How many atoms are in 3 C2H6

Answer:

To configure the electron sublevels and groups for each element, we can use the following format:

Element symbol: [Electron configuration] Sublevel: Group

V (Z=23): [Ar] 3d3 4s2 Sublevel: 3d, 4s Group: 5, 4

Ni (Z=28): [Ar] 3d8 4s2 Sublevel: 3d, 4s Group: 10, 4

Cu (Z=29): [Ar] 3d10 4s1 Sublevel: 3d, 4s Group: 11, 4

Zn (Z=30): [Ar] 3d10 4s2 Sublevel: 3d, 4s Group: 12, 4

Cr (Z=24): [Ar] 3d5 4s1 Sublevel: 3d, 4s Group: 6, 4

Mn (Z=25): [Ar] 3d5 4s2 Sublevel: 3d, 4s Group: 7, 4

The correct order according to the Cahn-Ingold-Prelog system is as follows: -COOH III > -CHO IV > -CH2OH II > -CH3 I.

The Cahn-Ingold-Prelog system is a set of rules for determining the priority of groups attached to chiral centers in a molecule. The priority of each group is determined by the atomic number of the atom directly attached to the chiral center, with higher atomic numbers taking precedence over lower ones.

To apply the Cahn-Ingold-Prelog system to the given groups, we need to identify the atoms attached to the chiral center and their corresponding atomic numbers. In this case, the chiral center is the carbon atom in each group, and the attached atoms are hydrogen (H), oxygen (O), carbon (C), and iodine (I).

The atomic numbers of these atoms are: H = 1, C = 6, O = 8, and I = 53.

Using these atomic numbers, we can rank the groups in order of decreasing priority as follows:-

COOH III (O = 8) > -CHO IV (O = 8) > -CH2OH II (O = 8) > -CH3 I (H = 1)

Therefore, the correct order according to the Cahn-Ingold-Prelog system is -COOH III > -CHO IV > -CH2OH II > -CH3 I.

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Sodium sulfide is created when the elements sodium (Na) and sulfur (S) combine. It is a combination reaction, and the balanced equation is 2Na(s)+S(s)fuseNa2S.

If the resulting compound is water-insoluble, a precipitate will form. For instance, a solution of magnesium bromide and silver nitrate (AgNO3) In these reactions, two compounds exchange ions to create two new compounds. As an illustration, consider Na2S + 2HCL + 2NaCl + H2S.

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The size, energy, and number of nodes of the 2p and 3p orbitals are different from one another. In an atomic orbital, a nodal surface is one where there is no chance of encountering an electron.

The 3d orbital in a multi-electron atom has a higher energy than the 3p orbital. 8 A 3p orbital has a radial node, while a 3d orbital has two. A 3p orbital can hold six electrons, but a 3d orbital can hold ten. Greater than 3p orbitals are 3d orbitals. The highest level of probability for an electron to occur is in an atomic orbital. The route that electrons travel while navigating the restrictions of the shell is referred to as a subshell.

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The maximum resulting tenacity of a cotton fiber after wet conditioning can be calculated using the following formula:

Maximum resulting tenacity = Dry tenacity / (1 + moisture regain)

Moisture regain is the amount of moisture absorbed by the fiber when it is fully saturated. For cotton, the moisture regain is around 8.5%.

Therefore, using the given dry tenacity of 5 g/den and a moisture regain of 8.5%, we can calculate the maximum resulting tenacity as:

Maximum resulting tenacity = 5 / (1 + 0.085) = 4.58 g/den

Therefore, the closest option to this answer is (a) 3.55 g/den.
After wet conditioning, a cotton fiber's tenacity usually increases. Given that the dry tenacity is 5 g/den, the maximum resulting tenacity, in g/den, that this fiber would achieve is: c. 6.20 g/den.

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Answer:

\(2.89 \times {10}^{20} \: \: molecules\)

Explanation:

The number of molecules of KCN can be found by using the formula

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 0.00048 × 6.02 × 10²³

We have the final answer as

\(2.89 \times {10}^{20} \: \: \: molecules\)

Hope this helps you

The carbonyl stretch frequency for mesityl oxide is one that is said to be expected to take place in the range of 1670-1750 cm^-1 in the infrared (IR) spectrum.

Mesityl oxide is an organic substance that has the molecular formula C6H10O and consists of a carbonyl functional group situated within the molecule. Upon exposure to infrared radiation, mesityl oxide takes in certain light frequencies that correspond to the movements of its atomic and chemical bond constituents.

The stretching of the C=O bond in the carbonyl functional group within mesityl oxide is likely to produce a significant absorption band in the IR spectrum.

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To solve this problem using an ice table, we first need to write out the balanced equation for the reaction between the acid HA and the base NaOH:

HA + NaOH → NaA + H2O
Next, we need to determine which species in the buffer solution will be affected by the addition of NaOH. Since NaOH is a strong base, it will react with the weak acid HA to form its conjugate base A-. Therefore, we can assume that the concentration of A- will increase and the concentration of HA will decrease.
We can set up an ice table as follows:

|           | HA          | A-         | NaOH       |
| --------- | -----------| -----------| -----------|
| Initial   | 0.2 M      | 0.15 M    | 0 M         |
| Change    | -x          | +x          | +0.0015 M |
| Equilibrium| 0.2 - x    | 0.15 + x  | 0.0015 M   |
Since HA is a weak acid, we can assume that x will be much smaller than the initial concentration of HA. Therefore, we can simplify the equilibrium concentrations to:

[HA] ≈ 0.2 M - x
[A-] ≈ 0.15 M + x
[NaOH] ≈ 0.0015 M

Next, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution before the addition of NaOH:

pH = pKa + log([A-]/[HA])

We can rearrange this equation to solve for the pKa of the acid:
pKa = pH - log([A-]/[HA])

Plugging in the given values, we get:
pKa = 3.35 - log(0.15/0.2) = 3.42

Now we can use the equilibrium concentrations of HA and A- to calculate the pH of the solution after the addition of NaOH:
[OH-] = [NaOH] = 0.0015 M
[HA] = 0.2 M - x ≈ 0.2 M
[A-] = 0.15 M + x ≈ 0.15 M

The reaction between NaOH and HA will consume some of the HA and produce some of the A-. We can calculate the amount of HA consumed and the amount of A- produced using the stoichiometry of the reaction:
0.0015 mol NaOH x (1 mol HA / 1 mol NaOH) = 0.0015 mol HA consumed
0.0015 mol NaOH x (1 mol A- / 1 mol NaOH) = 0.0015 mol A- produced

Now we can calculate the new concentrations of HA and A-:

[HA] = 0.2 M - 0.0015 mol = 0.1985 M
[A-] = 0.15 M + 0.0015 mol = 0.1515 M

Finally, we can use the Henderson-Hasselbalch equation to calculate the new pH of the buffer solution:

pH = pKa + log([A-]/[HA])
pH = 3.42 + log(0.1515/0.1985) = 3.37

Therefore, the pH of the solution after the addition of NaOH is 3.37, which matches the given answer.

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The pH of the buffer solution after the addition of NaOH is 3.37.

To solve this problem using an ICE table, we need to first write the balanced chemical equation for the reaction between the acid HA and NaOH. This reaction is;

HA + NaOH → H₂O + NaA

Next, we need to determine which species in the buffer system will react with the added NaOH. In this case, it is the acid HA. So, the initial concentration of HA is 0.2 M and the initial concentration of A- is 0.15 M.

We can then use the ICE table to determine the change in concentration of each species in the buffer solution after the addition of NaOH. The ICE table is;

HA NaOH H₂O A⁻

I 0.2 M 0 M 0 M 0.15 M

C -x -0.0015M +x +x

E 0.2-x -0.0015M x 0.15+x

where x is the change in concentration of HA and A⁻ after the addition of NaOH.

The equilibrium expression for the reaction between HA and A⁻ is;

Ka = [H⁺][A⁻]/[HA]

At equilibrium, the concentration of H⁺ is equal to the concentration of NaOH that was added (since NaOH is a strong base that completely dissociates in water), so we can write;

Ka = [NaOH][A⁻]/[HA]

Substituting the equilibrium concentrations from the ICE table, we have;

Ka = (0.0015 M)(0.15+x)/(0.2-x)

Simplifying and assuming that x is small compared to 0.15 and 0.2, we can approximate;

Ka = (0.0015 M)(0.15)/(0.2) = 1.125 x 10⁻⁴

The pH of the buffer solution will be calculated using the Henderson-Hasselbalch equation;

pH = pKa + log([A⁻]/[HA])

Substituting the given values, we get;

3.35 = pKa + log(0.15/0.2)

Solving for pKa, we get;

pKa = 3.35 - log(0.15/0.2) = 3.72

Finally, we can use the Henderson-Hasselbalch equation again to calculate the pH of the buffer solution after the addition of NaOH;

pH = pKa + log([A⁻]/[HA])

Substituting the equilibrium concentrations from the ICE table, we get;

pH = 3.72 + log(0.15/(0.2-0.0015))

= 3.37

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--The given question is incomplete, the complete question is

"A buffer that contains 0.2M of acid HA and 0.15M of its conjugate base A-, has a pH of 3.35. What is the pH after 0.0015mol of NaOH is added to 0.5L of this solution?"--

Answer:

Explanation:

Look up the density of the material the object is made of in kg/m³.

Measure the volume of the object in m³.

Multiply the density by the volume.

You will then have the mass of the object in kg.

Answer:

How do you think plants would grow in space? Which tropism(s) would be affected and why?

In the absence of gravity, plants use other environmental factors, such as light, to orient and guide growth. A bank of light emitting diodes (LEDs) above the plants produces a spectrum of light suited for the plants' growth. Enjoy don't  fail :)

Explanation:

In an ecosystem like space where there is no gravity, geotropism would be affected .

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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The process that converts straight-chain alkanes into branched hydrocarbons is called: B. cracking.

The process that converts straight-chain alkanes into branched hydrocarbons is called option B: cracking. Cracking is a chemical process widely used in the petroleum industry to break down larger hydrocarbon molecules into smaller ones. It involves the thermal decomposition of hydrocarbons under high temperatures and pressures, often in the presence of a catalyst.

During cracking, long-chain alkanes are subjected to heat and pressure, causing the carbon-carbon bonds to break. This results in the formation of shorter hydrocarbon chains, including branched hydrocarbons. The process can occur in various forms, such as thermal cracking, catalytic cracking, or hydrocracking, depending on the specific conditions and desired products.

By converting straight-chain alkanes into branched hydrocarbons, cracking enhances the volatility, stability, and reactivity of the resulting hydrocarbon products. It is an essential process in the production of gasoline, diesel fuel, and other valuable hydrocarbon derivatives, as branched hydrocarbons often exhibit improved combustion characteristics and higher octane ratings.

Overall, cracking plays a vital role in the petroleum refining industry, enabling the transformation of long-chain alkanes into a more diverse range of hydrocarbon products with desired properties and applications.

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Answer:

Falso.

Explanation:

El modelo cuántico (no relativista) de Schrodinger, piensa a los electrones como ondas estacionarias, tal que la amplitud de dichas ondas decae muy rápidamente cuando se "alejaban" del radio atómico. (Es decir, habia poca probabilidad de que los electrones escaparan del radio atomico, y era por ello que se quedaban orbitando cerca del nucleo). Es decir, podríamos pensar en esto como una "nube de probabilidades" más que una nube electronica

La idea de nube electrónica realmente viene de modelos previos, como el de Rutherford o el de Bhor, los cuales eran modelos clásicos.

Answer:

a globe

snow rain clouds

Answer:

The invention of the wheel had an important impact on farming.

Answer:

C

Explanation:

Just took the test

Electrons are well described as waves in the context of quantum mechanics because their behavior and properties are fundamentally different from classical objects, such as balls.

Classical objects are described by classical mechanics, which assumes that they have a definite position and momentum at any given time. In contrast, the wave-particle duality of electrons in quantum mechanics implies that they can exhibit both wave-like and particle-like behavior, depending on the context.

In the double-slit experiment, electrons are shown to exhibit interference patterns that are characteristic of wave behavior. This pattern arises due to the interaction of the electrons' wave functions with the slits and each other, which produces regions of constructive and destructive interference.

Furthermore, the uncertainty principle in quantum mechanics states that there is a fundamental limit to the precision with which one can simultaneously know the position and momentum of a particle. This uncertainty arises due to the wave-like nature of particles, and is not present in classical mechanics.

Therefore, the wave-like behavior of electrons in the double-slit experiment and their fundamental differences from classical objects make it appropriate to describe them as waves in this situation.

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When potassium permanganate (KMnO4) oxidize bromide ions (Br-) in an aqueous solution, it produces a new form of bromine called bromine water (Br2(aq)). Bromine water is a pale yellow solution that contains diatomic bromine molecules (Br2). The reaction can be represented by the following chemical equation:

2KMnO4 + 6H2O + 10Br- → 2MnO2 + 2KOH + 3Br2 + 8H+

The reaction involves the transfer of electrons from bromide ions to the manganese atoms in KMnO4, causing the oxidation of Br- to Br2.

Now, coming to the second part of your question, non-polar mineral oil is not a suitable solvent for bromine water or bromide ions as they are both polar substances. However, if we consider the solubility of Br- and Br2 in non-polar solvents, Br2 would be more soluble in non-polar mineral oil due to its non-polar nature. Bromide ions, on the other hand, are polar and would not dissolve readily in non-polar solvents. In general, polar substances dissolve in polar solvents, while non-polar substances dissolve in non-polar solvents.

In conclusion, when KMnO4 oxidizes Br-, it produces bromine water (Br2(aq)). If we consider the solubility of Br2 and Br- in non-polar mineral oil, Br2 would be more soluble due to its non-polar nature.

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Answer:

the last 2

Explanation:

pls mark me brainliest

Answer:

D. Molecules, atoms, electrons.

Explanation:

I did a quiz on this and got it right

Answer:

D. Molecules, atoms, electrons.

Explanation:

i did the test trust i got it right

The rate law for the given reaction is k[A][B]^3.

Given reaction is 2A + 3B → Products which is first order in A and third order overall.

To find the rate law we need to use the equation for the third-order reaction which is given as :

k[A]^m[B]^n

where, k is the rate constant

m is the order of the reaction with respect to A (i.e., first order in A)

n is the order of the reaction with respect to B (i.e., third order in B)

By substituting the given values , k[A]^1[B]^3

The order of the reaction with respect to A is 1, so we include [A] to the power of 1, which we can write as [A]^1 or simply as [A]. The order of the reaction with respect to B is 3, so we include [B] to the power of 3, which we can write as [B]^3.

The overall rate law is : k[A][B]^3

Thus, the rate law for the given reaction is k[A][B]^3.

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The pressure of the gas when the volume is 1.76 L and the temperature is 308 K is approximately 1.77 atm using the combined gas laws.

To calculate the pressure of an ideal gas using the combined gas law, we can use the formula:

P1 * V1 / T1 = P2 * V2 / T2

Where:

P1 = Initial pressure of the gas (1.02 atm)

V1 = Initial volume of the gas (2.32 L)

T1 = Initial temperature of the gas (285 K)

P2 = Final pressure of the gas (to be determined)

V2 = Final volume of the gas (1.76 L)

T2 = Final temperature of the gas (308 K)

Plugging in the values:

(1.02 atm) * (2.32 L) / (285 K) = P2 * (1.76 L) / (308 K)

Now, solve for P2:

P2 = (1.02 atm) * (2.32 L) * (308 K) / (1.76 L) / (285 K)

= 1.77 atm

Therefore, the pressure of the gas when the volume is 1.76 L and the temperature is 308 K is approximately 1.77 atm.

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Answer: If a pinch of mercury is consumed you will die from it so using a mercury thermometer is very unsafe if its breaks and a child consumes that mercury.

Explanation:

Don't use one

If the glass breaks and the mercury is not properly cleaned up, the little silvery ball within a mercury thermometer might be hazardous. As the mercury evaporates, it may pollute the air around and turn dangerous to both people and animals.

Children that have been exposed to mercury have lower IQs, hearing impairments, and worse coordination.

Long-term exposure worsens and exacerbates symptoms, which may lead to personality changes or even coma.

Mercury thermometers can be replaced by a number of things:

These non-mercury fever thermometers are significantly safer and equally accurate as mercury thermometers.

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Answer:

ELECTROMAGNETIC SPECTRUM. Radios, televisions, mobile phones, and radar use signals made up of electromagnetic waves. These are waves that carry energy as electricity and magnetism at the speed of light. Light we can see is also an electromagnetic wave, but other types of electromagnetic wave are invisible.

Explanation:

~Hope this helps

Answer:

Taste (gustation) and smell (olfaction) are called chemical senses because both have sensory receptors that respond to molecules in the food we eat or in the air we breathe. There is a pronounced interaction between our chemical senses.

Explanation:

hope this helps

Answer:

gas and liquid

Explanation:

Answer:

By using a filter.

Explanation:

The American Heart Association recommends that linoleic acid should make up 5-10% of total daily calories.

linoleic acid is an essential omega-6 fatty acid that the body cannot produce on its own and must be obtained through the diet. It plays a crucial role in maintaining overall health, particularly in relation to heart health.

The American Heart Association (AHA) recommends that linoleic acid should make up 5-10% of total daily calories. This recommendation is based on the beneficial effects of linoleic acid on heart health. Studies have shown that consuming an adequate amount of linoleic acid can help lower the risk of cardiovascular diseases.

Linoleic acid is found in various plant-based oils, such as soybean oil, sunflower oil, and corn oil. These oils can be used in cooking or as dressings for salads and other dishes.

It is important to note that while linoleic acid is beneficial, the overall balance of fatty acids in the diet is also crucial for optimal health. It is recommended to consume a variety of healthy fats, including omega-3 fatty acids, in addition to linoleic acid.

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The atomic mass X of Kr in the fission reaction is: C. 89u.

To determine the atomic mass X of Kr in the fission reaction 235U + n -> 144Ba + Kr + 3n, we need to apply the principle of conservation of mass:

Initial mass = Final mass

Initially, we have 235U and one neutron (n), so the total initial mass is 235 + 1 = 236 units. After the fission reaction, we have 144Ba, Kr, and 3 neutrons. The total final mass should also be 236 units.

Now, let's find the atomic mass X of Kr:
Initial mass = 144Ba + Kr + 3n
236 = 144 + X + 3

Since the mass of a neutron (n) is 1 unit, the equation becomes:
236 = 144 + X + 3

Solve for X:
X = 236 - 144 - 3
X = 89

Therefore, the atomic mass X of Kr in the fission reaction is 89 units, making the correct answer (c) 89 u.

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