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Katena32 [7]
3 years ago
13

5,500 cm scientific notation

Chemistry
2 answers:
Alex73 [517]3 years ago
7 0

5,500 written as 5.5 x 103 in scientific notation. hope i helped

Vikentia [17]3 years ago
5 0
5.5 x 10^3

Explain:it is 5.5 * 10 with the exponer of 3
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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
4 years ago
Can someone help me balance this equation please?<br> Br2 + S2O32– + H2O → Br1– + SO42– + H+
tamaranim1 [39]

Explanation:

Br2 + S2O32- + 5H2O –> 2Br- + 2SO4 + 10H+ + 6e

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QveST [7]

Answer:

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Explanation:

The elements in Group 18 of the periodic table are called "inert", or noble gases.

These elements have their valence shell (the outermost shell of the atom) full of electrons, so they do not gain/give off electrons, and therefore, they do not react with other elements, so they do not form compounds.

In fact, normally the elements try to gain/give off electrons in order to fullfill their outermost shell (the valence shell). For instance, an atom that has 1 electron only in its valence shell, try to "give away" this electron in order to have its outermost shell completed. On the other hand, an atom which has 7 electrons in its valence shell tries to "gain" one electron in order to fullfill the valence shell.

Noble gases, instead, have already 8 electrons in their valence shell, so their valence shell is already completed, therefore they do not react with other elements, and therefore they are called "inert".

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