Chemistry, 29.06.2019 17:00 Weser17

Alkali metals react with water to produces what product?

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Answer from: only1cache

Alkali metals react with water to produce hydroxide and hydrogen gas.

Explanation:

That is a chemical property of alkali metals: they are highly reactive and react vigorously with water to produce the correspondant hydroxide and hydrogen gas.

These are some of those reactions:

2Li (s) + 2H₂O (l) → 2LiOH (aq) + H₂(g) ↑2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂(g) ↑2K (s) + 2H₂O (l) → 2KOH (aq) + H₂(g) ↑

The alkali metals are the elements of group 1 of the periodic table: Li, Na, K, Rb, Cs, and Fr. They have one valence electron which may lose easily to form an ion with charge +1.

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

a chemical reaction

explanation:

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