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3 years ago

Cracking the periodic table code why aren't the elements listed in alphabetical order answer key

2 answers:

8 0

<span>They could be arranged in alphabetical order but that wouldn't tell you much about their chemistry. In the periodic table, the elements are arranged according to their atomic number. Notice how the atomic number increases as you move across the table. Normally, the higher the atomic number, the heavier the element. So lead (Pb), atomic number 82 is heavier than tin (Sn) atomic number 50.</span>

finlep [7]3 years ago

4 0

The elements in the periodice table are not listed in alphabetical order, because the arragement in rows (periods) and columns (groups or familes), in increasing order of atomic number (number of protons of the atoms) permits to explain similarities among the elements, trend in some properties, and even predict properties of unknown elements.

For example, the elements of the first group (family), called alkaline metals, all have 1 valence electron, have similar physical properties (ductibility, malleability, luster, thermal and electricity conductivity), react in similar way with water, show a trend in the atomic radii and in the ionization energy.

You can tell similar stories for other groups like, alkalyne earth metals, halogens and noble gases.

You can also tell trends in electroneativities, and atomic radii, for a row of elements, as per the order they are in the row.

So, the current array resulted very helpul for chemists to explain and predict the behavior and properties of the elements.

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What particle is J.J Thomson credited with discovering

Ilya [14]

He was credited with discovering the subatomic particle also known as the electron in 1897.

6 0

3 years ago

Methane reacts with chlorine in the presence of ultraviolet light

earnstyle [38]

Answer:

When a mixture of methane and chlorine is exposed to ultraviolet light - typically sunlight - a substitution reaction occurs and the organic product is chloromethane. CH 4 + Cl 2 → CH 3 Cl + HCl However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms.

Explanation:

hope it help i try best

7 0

3 years ago

What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?

Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is $O_{2}$ molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

First we have to calculate the moles of methane.

$\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}}$ = $\frac{23g}{16.04g/mole}$ = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give $\frac{2moles\times 1.434moles}{1moles}$ moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is $O_{2}$ molecule.

6 0

3 years ago

The change in entropy, Δ S ∘ rxn , is related to the the change in the number of moles of gas molecules, Δ n gas . Determine the

rewona [7]

Answer:

The entropy decreases.

Explanation:

The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where

Δngas = n(gaseous products) - n(gaseous reactants)

If Δngas > 0, the entropy increases
If Δngas < 0, the entropy decreases.
If Δngas = 0, there is little or no change in the entropy.

Let's consider the following reaction.

2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)

Δngas = 0 - 3 = -3, so the entropy decreases.

8 0

3 years ago

Find the potentials of the following electrochemical cell:

melomori [17]

Answer: 0.18 V

Explanation:-

$Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni$

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0_(Cd^{2+}/Cd)$ =-0.40V[/tex]

$E^0_(Ni^{2+}/Ni)$ =-0.24V[/tex]

$Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni$

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}$

$E^0=-0.24-(-0.40)=0.16V$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential = 0.16 V

$E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}$

$E_{cell}=0.18V$

Thus the potential of the following electrochemical cell is 0.18 V.

6 0

3 years ago