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2 years ago

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The distance between two adjacent peaks on a wave is called the wavelength. (2pts) a. The wavelength of ultraviolet light is 255

nm. What is the wavelength in meters? b. The wavelength of a beam of red light is 683nm. What is its wavelength in angstroms?

1 answer:

kicyunya [14]2 years ago

6 0

Answer:

a.2.55e-7

b.6830

Explanation:

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Assertion (A): An orbital cannot have more than two electrons, moreover, if an orbital has two electrons they must have opposite

VMariaS [17]

Answer: Its option "A" Both A and R are true and R is the correct explanation of A.

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3 years ago

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A cloud that forms near the ground

stiks02 [169]

Hey there!
Great question=)

Answer:Fog, these are clouds that form near the ground.

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3 years ago

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Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?

maria [59]

Answer:

An alkali metal present in period 2 have larger first ionization energy.

Explanation:

Ionization energy:

The amount of energy required to remove the electron from the atom is called ionization energy.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.

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3 years ago

2. Choose the atom from each of the following pairs with the greater ionization energy: a. Be and Ba b. Al and Ar c. Ca and Cl

beks73 [17]

Considering the definition of ionization energy, the highest ionization energy belongs to the element:

a. Be

b. Ar

c. Cl

Electrons are held in atoms by their attraction to the nucleus, which means that energy is needed to remove an electron from the atom.

You should keep in mind that the electrons of the last layer are always lost, because they are the weakest attracted to the nucleus.

Ionization energy, also called ionization potential, is the necessary energy that must be supplied to a neutral, gaseous, ground-state atom to remove an electron from an atom. When an electron is removed from a neutral atom, a cation with a charge equal to +1 is formed.

In a group, the ionization energy increases upwards because when passing from one element to the bottom, it contains one more layer of electrons. Therefore, the valence layer electrons, being further away from the nucleus, will be less attracted to it and it will cost less energy to pluck them.

In the same period, in general, it increases as you shift to the right. This is because the elements in this way have a tendency to gain electrons and therefore it will cost much more to tear them off than those on the left which, having few electrons in the last layer will cost them much less to lose them.

Considering all the above, from each of the pairs, the highest ionization energy belongs to the element:

a. Be

b. Ar

c. Cl

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6 0

2 years ago

You need to prepare at 2.0 mL sample of a diluted drug for injection. The total amount of the drug to be injected in this 2.0 mL

pav-90 [236]

Answer:

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

Let´s convert it to molarity:

0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l

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3 years ago