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2 years ago

10

Which elements have the greatest tendency to behave as oxidizing agents?

1 answer:

son4ous [18]2 years ago

8 0

The force acting between two charged particles A and B is 5.2 x 105 newtons. Charges A and B are 2.4 x 102 meters apart. If the charge on particle A is 7.2 x 108 coulombs, what is the charge of particle B? (k 7 9.0 x 109 newton meters?/coulomb)

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2 years ago

Does water added to a flask containing chlorine gas and solid sodium react in the chemical reaction to make sodium chloride

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Explanation:

Chemically, we can have a reaction between chlorine gas and solid sodium

This reaction is actually explosive and would produce fine powder of sodium chloride

We have the reaction as follows:

$2Na_{(s)}+Cl_{2(g)}\rightarrow2NaCl_{(s)}$

Essentally, what we can deduce from here is that we do not need to add water to the flask. Except for the reason that we would want the sodium chloride solid in the solution form, there is absolutely no reason to add water to the flask as the reaction would proceed normally

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1 year ago

What is formed by the hydrolysis of Starch?

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3 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

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2 years ago