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Tema [17]
3 years ago
12

Let us write the appropriate equilibria and associate the correction

le="K_b" alt="K_b" align="absmiddle" class="latex-formula"> values. Remember, we will want to calculate the concentrations of all species in a 0.390 M Na ₂SO₃ (sodium sulfite) solution. The ionization constants for sulfurous acid are K_a_1 = 1.4 × 10⁻² and K_a_2 = 6.3 × 10⁻⁸.
Chemistry
1 answer:
rosijanka [135]3 years ago
3 0

Explanation:

The relation between K_a\&K_b is given by :

K_w=K_a\times K_b

Where :

K_w=1\times 10^{-14} = Ionic prodcut of water

The value of the first ionization constant of sodium sulfite = K_{a1}=1.4\times 10^{-2}

The value of K_{b1}:

1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}

K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}

The value of the second ionization constant of sodium sulfite = K_{a2}=6.3\times 10^{-8}

The value of K_{b2}:

1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}

K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}

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The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0
vodomira [7]
According to the reaction equation:

and by using ICE table:

              CN-  + H2O ↔ HCN  + OH- 

initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

               = (1 x 10^-14) / (4.9 x 10^-10)

               = 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013 

∴ POH = -㏒[OH]

            = -㏒0.0013

            = 2.886 

∴ PH = 14 - POH

         = 14 - 2.886 = 11.11
5 0
3 years ago
If a scientist creates a new rock-like substance in a laboratory, why wouldn’t this type of material be classified as a true min
miv72 [106K]
Because it’s not a naturally occurring substance
3 0
3 years ago
Read 2 more answers
U copy and pasted what he said bassically...
OlgaM077 [116]
What it said chemistry and i was looking to answer something
3 0
4 years ago
The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the con
gtnhenbr [62]

Answer:

Product A: cis; no

Product B: cis: no  

Explanation:

Two common methods of forming oxiranes from alkenes are:

  • Reaction with peroxyacids
  • Formation of a halohydrin followed by reaction with base

1. Reaction with peroxyacids

(a) Stereochemistry

The reaction with a peroxyacid is a syn addition, so the product has the same stereochemistry as the alkene.

The starting alkene is cis, so the product is <em>cis</em>-2,3-diethyloxirane.

(b) Configuration

The product is optically inactive because it has an internal plane of symmetry.

It will not rotate the plane of polarized light.

2. Halohydrin formation

(a) Stereochemistry

The halogenation of the alkene proceeds via a cyclic halonium ion.

The backside displacement of halide ion by alkoxide is also stereospecific, so a cis alkene gives a cis epoxide.

The product is <em>cis</em>-2,3-diethyloxirane.

(b) Configuration

The cyclic halonium ion has an internal plane of symmetry, as does the product (meso).

The oxirane will not rotate the plane of polarized light.

 

8 0
4 years ago
Question 2: you have two systems: 1) solid gold and water 2) solid sodium and waterWhich one can spontaneously release energy an
Valentin [98]

Hey there!

Solid Sodium and water will react spontaneously and release energy.  This is based on the reactivity series. Sodium is a highly reactive metal and hence, it is placed at the top of the reactivity series. This is because it loses its outermost electron very readily. When it comes in contact with water, it reacts with it violently to form sodium hydroxide and hydrogen gas. This reaction is exothermic and hence, accompanied with a release of energy.  Gold lies at the bottom of the reactivity series as it is very stable and does not give away its outermost electrons easily. Therefore, when it comes in contact with water, there is no reaction and no release of energy.

7 0
4 years ago
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