_____ is a halogen used in gaseous and liquid form for large scale disinfection of drinking water and sewage. A. Iodine B. Chlorine C. Bromine D. Fluorine E. Betadine

Answers

Answer 1
Answer:

Answer:

Chlorine

Explanation:

Chlorine is a halogen that is a strong oxidizer (it takes electrons from nearby compounds).  In so doing, it kills bacteria, viruses, and other microorganisms.  The chlorine reacts with cell walls or other vital organic compounds (e.g., proteins)  to render them useless.  Chlorine is relatively inexpensive and generally easy to handle, but it is dangerous in gaseous form and highly alkaline in solution, so must be stored and handled properly.


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According to VSEPR theory, what causes water molecules to have a bent shape? A. the unusual location of the free electrons. . B. repulsive forces between specific lone pairs of electrons. . C. attraction between the fixed orbitals of the unshared pairs of oxygen electrons. . D. ionic attraction and repulsion

Answers

According to VSEPR theory, what causes water molecules to have a bent shape?

Answer: Out of all the options presented above the one that best represents what causes water molecules to have a bent shape according to VSEPR theory is answer choice B) repulsive forces between specific lone pairs of electrons. the bond angle in a water molecule is bent.

I hope it helps, Regards.

The correct answer is \boxed{{\text{option }}\left( {\text{B}} \right)} that is repulsive forces between specific lone pairs of electrons.

Further Explanation:

The total number of valence electrons of {{\text{H}}_2}{\text{O}} is calculated as,

Total valence electrons (TVE) = [(1) (Valence electrons of O) + (2) (Valence electrons of H)]

\begin{aligned}{\text{Total valence electrons}}\left( {{\text{TVE}}}\right)&=\left[{\left({\text{1}}\right)\left({\text{6}}\right)+\left({\text{2}}\right)\left({\text{1}}\right)}\right]\n&=8\n\end{aligned}

In {{\text{H}}_2}{\text{O}}, the total number of valence electrons is 8. Here, oxygen forms single bond with the hydrogen atom and therefore, 2 pair of electrons are used in the formation of two single bonds with hydrogen atom. Remaining 2 pair of electrons are used to complete the octet of oxygen atom. Therefore, {{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}contains two bond pairs and two lone pairs (refer to the image attached).

According to the VSEPR theory, central atom oxygen has two bond pair with two lone pair, therefore, {{\text{H}}_2}{\text{O}} has an {\text{A}}{{\text{B}}_2}{{\text{E}}_2} arrangement. Therefore, these four pairs of electrons spread out as tetrahedral arrangement to minimize lone pair-lone pair and bond pair-bond pair repulsion. Since lone pairs are not considered in the shape of the molecule and therefore, the final shape of water molecule is bent-shape.

Learn more:

1. Molecular shape around the central atom in the amino acid glycine: brainly.com/question/4341225

2. Balanced chemical equation: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Covalent bonding and molecular structure

Keywords: hybridization, water, geometry, sp3, steric number, 3, p orbital, s orbital, Lewis structure, hybridization of water, central atom, carbon, hydrogen, oxygen, shape, double bond, h2o.

What is the percent by mass of chlorine in NaCl?

Answers

The percent  by mass  of chlorine  in NaCl is 60.68 %  of chlorine

 calculation


mass %  of chlorine =mass  of chlorine /mass  of NaCl  x 100


from periodic table the  mass of cl= 35.5  g/mol  

                                       that of   of Na  = 23  g/mol

The mass of NaCl is therefore  = 35.5 g/mol + 23 g/mol  =58.5  g/mol


mass %  is therefore=  (35.5 / 58.5   x 100) =60.68 %

In the given question, 60.66% is the percent by mass of chlorine in NaCl.

Mass is a physical property of matter that measures the amount of substance in an object. It is typically measured in grams (g) or kilograms (kg).

The molar mass of NaCl is 58.44 g/mol. The molar mass of chlorine is 35.45 g/mol. The formula of NaCl indicates that there is one chlorine atom for every sodium atom. Therefore, the mass of chlorine in NaCl is 35.45 g/mol.

To calculate the percent by mass of chlorine in NaCl, we need to divide the mass of chlorine by the total mass of NaCl and multiply by 100. This can be expressed as follows:

Percent by mass of chlorine = (mass of chlorine/mass of NaCl) × 100

Substituting the values, we get:

Percent by mass of chlorine = (35.45 g/mol/58.44 g/mol) × 100

= 60.66%

Therefore, the percent by mass of chlorine in NaCl is 60.66%.

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A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the percent yield of oxygen in this chemical reaction?Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..

Answers

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

Answer:

B. %73.40 on edge 2020

Explanation:

Took the quiz and got it right, hope this helped:)

I am in desperate need of help will give 40 points to whoever helps me!!!!! I have a quiz tomorrow but I have no idea how to do these practice problems. I know u have to use the boiling and freezing pint formula but i don't know how to plug them in because these problems all look different and asks for different things

Answers

All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.
Freezing point depression or Boiling point elevation:

ΔT = -K (m) (i)

ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent. 

K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent. 

m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent. 

i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.

Which rule should be followed when naming binary acids

Answers

 Here you are this will explain it it is one of my own persoal papers    

Acids are divided into two groups: Binary and Oxyacids.  NAMING BINARY ACIDS: The name of the binary acid consists of two words. The first word has three parts: the “hydro” prefix the root of the nonmetal element the “ic” ending The second word is always “acid” Examples: HCl = hydro chlor ic acid = hydrochloric acid HBr = hydro brom ic acid = hydrobromic acid HF = hydro fluor ic acid = hydrofluoric acid 2. NAMING OXYACIDS: These are more difficult to name because these acids have hydrogen, a nonmetal, and may have varying numbers of oxygen atoms. For example, H2SO5, H2SO4, H2SO3, and H2SO2 are all acids. How do we name them? To begin, we need a point of reference. Our reference point is this: The “ate” ions (sulfate, nitrate, etc) make the “ic” acids (sulfuric acid, nitric acid) Examples: SO4 2- = sulfate ion H2SO4 = sulfuric acid NO3 - = nitrate ion HNO3 = nitric acid Once we have our point of reference, the acid with one more oxygen than the -ic acid is called the per-_________-ic acid. The acid with one less oxygen then the -ic acid is called the ___________-ous acid. If the acid has one less oxygen than the -ous acid, it is called the hypo-____________-ous acid. Examples: H2SO5 = persulfuric acid HNO4 = pernitric acid H2SO4 = sulfuric acid HNO3 = nitric acid H2SO3 = sulfurous acid HNO2 = nitrous acid H2SO2 = hyposulfurous acid HNO = hyponitrous acid The KEY: 
Make sure that a hydrogen is bonded with a nonmetallic element, like HBr, HCl, or HF. 

How do you determine formula units in a unit cell?

Answers

In order to determine the number of formula units of a substance in a cell, you need to know the density of the substance and the edge length of the unit cell. Suppose that you are given the following problem:The density of TlCl(s) is 7.00 g/cm³, and the length of an edge of a unit cell is 385 pm. How many formula units of TlCl are in a unit cell?
Here’s how you solve the problem.
Calculate the volume of the unit cell.
If the length of an edge of the cell is a, then V = a³.
V = (385 × 10⁻¹² m)³ = 5.71 × 10⁻²⁹ m³Use the density to calculate the mass of the unit cell (The density of TlCl is the same, no matter what volume of TlCl we have).
Mass = 5.71 × 10⁻²⁹ m³ × (7.00 g/1 cm³) × (100 cm/1 m)³ = 3.99 × 10⁻²² gUse the molar mass to convert grams to moles.
3.99 × 10⁻²² g × (1 mol/239.8 g) = 1.67 × 10 ⁻²⁴ molUse Avogadro’s number to calculate the number of formula units.
1.67 × 10 ⁻²⁴ mol × (6.022 × 10²³ formula units/1 mol) = 1.00 formula units
Therefore, 1 unit cell of TlCl contains 1 formula unit.

Final answer:

To determine the formula units in a unit cell, understand the type of the unit cell structure (simple cubic, body-centred cubic, or face-centred cubic) and apply the appropriate calculation for that type.

Explanation:

To determine the number of formula units in a unit cell, one must first understand what these terms mean. A unit cell is the smallest repeating unit of a crystal lattice. There exist several different types that affect formula unit arrangement and quantity, including simple cubic, body-centred cubic, and face-centred cubic.

A crystal lattice is a three-dimensional framework of atoms, ions, or molecules arranged in a repeating pattern. The formula unit is the lowest whole-number ratio of ions represented in an ionic compound. In a simple cubic unit cell, there's one formula unit, in a body-centred cubic cell, there are two, and in a face-centred cubic cell, there are four.

For instance, if you are dealing with a simple cubic arrangement the unit cell is composed of 8 atoms forming the vertices and each one is shared amongst 8 neighbouring cells, you have (8 x 1/8 = 1) formula unit per cell. Similarly, in a body-centred cubic, there is 1 atom in the centre fully inside, and 8 on the corners shared with neighbouring cells, hence (1 + 8 x 1/8 = 2) formula units per unit cell.

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