Describe the range of radii of most atoms in nanometers (nm)

Answers

Answer 1
Answer:

Final answer:

The range of radii of most atoms is typically in the nanometer scale (nm) and can be measured using the covalent radius. The size of an atom's nucleus is much smaller than the atom itself. The Bohr model provides a formula to calculate the radius of hydrogen-like atoms.

Explanation:

The range of radii of most atoms is typically in the nanometer scale (nm). The covalent radius, which is defined as half the distance between the nuclei of two identical atoms when they are joined by a covalent bond, provides a practical way to measure the size of atoms. As we move down a group in the periodic table, the covalent radius generally increases, indicating a larger size of the atom. For example, the covalent radius of the halogens increases as we move from fluorine to iodine.

The size of an atom's nucleus, on the other hand, is much smaller than the atom itself. The nucleus has a diameter of about 10-15 meters, while the typical atom has a diameter of the order of 10-10 meters. This difference in size illustrates the emptiness of atoms, with the distance from the nucleus to the electrons being typically 100,000 times the size of the nucleus.

The Bohr model provides a formula to calculate the radius of hydrogen-like atoms, which depends on the principal quantum number (n) and the atomic number (Z). The calculated radii of the orbits of the hydrogen atom have been experimentally verified to have a diameter of a hydrogen atom.

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Answer 2
Answer:

Final answer:

The range of radii of most atoms is typically measured in nanometers (nm). Covalent radius and hydrogen-like orbits are two methods used to estimate the size of atoms. The size of an atom can vary depending on the element and measurement technique, but most atoms have radii on the order of nanometers (nm).

Explanation:

The range of radii of most atoms is typically measured in nanometers (nm). The size of an atom can be estimated using various techniques. One commonly used measure is the covalent radius, which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond. The covalent radii of different elements can be found in tables and can vary depending on the element and its position in the periodic table.

Another way to estimate the size of atoms is by looking at the sizes of their orbits in hydrogen-like atoms. These orbits are given in terms of their radii by a mathematical expression that includes a constant called the Bohr radius, which is approximately 5.292 × 10-11 m.

Overall, the size of an atom can vary depending on the element and the specific measurement technique used, but most atoms have radii on the order of nanometers (nm).

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What is the molarity of a solution prepared from 25.0 grams of methanol (CH3OH, density = 0.792 g/mL) with 100.0 milliliters of ethanol (CH3CH2OH)? Assume the volumes are additive.

Some people argue that a policy of putting out wildfires is having a negative long-term effect on the number and strength of wildfires. Suggest why this could be the case.

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Answer:

The reason for this is that putting out the fires only postpones the fire outbreak to a later date, and there is the fear of the fire outbreak being more sever when it actually comes.

When wildfires burn, they clean up the forest off dead trees and falling logs. Also dried leaves and twigs and unnecessarily dense vegetation is cleared up by the fire. These materials are the main fuel of these wildfires. Putting out these fires, especially those that start naturally means that these fuel that should be cleaned up are allowed to accumulate so that when the fire actually happens, it does so with an unnatural intensity. Also, when fire burns, the ashes that are left act as nutrition for the forest, and the forest is allowed to regrow; reborn from the ashes. The only cases that might need human intervention is when the fire is human caused or due to human activities. Natural causes of fire can be due to a very high temperature, lightning striking a tree, etc.

In a 66.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.290. What is the mass of each component? g

Answers

Answer:

23.84g CH30H

32.81g H2O

Explanation:

We will be using the definition of mole fraction to determine the relationship between the number of moles of methanol,

CH3OH , and the number of moles of water.

But mole fraction gives the ratio between the number of moles of a component i of a solution to the total number of moles present in that solution.

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

Final answer:

In a 66.0g aqueous solution of methanol with a mole fraction of 0.290, the mass of the methanol is approximately 19.14g and the mass of the water is approximately 46.86g.

Explanation:

In this aqueous solution of methanol (CH4O), we know that its mass is 66.0g and the mole fraction of methanol is 0.290. The mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in the solution.

In order to find the mass of each component, namely the methanol and the water, we first need to establish that if the mole fraction of methanol is 0.290, the mole fraction of water must be 0.710 (because the total of all mole fractions in a solution is always equal to 1).

We then can set up the following proportion: mass of methanol/mass of water = mole fraction of methanol/mole fraction of water. After solving this equation, the mass of methanol will be approximately 19.14g and the mass of the water will be approximately 46.86g.

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How do you fight off ADHD medication

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Answer:A medication break can ease side effects. A lack of appetite, weight loss, sleep troubles, headaches, and stomach pain are common side effects of ADHD medication.

Explanation: It may boost your child’s growth. Some ADHD medications can slow a child’s growth in height, especially during the first 2 years of taking it. While height delays are temporary and kids typically catch up later, going off medication may lead to fewer growth delays.

It won’t hurt your child. Taking a child off ADHD medication may cause their ADHD symptoms to reappear. But it won’t make them sick or cause other side effects.

When a solid compound is formed from chemicals that are in solution, it is called aprecipitate
condensate
crystal

Answers

Answer:

Precipitate

Explanation:

Please answer this question

Answers

Answer:

I'm not an expert at this, but I assume its mercury.

Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V

An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:

Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)

Calculate the cell potential under these nonstandard concentrations.

Express the cell potential to two decimal places and include the appropriate units.

Answers

Answer:

Cell potential under non standard concentration is 4.09 v

Explanation:

To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions:  1 M , 1 atm).

Nernst equation at 25ºC is:

E = E^(0) - [((0.0592)/(n) · log Q)]

where

E: Cell potential (non standard conditions)

E^(0) = Cell potential (standard conditions)

n: Number of electrons transfered in the redox reaction

Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)

For example, consider the following general chemical reaction:

aA + bB --> cC + dD

where

a, b, c, d: coefficient of balanced chemical reaction

A,B,C,D: chemical compounds in the reaction.

Using the previous general reaction, expression of Q is:

Q = (C^(c) * D^(d) )/(A^(a)*B^(b)  )

Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: E^(0), n and Q

Let´s  calcule potential in nomal conditions  (E^(0)):

1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:

Half-reactions:                               Eo (v):

(VO_(2))^(2+) + e- --> (VO_(2))^(+)                -0.23

Ni --> Ni^(2+)  + 2 e-                   +0.99    

Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:

Half-reactions:                                                                Eo (v):

2 * [ (VO_(2))^(2+) + e- --> (VO_(2))^(+)]                                          -0.23

1 * [Ni --> Ni^(2+)  + 2 e-     ]                                         +0.99  

-------------------------------------------------------------------         -------------

2 (VO_(2))^(2+) + 2e- + Ni  --> 2(VO_(2))^(+) + Ni^(2+) + 2e-              0.76 v  

               

Then global balanced chemical reaction is:

2 (VO_(2))^(2+) + Ni  --> 2(VO_(2))^(+) + Ni^(2+)      

 and the potential in nomal conditions is:

E^(0) = 0.76 v

Also from the balanced reaction, we got number of electons transfered:

n = 2

2. Calculate Q:

Now using previous information, we can establish Q expression and we can calculate its value:

Q = ([(VO_(2)+]^(2)* [Ni^(2+)   )/([VO_(2+)]^(2) * Ni )]

From the exercise we know:

[VO_(2) ^(2+)] = 2.5 M

[VO_(2)+] = 0.083 M

[Ni^(2+)] = 2.5 M

[Ni] = 1 M, it is a pure solid, so its activity in Q is unit (1). It is also applied for pure liquids.

Q = ((2.5)^(2)* 2.5 )/((0.083)^(2) * 1) = 2,268.11

3. Use Nernst equation:

Finally, we replace all these results in the Nernst equation:

E = E^(0) - ((0.0592)/(n) - log Q)\n  \nE = 0.76 - ((0.0592)/(2)-log (2,268.11) \nE = 0.76 - (0.0296 - 3.36)\nE = 4.09 v

Cell potential under non standard concentration is 4.09 v

Final answer:

To calculate the cell potential under nonstandard conditions, we need to apply the Nernst Equation. This involves finding the reaction quotient (Q) from the given concentrations and then subtracting a value derived from Q and the number of electrons transferred, from the cell potential under standard conditions.

Explanation:

For calculating the cell potential under nonstandard conditions for an electrochemical cell, we need to use the Nernst equation. In this case, the Nernst Equation is Ecell = E∘cell - (0.0592/n) * logQ, where Q, the reaction quotient, is the ratio of the concentrations of the products to the reactants raised to their stoichiometric coefficients.

Given the half-cell reduction potentials, we can calculate the cell potential under standard conditions (E°cell) by subtracting the potential of the anode from the potential of the cathode (E°cell = Ecathode - Eanode = 0.99V - (-0.23V), resulting in E°cell = 1.22V.

Next, Q = [Ni2+]/([VO2+]×[H+]²), substituting the given concentrations, Q = (2.5)/(0.083×1.1²).

After calculating Q, we substitute all known values into the Nernst Equation and solve for Ecell. Hence, the cell potential under these nonstandard conditions can be calculated.

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