I don’t know how to do this

Answer:

the answer is C.

Explanation:

Bonds form when atoms share or transfer valence electrons. Valence electrons are the electrons in the outer energy level of an atom that may be involved in chemical interactions. Valence electrons are the basis of all chemical bonds.

Answer:

Explanation

Boron (B)

Silicon (Si)

Germanium (Ge)

Arsenic (As)

Antimony (Sb)

Tellurium (Te)

Polonium (Po

Answer:

his mother died i think

Explanation:

Although he was not accepted, his mother was determined to give her son a first-class education and they walked to St. Petersburg. Dmitri was accepted into the Institute of Pedagogy on a full scholarship. That same year his mother died and at the age of sixteen, Dmitri was orphaned.

Answer:

a lower temperature to liquefy

Explanation:

This question is asking for the number of particles that one formula unit of magnesium bromide will produce when dissolved in solution. At the end, the result turns out to be three (3) particles as shown below:

In chemistry, we understand ionization as a process whereby ionic substances split into ions as they dissolve in water. In this case, for magnesium bromide, MgBr₂, we see one Mg²⁺ and two Br⁻ ions are produced as shown below:

\(MgBr_2\rightarrow Mg^{2+}+2Br^-\)

Which means we will have three (3) particles when one formula unit of this salt dissolves.

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20.0 moles of oxygen react with Antimony to form 13.3 moles of Antimony (III) Oxide. We want to calculate how many moles of Antimony (III) Oxide will be formed from 20.0 moles of oxygen. This is a stoichiometry problem.

The link between the proportional amounts of components participating in a reaction or generating a compound is known as stoichiometry, and it is often expressed as a ratio of whole integers.

Assuming a balanced chemical equation, the stoichiometric ratio between Antimony (III) Oxide  and oxygen can be used to determine the number of moles of Antimony (III) Oxide  formed.

For example, the balanced equation for the reaction of Antimony with O2 to form Antimony (III) Oxide  is:

4 Antimony + 3 O2 → 2 Antimony (III) Oxide

From this equation, it can be seen that 3 moles of oxygen react with 2 moles of Antimony (III) Oxide . Therefore, if there are 20.0 moles of O2, then the number of moles of Antimony (III) Oxide  formed would be:

20.0 moles oxygen × (2 moles Antimony (III) Oxide  / 3 moles oxygen) = 13.3 moles Antimony (III) Oxide.

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Derivate planar density expressions for FCC(100) and (111) is 1/4R² and 4R²√3, number of  atoms are centered on the [100] in FCC is 4.

Atoms can be arranged either closely or widely together. A periodic three-dimensional pattern of atoms is packed into crystalline solids like metals. Atoms are not subject to periodic packing in non-crystalline materials like silicon oxide. A unit cell is a crystal structure's fundamental building block. The packing density of crystals is measured by planar density.

For (100) plane one atom at each of  four cube corners

,each of which is shared with four adjecent unit cells.

Centre atom lies with in the unit cell.

Thus there is the equivalence of 2 atoms associated with this FCC(100) plane.

The plannar section in the above figure is square.

Where

The side lengths are equal to unit cell edge length = 2R√2

Area of the square =( 2R√2 )2

                            =8R2

a) So plannar density of (100) plane

PD (100)plane =(number of atoms centerd at (100) plane)  / (area of plane)

                    = (2) / 8R2

                    =1 / 4R2

∴Plannar density of  FCC(100) is 1 / 4R2.

b) There are 6 atoms whose centre lie on this plane .

Total  equivalence is 2 atoms associated with this FCC(111) plane at centered.

In above diagram

 (2R)2 + h2 = 4R2

which leads to h = 2R√3

Thus Area =4(R) (h) /2

                 =4(R) (2R√3) /2

                 =4R2√3

PD (111)plane =(number of atoms centerd at (111) plane)  / (area of plane)

                    =(2) / 4R2√3

                     =1/ (2R2√3)

∴Plannar density of  FCC(111) is 1/ (2R2√3).

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Answer:

50.05 g.

Explanation

you welcome ;)

The properties of a natural resource that make it useful for humans as a material or energy source is the ability to convert mass into energy and vice versa.

The expression natural resources make reference to all types of matter and energy extracted from nature that can be used to produce goods and services.

Some examples of natural resources include for example irreversible resources such as fossil fuels (i.e., oil, or coal, gas, minerals such as metals, rocks, etc) as well as those based on the use of reversible energy such as eolic air energy, solar radiation or sunlight, soil and hydric resources or water.

Therefore, with this data, we can see that natural resources can be defined as any material and or energy obtained from nature that may be irreversible or reversibly used to produce goods and services.

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milk, cheese and other dairy foods.

green leafy vegetables – such as curly kale, okra but not spinach (spinach does contain high levels of calcium but the body cannot digest it all)\

and the question is not finished and ... what?

Answer:

acceleration

Explanation:

Answer:

28.55 mL or 0.02855 L

Explanation:

Applying,

Density(D) = mass(m)/Volume(V)

D = m/V............... Equation 1

Where D = Density of the Bromine, m = mass of the bromine, V = volume of the bromine

make V the subeject of the equation

V = m/D............. Equation 2

From the question,

Given: m = 88.5 g, D = 3.10 g/mL

Substitute these values into equation 2

V = 88.5/3.10

V = 28.55 mL

Hence the volume occuppied by the Bromine gas is  28.55 mL or 0.02855 L

Answer:

A) Using the ideal gas law, we can calculate the number of moles of gas present:

```

PV = nRT

```

where:

* P = pressure (atm) = 1 atm

* V = volume (L) = 5.6 L

* n = number of moles of gas

* R = ideal gas constant = 0.08206 L atm / mol K

* T = temperature (K) = 273.15 K

Solving for n, we get:

```

n = (P * V) / RT

```

```

n = (1 atm * 5.6 L) / (0.08206 L atm / mol K * 273.15 K)

```

```

n = 0.25 mol

```

Therefore, there are 0.25 moles of gas present.

B) The molar mass of the gas can be calculated by dividing the mass of the gas (8.0 g) by the number of moles of gas (0.25 mol):

```

Molar mass = Mass / n

```

```

Molar mass = 8.0 g / 0.25 mol

```

```

Molar mass = 32 g/mol

```

The molar mass of the gas is 32 g/mol.

C) The common atmospheric gas with a molar mass of 32 g/mol is oxygen (O2). Therefore, the gas that was collected is oxygen.

Explanation:

The no. Of moles of carbon is found to be 7.5×10⁻⁴.

No. of moles are give by = Given mass/molar mass.

Given mass is 0.009g and the molar mass of carbon atom is 12g.

So, the no. Of moles are given by ,

No. Of moles = 0.009/12

= 0.00075 = 7.5×10⁻⁴mol.

The value of 6.023 x 10²³ is equal to one mole of any substance (Avogadro number). It can be used to quantify the chemical reaction’s by-products. The symbol for the unit is mol.

Carbon element is insoluble in water, diluted acids and bases, as well as organic solvents, carbon is an inert material. It combines with oxygen at high temperatures to generate carbon monoxide or dioxide. Diamond and graphite are two distinct allotropes, crystalline forms of the carbon atom.

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Answer:

12.5 grams

Explanation:

The strict regulations established by USP Chapter <797> regarding hood-cleaning procedures indicate the critical importance of hood cleaning to patient health and safety.

The purpose of these guidelines is to ensure that compounding pharmacies maintain a clean and sterile environment when preparing medications, especially those that will be administered to patients. The guidelines specify the frequency of cleaning and the level of detail required to ensure that the hoods are free of contaminants, which could compromise the quality and efficacy of the compounded products.

Therefore, Failure to comply with these regulations could result in contamination of medications and subsequent harm to patients. Thus, it is crucial to adhere to these guidelines to maintain a safe and sterile environment for compounding medications.

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Answer:

24.31 g/mol.

Explanation:

moles =mass/molar mass

n=w/m

0.6 is the equilibrium constant for the given reaction.

To calculate the equilibrium constant (K) for the given reaction, we need to use the molar concentrations of the reactants and products at equilibrium. The equilibrium constant expression is given by:

\(K= [H_{2}O]^{2} / ([H_{2}^{2} * [O_{2}])\)

Given the moles of H2, O2, and H2O in the 8.1 L container, we can convert them to molar concentrations by dividing the number of moles by the volume:

[H2] = 1.83 moles / 8.1 L

[O2] = 2.33 moles / 8.1 L

[H2O] = 3.95 moles / 8.1 L

Substituting these values into the equilibrium constant expression, we have:

K = \((3.95 / 8.1)^{2}\) / (\((1.83 / 8.1)^{2}\) * (2.33 / 8.1))

Evaluating this expression and rounding to one decimal place, we find the equilibrium constant to be:

K ≈ 0.6

Therefore, the equilibrium constant for the given reaction is approximately 0.6.

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3.125 × 10^8 atoms of calcium-41 and 4.6875 × 10^9 atoms of potassium-41.

Option D is correct

A rare and long-lived radioactive isotope of calcium is calcium-41 (41Ca).

# Given-

Half life = 1.03×10^5 years

- After 1.03×10^5 years (1 half life)

calcium-41 will be 50%

potassium-41 will be 50%

- After 2.06×10^5 years (2 half lives)

calcium-41 will be 25%

potassium-41 will be 75%

- After 3.09×10^5 years (3 half lives)

calcium-41 will be 12.5%

potassium-41 will be 87.5%

- After 4.12×10^5 years (4 half lives)

calcium-41 will be 6.25%

potassium-41 will be 93.75%

After 4.12×10^5 years,

Calcium-41  = 6.25/100 × 5 × 10^9  =  3.12×10^8 atoms

Potassium-41  = 93.75/100 × 5 × 10^9  =  4.69×10^9 atoms

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Answer:

a.) (aq)

Explanation:

(aq) means aqueous

so its in a solvent

Answer:

Remove 1 electron

Explanation:

In the atom of each element, there are three subatomic particles viz: proton, neutron and electron. The number of proton (positively charged) and electron (negatively charged) determines the charge of that element. The more the proton, the more positively charged an ion is and vice versa for electron.

According to this question, a neutral atom of lithium (Li) with atomic no. 3 is given i.e. a lithium atom with charge 0. To make the lithium atom's charge change to +1, ONE ELECTRON MUST BE REMOVED OR LOST.

Note that, the proton number (atomic number) of an element does not change, rather the electron number changes in relation to the no. of protons.

This representation suggests that the element is phosphorus (P) with a mass number of 30, which is incorrect. The correct mass number for phosphorus is approximately 30.97. The incorrect representation for a neutral atom is 36Li

To determine the correct representation for a neutral atom, we need to consider the atomic number (Z) and mass number (A) of the element. The atomic number represents the number of protons in the nucleus, while the mass number represents the sum of protons and neutrons.

Let's analyze the given representations:

36Li:

This representation suggests that the element is lithium (Li) with a mass number of 36, which is incorrect. The correct mass number for lithium is approximately 6.94.

613C:

This representation suggests that the element is carbon (C) with a mass number of 13, which is correct. Carbon has different isotopes, and 13C represents one of its stable isotopes.

3063Cu:

This representation suggests that the element is copper (Cu) with a mass number of 63, which is correct. Copper has different isotopes, and 63Cu represents one of its stable isotopes.

1530P:

This representation suggests that the element is phosphorus (P) with a mass number of 30, which is incorrect. The correct mass number for phosphorus is approximately 30.97.

Therefore, the incorrect representation for a neutral atom is 36Li, as it does not match the known properties of lithium.

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The compound name of Al(ClO₂)O₃ is aluminium chlorite.

A chemical compound is a substance formed by the union of two or more chemical elements in a fixed ratio, the union being a chemical bond.

A chemical formula is a notation indicating the number of atoms of each element present in a compound. It shows the types and number of each atom of element that makes up a chemical compound.

According to this question, a chemical compound is given with the chemical formula; Al(ClO₂)O₃. This suggests that this compound is made up of the following elements:

The chemical compound is called aluminium chlorite.

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Answer:

301.15 K

Explanation:

This exercise is related to the subject of Structured Thinking. See the definition of structured thinking below.

The process of developing a structured framework to address an unstructured situation is known as structured thinking.

Structured thinking is a strategy of problem-solving that entails breaking a huge problem into smaller ones in order to tackle the larger problem more quickly and effectively.

Categorizing things and identifying themes are two subjects under structured thinking.

To categorize anything requires:

Identifying themes entails:

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Albert Einstein's work focused on relativity theory and the as it could be applied to gravity, quantum mechanics and the theory of light.

The physicist Albert Einstein was born in Ulm Württemberg Germany in 1875. He migrated to the United States at the height of the Nazi regime in Germany where he became a professor of physics at Princeton University.

His work focused on the relativity theory and the as it could be applied to gravity, quantum mechanics and the theory of light. These ideas still remain fundamental in science today.

I became interested in the work of Albert Einstein because of the fact that is work is fundamental to the understanding of quantum mechanics and the idea of gravity. He laid the foundation upon which modern physics is built.

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From the solution that we have in the question;

The equilibrium constant, denoted as K, is a value that quantitatively represents the ratio of the concentrations of products to reactants at equilibrium in a chemical reaction.

It is a fundamental concept in chemical equilibrium.

The value of the equilibrium constant provides valuable information about the position of equilibrium and the relative concentrations of species involved in a chemical reaction.

Kc = [X] [Y]/[XY]

\(0.25 = (0.1 + x)^2/(0.5 - x)\)

\(0.25(0.5 - x) = (0.1 + x)^2\)

\(0.125 - 0.25x =0.01 + 0.2x + x^2\\ x^2 + 0.45x - 0.115 = 0\)

x = 0.18 M

The equilibrium amount of X and Y=  0.28 M and the equilibrium concentration of XY = 0.32 M

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Based on the answer to the question that we have;

The ratio of the product to reactant concentrations in a chemical reaction at equilibrium is represented quantitatively by the equilibrium constant, abbreviated as K.

It is a cornerstone of the theory of chemical equilibrium.

A chemical reaction's equilibrium position and the relative concentrations of the species involved can both be learned from the equilibrium constant's value.

Kc = [X][Y]/[XY]

\(0.25 = (0.1 + x)^2/(0.5 - x)\\0.25(0.5 - x) = (0.1 +x)^2\\0.125 - 0.25x = 0.01 +0.2x +x^2\\= 0.18 M\)

The equilibrium concentration of;

XY =0.5 - 0.18

=0.32 M

Then the equilibrium amount of

X and Y is

0.1 + 0.18= 0.28 M.

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The system developed to extinguish fires in commercial cooking operations, reducing the temperature of the burning liquid is the "wet chemical extinguishing system".

This system is specifically designed to suppress fires that start in deep fryers and other cooking appliances commonly found in commercial kitchens. The wet chemical extinguishing system is a specialized fire suppression system that uses a specially formulated liquid agent to extinguish fires by cooling the fuel and preventing re-ignition. The liquid agent is discharged in a fine mist, which blankets the burning fuel and cools it to below its ignition temperature. This helps prevent the fire from spreading and re-igniting, and reduces the potential for injury or property damage. Other fire suppression systems temperature that are commonly used include the combination system, which combines wet chemical and dry chemical agents, the dry pipe system, which is a type of sprinkler system that uses pressurized air to prevent water from entering the pipes until a fire is detected.

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