressure (which is the same as the initial pressure in this case)

V2 = final volume (what we're trying to find)

T2 = final temperature

Plugging in the given values, we get:

(1.00 atm x 3.00 L) / (298.15 K) = (1.00 atm x V2) / (318.15 K)

Solving for V2, we get:

V2 = (1.00 atm x 3.00 L x 318.15 K) / (298.15 K x 1.00 atm)

V2 = 3.20 L

Therefore, the gas will occupy a volume of 3.20 L at 45.0°C and 1.00 atm.

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The statement that is true about the reaction is that two moles of Fe III are formed.

We define a redox reaction as one that involved the oxidation and the reduction of a given chemical specie. The key thing that we must have at the back of our minds is that oxidation and reduction are two processes that occur at the same time.

In this case, we can see that the redox reaction that is taking place here has to do with the fact that electrons would have to be transferred from the iron III ion to the bromine molecule hence the iron III ion and the bromide ions are obtained at the end of the reaction.

The reaction equation can be written as;

\(Br_{2} + 2Fe^{2+} ---- > 2Br^- + 2Fe^{3+}\)

We can see that two moles of Iron III ion is formed.

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b. Ethylene is symmetrical; all bond dipole moments cancel out.

In ethylene (C₂H₄), the molecule is symmetrical with a linear structure. Each carbon atom is bonded to two hydrogen atoms and shares a double bond with the other carbon atom. The electronegativity of carbon and hydrogen is similar, resulting in an equal sharing of electrons in the bonds. As a result, the bond dipole moments in ethylene are equal in magnitude but opposite in direction. The dipole moments cancel each other out, resulting in a net dipole moment of zero for the entire molecule.

In propene (C₃H₆), there is an additional methyl (CH₃) group attached to one of the carbon atoms. The methyl group has a slight electron-donating effect due to the presence of the carbon-hydrogen bonds. This creates a small imbalance in electron distribution within the molecule, resulting in a slight dipole moment. The dipole moment in propene is approximately 0.3 D, indicating a small separation of charges due to the presence of the methyl group.

Therefore, option b is correct as it explains that the zero dipole moment in ethylene is due to its symmetrical structure, while the small dipole moment in propene arises from the electron-donating ability of the methyl group.

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The complete question is:

Select all the statements that correctly explain why ethylene has a zero dipole moment, whereas propene has a small dipole moment of 0.3 d.

a. small dipole moment in propene arises from electron-donating ability of methyl group

b. ethylene is symmetrical; all bond dipole moments cancel out

c. small dipole moment in propene arises from electron-withdrawing ability of methyl group

The number of moles of H₂O that could be produced is 3.3 moles

From the question, we are to determine the number of moles of H₂O that could be produced

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

4NH₃ + 5O₂ → 4NO + 6H₂O

This means,

4 moles of NH₃ reacts with 5 moles of O₂ to produce 4 moles of NO and 6 moles of H₂O

Now,

If 4 moles of NH₃ reacts with 5 moles of O₂ to produce 6 moles of H₂O

Then,

Excess NH₃ will react with 2.75 moles of O₂ to produce x moles of H₂O

x = \(\frac{2.75 \times 6}{5}\)

x = 3.3 moles

Hence, the number of moles of H₂O that could be produced is 3.3 moles

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Answer:

The answer is Aluminum

Hope this helps you

Answer:

P₂ = 80 KPa

Boyle's law is applied.

Explanation:

Given data:

Initial volume = 180 L

Initial pressure = 35.5 KPa

Final pressure = ?

Final volume = 80.0 L

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

35.5 KPa × 180 L = P₂ × 80.0 L

P₂ = 6390 KPa. L/ 80.0 L

P₂ = 80 KPa

Answer:

A

Explanation:

He did not meet his goal because 25% is lesser than 30% and he needs 30% or more Osvoldo's objective was not achieved.

If he had 220 grams of carbohydrates today, and 55 of those grams came from whole grains, then the proportion of his total carbohydrates that were made up of whole grains was.

It is assumed that Osvoldo aims to consume at least 30 percent of his daily grams of carbohydrates from whole grains.

He consumed 220 grams of carbohydrates today.

55 grams of whole-grain carbohydrates

We must ascertain whether Osvoldo achieved his objective or not.

We must determine if 55 is at least 30% of 220.

He only obtains 25% of his daily grams of carbohydrates from whole grains because 55 is 25% of 220.

55/220 × 100 = 25%

he has a goal of getting at least 30% percent of his grams of carbohydrates each day from whole grains, Hence, Osvoldo did not meet his goal.

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The force of attraction between the two nuclei in the CO molecule is zero.

To calculate the force of attraction between the two nuclei in the carbon monoxide (CO) molecule, we can use Coulomb's law. Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Let's denote the charge of the carbon nucleus as Qc and the charge of the oxygen nucleus as Qo. The force of attraction between them (F) can be calculated using the equation:

F = k * (Qc * Qo) / r^2

Where:

k is the Coulomb's constant (9 × 10^9 N·m^2/C^2)

Qc is the charge of the carbon nucleus

Qo is the charge of the oxygen nucleus

r is the distance between the nuclei

Since the carbon and oxygen nuclei are electrically neutral, their charges (Qc and Qo) cancel each other out. Therefore, the net charge for each nucleus is zero.

Hence, the force of attraction between the two nuclei in the CO molecule is zero.

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The two nuclei in the carbon monoxide (CO) molecule are approximately 0.1128 nm (or 1.128 × 10⁻¹⁰ m) apart.

Given:

Distance between the nuclei = 0.1128 nm = 1.128 × 10⁻¹⁰ m

Mass of carbon atom (C) = 1.993 × 10⁻²⁶ kg

Mass of oxygen atom (O) = 2.656 × 10⁻²⁶ kg

The carbon monoxide molecule (CO) consists of a carbon atom bonded to an oxygen atom. To determine the distance between the nuclei, we consider the bond length between the two atoms.

The given distance of 0.1128 nm represents the bond length or the distance between the nuclei of the carbon and oxygen atoms in the CO molecule.

The mass of the carbon atom is 1.993 × 10⁻²⁶ kg, and the mass of the oxygen atom is 2.656 × 10⁻²⁶ kg. These values indicate the relative masses of the atoms involved in the CO molecule.

It's important to note that the bond length and atomic masses provided are approximations based on the most common isotopes of carbon and oxygen.

Therefore, the two nuclei in the carbon monoxide molecule are approximately 0.1128 nm (or 1.128 × 10⁻¹⁰ m) apart.

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Answer:

The noble gases, group 8A

Noble gas are present in the group are non reactive in nature.

Noble gases is defined as group 18 (VIIIa) of the periodic table is made up of seven different chemical elements. The noble gases are employed to create inert atmospheres, usually for arc welding, to shield specimens and prevent chemical reactions. The substances are utilized in lasers as well as lamps like neon and krypton headlamps.

Gases that are noble or inert. Helium, neon, argon, krypton, xenon, and radon are the noble gases, or inert gases, that make up group 8A (or VIIIA) of the periodic table (Rn). The name of these elements refers to how little they react with other elements or compounds. Non-reactive elements do not mix with other elements readily.

Thus, noble gas are present in the group are non reactive in nature.

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Answer:the water starts to boil and creates steam

Explanation:

The pH of a solution that has a concentration of \(2.78 * 10^{-3}\)M is 2.56. The aqueous solution is acidic because it is less than 7 on the pH scale.

pH is a figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values more alkaline.

The pH is equal to −log {H}

According to this question, \(2.78 * 10^{-3}\) of hydrogen bromide is given. The pH can be estimated as follows:

pH = \(- log\) {\(2.78 * 10^{-3}\)}

pH = 2.56 (acidic)

The complete question is as follows:

What is the hydronium ion concentration for an aqueous solution with a hydroxide ion concentration of 2.78x10^-10 M? Is the solution acidic basic or neutral?

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The illustration of Thomson's atom represents a neutral atom. In this case, the total amount of positive charge and the total amount of negative charge must be equal. This means that there are equal numbers of protons and electrons in the atom. This is what makes the atom neutral.

A neutral atom is an atom that has no electrical charge. An atom is neutral because it has the same amount of positively charged protons and negatively charged electrons. The nucleus of an atom contains protons, which are positively charged particles. Electrons, which are negatively charged particles, are located in the atom's electron cloud around the nucleus.

Electrons, protons, and neutrons are the three components of atoms. Electrons are negatively charged, protons are positively charged, and neutrons have no charge. Electrons are found outside the nucleus of the atom and are continually moving at high speeds.

In summary, if the illustration of Thomson's atom represents a neutral atom, then the total amount of positive charge and the total amount of negative charge must be equal. This means that there are equal numbers of protons and electrons in the atom. This is what makes the atom neutral.

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Over time, the atomic size shrinks because as number of shells stays constant but more electrons are added, increasing the nuclear charge.

The distance between an atom's central nucleus and also its outermost shell is known just like its atomic size. The smallest distance here between atom's nucleus and its outermost shell is known as the atomic radius within basic chemistry.

The orbitals from higher primary energy levels have greater sizes than the orbitals form lower principal energy levels. Atomic radius grows down a group as a result of the stronger impact of the major energy levels outweighing a increase overall nuclear charge.

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Answer:

А  There are two peroxide molecules on the reactant side.

The statement is correct because the left hand side is the reactant side. We can read the reaction as;

  2 molecules of hydrogen peroxide decomposed  to given 1 molecule of oxygen gas and 2 molecules of water .

D Mass must be conserved so there are an equal number of molecules in the reactants and products.

E Mass must be conserved so there are an equal number of atoms in the reactants and products.

Since the chemical equation is balanced, choices D and E are correct. The reaction will have equal number of molecules and atoms on both sides of the expression.

Explanation:

The pH scale measures the acidity or basicity of a substance. A pH of 5 is one unit lower than a pH of 6, which means that a substance with a pH of 5 is ten times more acidic than a substance with a pH of 6.

This is because the pH scale is logarithmic, which means that a change of one unit on the scale represents a tenfold change in acidity or basicity. To understand this better, consider the example of lemon juice and orange juice. Lemon juice has a pH of around 2, making it highly acidic, while orange juice has a pH of around 4, making it less acidic than lemon juice but still acidic nonetheless. The difference in pH between these two juices is two units, which means that lemon juice is 100 times more acidic than orange juice. In summary, a substance with a pH of 5 is ten times more acidic than a substance with a pH of 6.

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Approximately 4 L of C₂H₂ are required to produce 8 L of CO₂ assuming the reaction is at STP.

To solve for the amount of C₂H₂ needed, we can use the stoichiometry of the balanced equation, combustion of hydrocarbons and the ideal gas law. According to the balanced equation, 2 moles of C₂H₂ react to produce 4 moles of CO₂. Therefore, we can set up the following ratio by assumimg x moles of C₂H₂ and then solving for x,

x = (2/4) * (8 L CO₂)

= 4 L C₂H₂

Since the problem states that the reaction is at STP (standard temperature and pressure), we can assume that the volume of gas is directly proportional to the number of moles of gas.

V = nRT/P

Solving for n at STP,

n = PV/RT = (4 L) x (1 atm) / (0.0821 L·atm/mol·K x 273.15 K) ≈ 0.167 mol

Therefore, approximately 0.167 mol (or 4 L) of C₂H₂ is required to produce 8 L of CO₂ assuming the reaction is at STP.

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The approximate resistance value at 45 degrees Celsius is around 5.8 ohms.

To calculate the value of the resistance temperature coefficient at 45 degrees Celsius using linear approximation, we can use the formula:

R(T) = R0 + α(T - T0),

where R(T) is the resistance at temperature T, R0 is the resistance at a reference temperature T0, α is the resistance temperature coefficient, and (T - T0) is the temperature difference.

Given that the resistance at 25 degrees Celsius is 5 ohms (R0 = 5) and the resistance at 50 degrees Celsius is 6 ohms (R(T) = 6), we can calculate the value of α.

6 = 5 + α(50 - 25),

Simplifying the equation:

1 = 25α,

Therefore, α = 1/25 = 0.04 ohm/degree Celsius.

Using the linear approximation, we can approximate the value of the resistance at 45 degrees Celsius:

R(45) = 5 + 0.04(45 - 25) = 5 + 0.04(20) = 5 + 0.8 = 5.8 ohms.

Therefore, the value of the resistance at 45 degrees Celsius is approximately 5.8 ohms.

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Step 1 - First, we need to transform kg into g. We need to multiply it by 1000:

158 kg = 158,000 g of phosphorus

Step 2 - Now we need to know the molar mass of phosphorus. Look for it at the periodic table. It is 30.97 g/mol.

Step 3 - We transform grams into moles:

30.97 g ---- 1 mol

158,000 g ---- x mol

x = 5,101.7 moles of phosphorus

Step 4 - We transform moles into atoms using Avogrado's constant:

6.022 x 10^23 ---- 1 mol

x ---- 5,101.7 moles of phosphorus

x = 3.072 x 10^27 atoms

If you want to calculate it in another way:

158000/30.97 x 6.022 x 10^23 = 3.072 x 10^27 atoms

Answer: There are 3.072 x 10^27 atoms of phosphorus

A BCC unit cell with a lattice constant, a, of 2.4 Å. The volume atomic concentration of the unit cell is 0.0722 atoms/ų.

To determine the volume atomic concentration of a Body-Centered Cubic (BCC) unit cell, we need to consider the number of atoms present in the unit cell and the volume occupied by the unit cell.

In a BCC unit cell, there is one atom located at the center of the cube and eight atoms at the corners, but each corner atom is shared among eight adjacent unit cells. Therefore, the total number of atoms present in the unit cell is 1.

The volume of a BCC unit cell can be calculated using the formula:

Volume = a³

where "a" is the lattice constant.

Given that the lattice constant, a, is 2.4 Å, we can calculate the volume of the unit cell as follows:

Volume = (2.4 Å)³

Converting the units to cubic angstroms:

Volume = 13.824 ų

Now, to determine the volume atomic concentration, we need to divide the number of atoms (1) by the volume of the unit cell:

Volume Atomic Concentration = Number of Atoms / Volume

Volume Atomic Concentration = 1 / 13.824 ų

The volume atomic concentration of the BCC unit cell is approximately 0.0722 atoms/ų.

Therefore, the volume atomic concentration of the unit cell in a BCC crystal with a lattice constant of 2.4 Šis approximately 0.0722 atoms/ų.

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The complete question is:

A BCC unit cell with a lattice constant, a, of 2.4 Å. Determine the volume atomic concentration of the unit cell.

Answer:

-205.7 kJ

Explanation:

if ur a connexus kid i just took this too lol

Answer:

A (-205.7)

Explanation:

The volume of concentrated aqueous ammonia must be diluted to 1.00 l to make a solution that has a molarity of 1.85 m. concentrated nh3 (aq) is 15 m. report the volume of concentrated aqueous ammonia in milliliters is 123 mL

The number of moles of solute must always remain constant while diluting solutions, which is the most crucial thing to remember.

In other words, the amount of solute in the concentrated sample must be identical to the amount of solute in the diluted solution.

The concentration is the number of moles per unit volume. So

c=n/V

n=CV

n= 1.85x1

n=1.85

Now your task is to determine what volume of the concentrated solution would contain this many moles of ammonia.

Plugging the values, we get

V=1.85/15

V=0.123 L

Now for dilution calculations, we have

C1V1=C2V2

V1=C2/C1.V2

V1=1.85/15x1.00= 123 mL

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