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MakcuM [25]
3 years ago
5

Which atom absorbs more energy, one in which an electron moves from the second shell to the third shell or an otherwise identica

l atom in which an electron moves from the first to the third shell?
Chemistry
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

identical atom in which an electron moves from the first to the third shell.

Atoms may occupy different energy states. The energy states are discrete, i.e. they occur at specific values only. Therefore an atom can only move to a new energy level if it absorbs or emits an amount of energy that exactly corresponds to the difference between two energy levels.

The lowest possible energy level that the atom can occupy is called the ground state. This is the energy state that would be considered normal for the atom.

An excited state is an energy level of an atom, ion, or molecule in which an electron is at a higher energy level than its ground state.

An electron is normally in its ground state, the lowest energy state available. After absorbing energy, it may jump from the ground state to a higher energy level, called an excited state.

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A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential
nika2105 [10]

Answer:

3.50*10^-11 mol3 dm-9

Explanation:

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.

Ag2C2O4 -->  2Ag+  +  C2O4 2-

So Ksp = [Ag+]^2 * [C2O42-]

In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.

So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)

= 3.50*10^-11 mol3 dm-9

7 0
3 years ago
In reverse osmosis, water flows out of a salt solution until the osmotic pressure of the solution equals the applied pressure. I
Vsevolod [243]

Answer : The final concentration of the seawater is, 2.909 mole/L

Explanation :

Formula used for osmotic pressure :

\pi=CRT

where,

\pi = osmotic pressure  = 70.0 bar = 70 atm

R = solution constant  = 0.0821 Latm/moleK

T= temperature of solution = 20^oC=273+20=293K

C = final concentration of seawater = ?

Now put all the given values in the above formula, we get the concentration of seawater.

70atm=C\times 0.0821Latm/moleK\times 293K

C=2.909mole/L

Therefore, the final concentration of the seawater is, 2.909 mole/L

4 0
4 years ago
The momentum of an object depends on which two quantities?
Maurinko [17]
The answer is D; Mass and Velocity
6 0
3 years ago
Read 2 more answers
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass %
11Alexandr11 [23.1K]

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

<em>Note: The question is stated more clearly below:</em>

<em>Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.</em>

<em>What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?</em>

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

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