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MrRissso [65]
3 years ago
13

Write a balanced equation to show the reaction of sulfurous acid with caesium hydroxide to form water and caesium sulfite.

Chemistry
1 answer:
Soloha48 [4]3 years ago
5 0

Answer:

H2SO3 + 2CsOH —> Cs2SO3 + 2H2O

Explanation:

When sulfurous acid react with caesium hydroxide, caesium sulfite and water are formed according to the equation:

H2SO3 + CsOH —> Cs2SO3 + H2O

Next, we balanced the equation by putting 2 in front of CsOH and 2 in front of H2O i.e

H2SO3 + 2CsOH —> Cs2SO3 + 2H2O

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According to the Octet Rule, an element is most stable with ______________ valence electrons
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What chemical process occurs in cells to release energy?
Lana71 [14]

Answer:

B. Respiration

Explanation:

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3 years ago
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
     NaA → Na+ + A- 
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
     Kb = [HA][OH-] / [A-] 
           = x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
     Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
7 0
3 years ago
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