Identify the compound with polar covalent bonding. identify the compound with polar covalent bonding. ar nh3 k kf h2

Answer:

c. palisade layer  

Answer:

Explanation:

Oxidation-reduction reaction is the simulaneous oxidation and reduction of the substances and is represented by two half-reactions.

The oxidation half-reaction is the loss of electrons, with the consequent increase in the oxidation state by the oxidized substance.

In this case, the process that shows the loss of electrons is:

That reaction shows:

On the other hand, the reduction half-reaction is the gain of electrons, with the consequent reduction of the oxidation state by the reduced substance.

In this case, the process that shows the gain of electrons is:

That reaction shows:

Summary:

Answer- The correct answer is ammonia.

Explanation- All other compounds that are formed do not have any essence of Nitrogen in it. Whereas, Methane is a compound  with carbon and hydrogen as elements while water has hydrogen and oxygen as its elements.

Lastly Hydrogen and chlorine combined together to form hydrochloric acid. Hence the only compound left with nitrogen is ammonia and has nitrogen and hydrogen as its basic elements.

Answer:

The answer is C. Ammonia

Explanation: I got it right on the test.

Have a good day :)

Final answer:

The process of sp³ hybridization involves the mixing of an s orbital with three p orbitals to form four sp³ hybrid orbitals. This type of hybridization occurs in molecules like methane. On the other hand, sp² hybridization involves the mixing of an s orbital with two p orbitals to form three sp² hybrid orbitals, which occurs in molecules like boron trifluoride and ethene. Lastly, sp hybridization is the mixing of an s orbital with a single p orbital to form two sp hybrid orbitals, which occurs in molecules like beryllium hydride.

Explanation:

sp³ hybridization is a process where an s orbital and three p orbitals mix to form a set of four sp³ hybrid orbitals. These hybrid orbitals are arranged in a tetrahedral geometry, with each lobe of the hybrid orbitals pointing towards one of the corners of the tetrahedron. This type of hybridization occurs in molecules such as methane (CH₄).

sp² hybridization is the mixing of an s orbital and two p orbitals to form a set of three sp² hybrid orbitals. These hybrid orbitals are arranged in a trigonal planar geometry, with each lobe of the hybrid orbital pointing towards one corner of the triangle. This type of hybridization occurs in molecules such as boron trifluoride (BF₃) and ethene (C₂H₄).

sp hybridization is the mixing of an s orbital with a single p orbital to form a set of two sp hybrid orbitals. These hybrid orbitals are oriented linearly, with each lobe of the hybrid orbital pointing in opposite directions. This type of hybridization occurs in molecules such as beryllium hydride (BeH₂).

In the case of ethene (C₂H₄), each carbon atom undergoes sp² hybridization to form three sp² hybrid orbitals. One of these hybrid orbitals forms a bond with the identical hybrid orbital on the other carbon atom, resulting in the formation of a double bond. The remaining two hybrid orbitals form bonds with the 1s orbitals of two hydrogen atoms. The unhybridized 2pz orbitals on each carbon atom form another bond by overlapping sideways with each other.

The concept of orbital hybridization allows atomic orbitals to combine and form hybrid orbitals that have different energy and orientation compared to the constituent orbitals. In the case of carbon, different combinations of s and p orbitals can produce four sp³, three sp², or two sp hybrid orbitals.

An sp³ hybrid orbital can also hold a lone pair of electrons. For example, in ammonia (NH₃), the nitrogen atom is surrounded by three bonding pairs and a lone pair of electrons. The nitrogen atom undergoes sp³ hybridization, with one hybrid orbital occupied by the lone pair.

Answer:

Al → Al³⁺ + 3 e⁻,     is an oxidation process.

S²⁻ → S + 2 e⁻,        is an oxidation process.

Cu → Cu²⁺ + 2 e⁻,    is an oxidation process.

Explanation:

So,

Al loses 3 electrons and produce Al³⁺c

S²⁻ loses 2 electrons and produce S.

Cu loses 2 electrons and produce Cu²⁺.

All the given processes (Al becoming Al₃+, S₂- becoming S, and Cu becoming Cu₂+) are examples of oxidation, as each involves the loss of electrons.

To determine whether each process is oxidation or reduction, we will look at the change in electron configuration for each reaction.

Thus, all the given processes are examples of oxidation, where each element or ion loses electrons.

Glucose is the most common disaccharide in our diet is the false statement regarding glucose.

The chemical formula of glucose is written as C₆H₁₂O₆.

Starch and glycogen is the highly branched compound or the polymers, which can be formed from the fundamental monomer unit of glucose. Glucose is also present in the blood of as a blood sugars and used as a source of energy or we can say that fuel for the human body. Glucose is a monosaccharide, it means it will not able to break down in following sub units.

Hence, option (c) is false because glucose is a monosaccharide.

To learn more about glucose, visit the below link:

https://brainly.com/question/461228

The false statement about glucose is (C), that is Glucose is the most common disaccharide in our diet. Glucose is actually a monosaccharide. Common disaccharides include sucrose, lactose, and maltose.

To determine which statement regarding glucose is FALSE, let's analyze the given options:

Answer: Oxygen

Explanation:

Answer: The substance which gets reduced in the hydrogen-oxygen fuel cell is oxygen.

Explanation:

A reduced substance undergoes reduction reaction. This is defined as the reaction in which an atom gains electrons.

A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.

The reactions which occur in hydrogen-oxygen fuel cell are:

At cathode:  [tex]H_2+2OH^-\rightarrow 2H_2O+2e^-[/tex]

At anode:  [tex]\frac{1}{2}O_2+H_2O+2e^-\rightarrow 2OH^-[/tex]

As, the oxygen is gaining electrons from the above reaction. Thus, it is undergoing reduction reaction and is getting reduced.

Hence, the substance which gets reduced in the hydrogen-oxygen fuel cell is oxygen.

Answer:

Explanation:

A solution in general is a homogeneous mixture in which a substance, named solute, is dissolved, in other substance, name solvent.

Solutions may be in solid, liquid or gas state. There are many kind of solvents. Usually, in a lab you work with liquid solutions. Some liquid solvents are: ethanol, glycerin, hexane, benzene, and water, among many others.

Aqueous solution is a solution where the solvent is water. Of course, the solute may be any one: NaCl, sugar, ethanol, an acid, a base, a salt.

What distinguish a solution in general and an aqueous solution is the solvent.

Answer:

The reduced atom is Br.

Explanation:

2Na + Br₂ → 2NaBr,

Na loses 1 electrons and is oxidized to Na⁺. (Na → Na⁺ + e).

Br₂ gains 2 electrons and is reduced to 2Br⁻. (Br₂ + 2e → 2Br⁻).

So, the reduced atom is Br.

Final answer:

The nuclear equation for the synthesis of rutherfordium-257 by bombarding californium-249 with carbon-12 is 24998Cf + 126C → 257104Rf + 410n. It shows the reactants, products, and the conservation of mass and atomic numbers in the reaction.

Explanation:

The student is asking for a nuclear equation for the synthesis of rutherfordium-257, which is created by bombarding californium-249 (Cf-249) with carbon-12 (C-12). The balanced nuclear equation takes into account the conservation of mass and atomic numbers. The californium-249, which has an atomic number of 98 and a mass number of 249, reacts with carbon-12, which has an atomic number of 6 and a mass number of 12.

The equation is:

24998Cf + 126C → 257104Rf + 410n

In this reaction, rutherfordium-257 has an atomic number of 104 and a mass number of 257. To conserve the mass number and atomic numbers on both sides of the equation, we need to add four free neutrons (n) with a mass number of 1 and an atomic number of 0 to the products side. This completes the balanced nuclear equation for the formation of rutherfordium-257.

Answer:

[tex]\boxed{\text{(a) 0.00 V; (b) 0.424 V}}[/tex]

Explanation:

We must look up the standard reduction potentials for the half-reactions.

                                                                    ℰ°    

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O     1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                       1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O             1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺                                      0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

                                                                                  ℰ°/V    

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻                  -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                                     1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻  + 6Cl⁻ + 14H⁺     0.00

(b) Fe²⁺/IO₃⁻

                                                                            ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻                                                -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O                      1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O     0.424

The ℰ° values for the cells are [tex]\boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}[/tex]

Answer:

Answer:

\boxed{\text{(a) 0.00 V; (b) 0.424 V}}

Explanation:

We must look up the standard reduction potentials for the half-reactions.

                                                                   ℰ°    

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O     1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                       1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O             1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺                                      0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

                                                                                 ℰ°/V    

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻                  -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                                     1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻  + 6Cl⁻ + 14H⁺     0.00

(b) Fe²⁺/IO₃⁻

                                                                           ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻                                                -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O                      1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O     0.424

The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}

Explanation:

its right trust

Answer:

A. Add a coefficient 4 before h2o and add a 2 before so2

Explanation:

Answer : The correct option is, Add a coefficient 2 before [tex]H_2O[/tex] and a 2 before [tex]SO_2[/tex].

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

If the amount of atoms of each type on the left and right sides of a reaction differs then to balance the equation by adding coefficient in the front of the elements or molecule or compound in the chemical equation.

The coefficient tell us about that how many molecules or atoms present in the chemical equation.

The given chemical reaction is,

[tex]2H_2S+3O_2\rightarrow H_2O+SO_2[/tex]

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen, sulfur and oxygen atoms are not balanced.

In order to balance the chemical equation, the coefficient '2' is put before the [tex]H_2O[/tex] and [tex]SO_2[/tex] and we get the balanced chemical equation.

The balanced chemical reaction will be,

[tex]2H_2S+3O_2\rightarrow 2H_2O+2SO_2[/tex]

trueee hope it helpsss

The given statement is true. Privacy and cleanliness are considered as two priorities for the guest quarters.

Quarters are place where people stay as guest for some duration of time. People consider many priorities while booking hotels or quarters for their stay because as they are going to live in it for some span of time.

The first two priorities include privacy and cleanliness. They look out for both the external as well as internal cleanliness. External cleanliness includes surroundings, laws, etc., whereas the inner cleanliness includes right from the articles present in the room till the toilet of the room. As they are living in another individual living area they obviously demand for some sort of privacy.

Answer:

By using a screen

Answer:

Use a screen

Explanation:

Answer:

Explanation:

A redox reaction is a reaction in which there is an exchange of electrons between different species. As one specie gains electrons the other loses it and this furnishes the reaction.

Redox reactions are made up of half-reactions which shows how a specie gains electrons and what happens to it. It also shows what happens to a specie that loses electrons.

To understand and be able to identify what an oxidation and reduction half reaction is, we need to know what oxidation and reduction entails.

Oxidation is the loss of electron by an atom. It also entails addition of oxygen to a specie, removal of hydrogen from a specie. This leads to increase in oxidation number of the specie.

Reduction is just the opposite of oxidation and electrons are gained by an atom here.

To identify an oxidation or reduction half in a redox reaction, we look at the species closely and we check for the following:

1. Changes in oxidation number

2. Wether a species loses or gains electrons.

3. Addition /removal of oxygen and hydrogen.

Let's take for example:

Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺

By inspection, we look at the species involved:

Fe: Fe²⁺ → Fe³⁺

The oxidation number increases from 2+ to 3+

There is a loss of an electron for this to occur

We see that this is an oxidation half

Mn: MnO₄⁻ → Mn²⁺

The oxidation number of Mn changes from +7 to +2:

X+ (-2x4) = - 1

X - 8 = - 1

X = 7

This implies that 5 electrons were gained.

There is also a loss of oxygen.

This makes the reaction a reduction half.

This simple way is used to identify reduction and oxidation halves.

To identify oxidation and reduction half-reactions in a redox equation, remember that oxidation involves the loss of electrons and reduction involves the gain of electrons. Balance the reaction by ensuring electrons lost equal electrons gained. Iron and hydrogen serve as examples of a reducing agent and an oxidizing agent, respectively.

To identify oxidation and reduction half-reactions in a redox equation, it's essential to understand the basic principle that oxidation involves the loss of electrons, while reduction involves the gain of electrons.

An oxidation half-reaction will show electrons as products (on the right side), indicating that the species is losing electrons. Conversely, a reduction half-reaction will have electrons as reactants (on the left side), signifying the gain of electrons by the species.

Example of Identifying Half-Reactions

Consider a redox process where Iron (Fe) is oxidized, and Hydrogen (H₂) is reduced. The oxidation half-reaction can be represented as Fe -> Fe²⁺ + 2e⁻, showing iron losing two electrons. For the reduction process, H⁺ + e⁻ -> H₂ illustrates hydrogen gaining electrons.

To ensure the reaction is balanced, the number of electrons lost in the oxidation process must be equal to those gained in the reduction process. This might require multiplying one or both half-reactions by a certain number to balance the electrons.

In the overall reaction, the species that gets oxidized (loses electrons) acts as the reducing agent, and the species that gets reduced (gains electrons) acts as the oxidizing agent. Therefore, in our example, iron acts as the reducing agent, and hydrogen acts as the oxidizing agent.

chlorine by twenty øne piløts

Answer:

Explanation:

1) Data:

a) Experimental value, m₁ = 105.2 amu

b) Known value, m = 107.5 amu

b) % error = ?

2) Formulae:

a) absolute error = | experimental value - known value|

b) % error = [ absolute value / known value ] × 100

3) Solution:

a) absolute error = | m₁ - m | = | 105.2 amu - 107.5 amu | = 2.3 amu

b) % error = [ 2.3 amu / 107.5 amu ] × 100 = 2.1% ← answer

Answer:

SEDIMENTARY ROCK

Explanation:

Answer:

[H⁺] = 1.0 x 10⁻¹² M.

Explanation:

∵ [H⁺][OH⁻] = 10⁻¹⁴.

[OH⁻] = 1 x 10⁻² mol/L.

∴ [H⁺] =  10⁻¹⁴/[OH⁻] = (10⁻¹⁴)/(1 x 10⁻² mol/L) = 1.0 x 10⁻¹² M.

∵ pH = - log[H⁺] = - log(1.0 x 10⁻¹² M) = 12.0.

∴ The solution is basic, since pH id higher than 7 and also the  [OH⁻] > [H⁺].

Answer:dispersed state

dispersed medium

colloid

Answer:

Label A points to the solid particles of pigment that give the paint its color. The solid particles are the (dispersed state.)

Label B represents the liquid surrounding the solid particles. This liquid is called the (Dispersed Medium)

This paint does not settle over time, so it is an example of a (Colloid)

Explanation:

Fernanda looked at a sample of paint under a microscope. At right is the sketch she made of what she saw.

Answer:

B.

Explanation:

Answer:

B) line is the correct answer.

Explanation:

When light shines on a sample, each element emits specific wavelengths producing a unique fingerprint called its line spectra.

The line spectrum is electromagnetic spectra consist of discrete spectra lines.

When the atoms are excited they emit ray of specific wavelengths that correspond to various colors. The emitted ray can be seen as a range of colored line, this range of colored lines is termed as line spectra.

lines spectra are usually utilized to recognize atoms and molecules. Each element has individual line emission spectra.

Answer:

235/92U+10n→144/54Xe+90/38Sr+2/10n

Explanation:

235/92U+10n→144/54Xe+90/38Sr+2/10n

Answer:

Fluorine; f2 + 2e– → 2f–e0 = +2.87 v

Explanation:

Answer:

137.6 mL.

Explanation:

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

(V₁T₂) = (V₂T₁)

V₁ = 125.0 mL, T₁ = 25°C + 273 = 298 K,

V₂ = ??? mL, T₂ = 55°C + 273 = 328 K,

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (125.0 mL)(328 K)/(298 K) = 137.6 mL.

Final answer:

To find the volume of the gas at a different temperature, we can use Charles's Law. By using the formula V1/T1 = V2/T2 and substituting the known values, we can determine that the volume of the gas at 55°C is 275 mL.

Explanation:

To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature, assuming the pressure remains constant. We can use the formula:

V1/T1 = V2/T2

Given that the initial volume (V1) is 125 mL and the initial temperature (T1) is 25.0°C, we can substitute these values into the formula. Rearranging the formula to solve for V2, we get:

V2 = (V1*T2)/T1

Substituting the known values, we have:

V2 = (125 mL * 55 °C) / 25.0 °C = 275 mL

Therefore, the volume of the gas at 55°C is 275 mL.

Exothermic Synthesis

Answer:

the answer is Exothermic Synthesis

Answer:

Explanation:

The coefficients of a chemical equation are the numbers that you put in front of each reactant and product. They are used to balance the equation and comply with the law of mass conservation.

By adjusting the coefficients you obtain the relative amounts (moles) of each product and reactant, i.e. the mole ratios.

Here an example.

The first information is what is called a word equation. E.g. nitrogen and hydrogen react to form ammonia:

        This equation shows the chemical formulae but it is not balanced. The law of mass conservation is not observed.

So, in order to comply with the law of mass conservation you adjust the coefficients as follow.

        As you see, it was necessary to modify the coefficients. Now the law of conservation of mass is observed and you get the mole ratios:

To curb the increase in Earth's temperature due to greenhouse gases, we need to reach 'net zero' carbon emissions by 2050 or sooner. The problem arises largely from burning fossil fuels and deforestation, leading to increased CO2 in the atmosphere. Continued trajectory could double Earth's CO2 levels from pre-industrial times by the end of the century.

To avoid a large increase in temperature due to greenhouse gas emissions, it is crucial that these emissions peak as soon as possible. According to the Union of Concerned Scientists, we should aim for 'net zero' carbon emissions by 2050 or even sooner. This would mean that no more carbon enters the atmosphere than is removed.

Greenhouse gases, primarily carbon dioxide, are largely a result of burning fossil fuels, such as coal and oil. Alongside this, the destruction of tropical forests which absorb CO2 from the atmosphere exacerbates the problem.

If this trajectory continues, Earth's CO2 levels are predicted to double from pre-industrial levels by the end of this century, leading to potentially catastrophic climactic changes. Hence, mitigation strategies to stabilize carbon emissions constitute a major part of our global strategy towards climate change.

https://brainly.com/question/34902503

#SPJ12

Answer:

C. Because people get sick from exposure to high levels of radiation

Explanation:

Answer:

4.32e+26

Explanation:

Answer:

4.32e+26

Explanation:

manganese The metal that is used as a sacrificial electrode to prevent the rusting of iron is manganese.