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3 years ago

why is it much easier for group 14 elements to become stable by sharing instead of transferring electrons9

2 answers:

5 0

Because they are closer to the farther end of the periodic table. Since they are closer to the farther end they don't want to give away their electrons because it would be easier for them to just steal them from other atoms.

Anvisha [2.4K]3 years ago

5 0

Answer:

Because of their intermediate electronegativity

Explanation:

First of all, any element of the periodic table interacts with other atoms to complete the octet in its last atomic level and achive stability, either losing or gaining electrons.

Group 14 elements have 4 electrons in the last level, so it would seem that for they its the same to lose or gain electrons. But actually its neither, for them is easier to share the electrons with other atoms.

This is because they don't have a very high electronegativity (making them able to steal electrons) nor a very low electronegtivity (making them anle to release electrons).

This intermediate electronegativity is the responsable of the sharing of electrons.

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vitfil [10]

Answer:

300K.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 300L

Initial temperature (T1) = 200K

Final volume (V2) = 450L

Final temperature (T2) =..?

Since the pressure is constant, the gas is obeying Charles' law.

Using the Charles' law equation, we can obtain the new temperature of the gas as follow:

V1/T1 = V2/T2

300/200 = 450/T2

Cross multiply to express in linear form

300 x T2 = 200 x 450

Divide both side by 300

T2 = (200 x 450)/ 300

T2 = 300K

Therefore, the new temperature of the gas is 300K.

6 0

3 years ago

Water is a solute. solvent.

Finger [1]

Answer:

solvent

Explanation:

3 0

3 years ago

Given the reaction:

dezoksy [38]

Answer:

V = 22.34 L

Explanation:

Given data:

Volume of O₂ needed = ?

Temperature and pressure = standard

Number of molecules of water produced = 6.0× 10²³

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of water:

1 mole contain 6.022× 10²³ molecules

6.0× 10²³ molecules × 1 mole / 6.022× 10²³ molecules

0.99 mole

Now we will compare the moles of oxygen and water.

H₂O : O₂

2 : 1

0.996 : 0.996

Volume of oxygen needed:

PV = nRT

V = nRT/P

V = 0.996 mol × 0.0821 atm.L/mol.K × 273.15 K / 1 atm

V = 22.34 L

3 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$