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2 years ago

Analysis

Chemistry

1 answer:

Karolina [17]2 years ago

6 0

The properties of substances can be used to put the into groups.

<h3>Grouping of substances</h3>

In chemistry, it is often necessary to put substances into groups based on similarity in their properties. This is what led to the idea of a periodic table of elements.

Similarly, when we have unknown substances, we can group them according to the similarities in their properties.

Learn more about properties of substances: brainly.com/question/19886211

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In the chemical equation 3c2h4 how many atoms of carbon are represented

Scorpion4ik [409]

An example.
water is H2O

2 hydrogen, 1 oxygen

so the number to the right means how much of what is on the left.

so it looks like 2, because C2, but look at the 3 at the beginning. that means
3 (c2h4)

so 6 carbons, 12 hydrogen

the ratio of c2 to h4 doesn't change it's always 1:2.

but the 3 at the front is a different number relating to how much you have

5 0

3 years ago

Laura jala un trineo una distancia de 2000 cm usando una cuerda, con una tensión constante de 80 N. La tarea requiere de 800 J d

Andre45 [30]

Respuesta:

60°

Explicación:

Paso 1: Información dada

Distancia recorrida (d): 2000 cm

Fuerza aplicada (F): 80 N

Trabajo realizado (W): 800 J

Paso 2: Convertir "d" a metros

Usaremos el factor de conversión 1 m = 100 cm.

2000 cm × 1 m/100 cm = 20 m

Paso 3: Calcular el ángulo entre la cuerda y el suelo (θ)

W = F × d × cosθ

800 J = 80 N × 20 m × cosθ

θ = 60°

5 0

3 years ago

When bonded with four hydrogen atoms, carbon has which shape?

jolli1 [7]

When carbon bonds with 4 hydrogen atoms which is methane having sp3 structure so it is tetrahedral. electricity is conducted by electrons so graphite conducts by free and delocalised sea of electrons.

3 0

3 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago

What was John Dalton's contribution to the development of the atomic theory?

Lorico [155]

John Dalton made some hypothesis about the structure of atom. He proposed that matter is composed of great number of indivisible particles called atoms they can neither be destroyed nor be created.

<h3>What is atomic theory?</h3>

There are different theories regarding the structure and electronic properties of an atom. Many scientists contributed to the modern theory of atomic structure in which John Dalton was first to mention the word atom.

According to Dalton' theory, matter is composed of indivisible particles called atoms. Atoms can neither be created nor be destroyed. All the atoms of the same element are identical in all aspects.

Atoms of different elements are different and the compounds are formed by the combination of atoms. Dalton's theory provided a sound basis for the laws of chemical combination and also several properties of gases and liquids known at that time.

However, he could not explain the reason for chemical combination of atoms and did not give any idea about the existence of isotopes and isobars.

Hence, the main aspects of Dalton's theory was the indivisibility of atoms and the chances of chemical combination.

To learn more about Dalton's theory, find the link below:

brainly.com/question/11855975

#SPJ1

6 0

10 months ago