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2 years ago

A. describe where to look on the periodic table to find Elements which have similar reactivity and other properties.

2 answers:

7 0

A.
Elements in the same group have similar properties.

B.
The similarity in their properties arises from the fact that they have an equal number of valence shell electrons.

C.
Fluorine, Chlorine, Bromine

Lana71 [14]2 years ago

6 0

The periodic table is the arrangement of elements according to their atomic numbers [modern periodic law]

A) The elements which have similar properties are placed in same group.

The periodic table is arranged in 18 different groups based on the elements with similar properties

B) the elements of a group have similar properties due to similar outer electronic configuration or same valence electrons.

For example

a) Alkali metals: they all have one electron in their outermost shell or they have one valence electron. due to this they can easily form cations by losing one electron and gain stable noble gas configuration.

Lithium, sodium and potassium

b) noble gases: they have completely filled outer shell or valence shell. So they are not much reactive and do not form compounds or molecules easily

Helium, neon, argon

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Which does not affect chemical equilibrium

Amanda [17]

An inert gas will not react with either the reactants or the products, so it will have no effect on the product/reactant ratio, and therefore, it will have no effect on equilibrium.

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2 years ago

Read 2 more answers

Can someone help Plz

Marrrta [24]

Answer: A. 1.60 liters

Explanation:

$2Na_2O_2+2CO_2\rightarrow 2Na_2CO_3+O_2$

As can be seen from the balanced chemical equation, 2 moles of $CO_2$ produce 1 mole of $O_2$ .

According to Avogadro's law, every 1 mole of the gas occupies 22.4 liters at STP.

Thus $2$ moles of $CO_2$ occupies $22.4\times 2=44.8L$ at STP and produce 1 mole of $O_2$ i.e. 22.4 L

44.8 L of $CO_2$ will produce 22.4 L of $O_2$

Thus $3.20L$ of $CO_2$ will produce $\frac{22.4}{44.8}\times 3.20=1.60L$

6 0

2 years ago

Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th

artcher [175]

Answer:

The equilibrium constant is:

$k_c=0.0030M^{-2}$

Explanation:

The correct equation is:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

$k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}$

Substituting:

$k_c=\dfrac{(0.105M)^2}{(1.1M)\cdot (1.50M)^3}$

$k_c=0.0030M^{-2}$

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3 years ago

Why there are only seven crystal systems??? why not other number

Anastasy [175]

In crystallography, the terms crystal system, crystal family, and lattice system each refer to one ... In total there are seven crystal systems: triclinic, monoclinic, orthorhombic, ..... A polar crystal possess a "unique" axis (found in no other directions) such ... lattices in three dimensions; each can apply in one lattice system only.

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3 years ago

Determine the LIMITING reactant in the following balanced equation:

padilas [110]

Answer:

KBr is limiting reactant.

Explanation:

Given data:

Mass of KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂ → 2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

Number of moles = 4 g/ 119 gmol

Number of moles = 0.03 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 6 g/ 70 gmol

Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

KBr : KCl

2 : 2

0.03 : 0.03

KBr : Br₂

2 : 1

0.03 : 1/2×0.03= 0.015

Cl₂ : KCl

1 : 2

0.09 : 2/1×0.09 = 0.18

Cl₂ : Br₂

1 : 1

0.09 : 0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.

5 0

2 years ago