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AveGali [126]
3 years ago
15

For this heterogeneous system 2 A ( aq ) + 3 B ( g ) + C ( l ) − ⇀ ↽ − 2 D ( s ) + 3 E ( g ) 2A(aq)+3B(g)+C(l)↽−−⇀2D(s)+3E(g) th

e concentrations and pressures at equilibrium are [ A ] = 5.66 × 10 − 2 M [A]=5.66×10−2 M , P B = 5.69 × 10 3 Pa PB=5.69×103 Pa , [ C ] = 9.14 M [C]=9.14 M , [ D ] = 15.78 M [D]=15.78 M , and P E = 9.33 × 10 4 torr PE=9.33×104 torr . Calculate the thermodynamic equilibrium constant, K K .

Chemistry
1 answer:
Archy [21]3 years ago
4 0

Answer:

K = 1.08 X 10¹²

Explanation:

The solved solution is in the attach document

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Question 20 (1 point)
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Answer:

A. The ratio of the element in glucose is different than in ethanol

Explanation:

The compounds given are;

      Glucose      C₆H₁₂O₆

     Ethanol         C₂H₆O

The two compounds given are different from one another. But they have the same elemental composition.

   Ethanol is made up of Carbon, hydrogen and Oxygen and so also, Glucose

Now,

  The number of atoms in both compounds are different;

                                   Number of atoms

                        C                             H                     O

Glucose           6                             12                      6

Ethanol            2                              6                       1

8 0
3 years ago
Which element has the highest 2nd ionization energy? <br> A. Be <br> B. C<br> C. Li<br> D. B
dlinn [17]

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The answer is c.........

6 0
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What are the products of photosynthesis?
kakasveta [241]

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2 years ago
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The chemical formula for disulfur monoxide is:<br> A. S2O<br> B. 2SO<br> C. SO2<br> D. S2O2
Delvig [45]
The answer is (A).
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3 0
2 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
2 years ago
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