How do we detect the change in energy?

Chemistry

1 answer:

IrinaVladis [17]3 years ago

3 0

It’s properties unless we can look at the pattern in which it changes matter.
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atroni [7]

Answer:

(1) atomic numbers

Explanation:

The observed regularities in the properties of the elements on the periodic table are periodic functions of their atomic numbers.

Atomic number is the number of protons in an atom.
The periodic law states that "the properties of elements are a periodic function of their atomic number".
Elements on the periodic table are arranged based on the atomic numbers they contain.
The number of positively charged particles in an atom is the atomic number.

6 0

3 years ago

What is the mass of 4.85 X 10^22 atoms of iron (Fe)?

ludmilkaskok [199]

Answer:

Mass = 4.5 g

Explanation:

Given data:

Number of atoms of iron = 4.85 ×10²² atoms

Mass of iron = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

4.85 × 10²² atoms of iron / 6.022 × 10²³ atoms × 1 mol

0.081 mol

Mass of iron:

Mass = number of moles × molar mass

Mass = 0.081 mol × 55.85 g/mol

Mass = 4.5 g

8 0

3 years ago

Which of the three has the largest ei1? which of the three has the largest ? 1s22s22p63s23p64s1 1s22s22p63s23p5 1s22s22p63s23p1?

coldgirl [10]

$E_i1$ will be largest for $1s^22s^22p^63s^23p^5$ .

Explanation: Ionization energy is the energy to knock off an electron from a gaseous atom of ion. First ionization energy or $E_i1$ is the energy required to remove 1 loosely held electron from 1 mole of gaseous atoms to produce 1 mole of gaseous ion carrying (+)1 charge.

$M(g)\rightarrow M^+(g)+e^-$

The electrons are filled according to Aufbau's rule and the orbitals which are strongly held to the nucleus follows the order $s>p>d>f$ .

Electron is released from the outermost shell that is from the electrons which are loosely held to the nucleus, this follows the pattern $s$ .

In configurations, $1s^22s^22p^63s^23p^64s^1,1s^22s^22p^63s^23p^5\text{ and } 1s^22s^22p^63s^23p^1$

The loosely held orbital is 4s, therefore electron will be lost from that easily.

Now, in 3p orbital, one configuration has 5 electrons and one has 1 electron.

The configuration having 5 electrons will be more tightly held by the nucleus because it has more electrons that the one having only 1 electron. Hence, the electron will be lost easily from the configuration having $3p^1$ as the valence shell.