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suter [353]
3 years ago
13

The statements in the tables below are about two different chemical equilibria. The symbols have their usual meaning, for exampl

e AG Gibbs free energy of reaction and stands for the equilibrium constant. stands for the standard In each table, there may be one statement that is false because it contradicts the other three statements. If you find a false statement, check the box next to It. Otherwise, check the "no false statements box under the table. statement false? statement false? Ink>0
AH° R<1
AG'>0
AG'>0
In > AH">TAS
no false statements

Chemistry
1 answer:
Talja [164]3 years ago
4 0

Answer:

see explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

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Fill in the blanks with the correct term.
zzz [600]

Answer:

Fill in the blanks with the correct term.

a. a liquid that dissolves another substance.

b. a chemical that is dissolved.

c. a value used to describe the amount of one substance dissolved in another.

d. a liquid consisting of one substance dissolved in another.

Explanation:

a. A liquid that dissolves another substance is called the solvent.

b. A chemical that is dissolved solute.

c. A value used to describe the amount of one substance dissolved in another is called concentration.

d. A liquid consisting of one substance dissolved in another is called a solution.

4 0
3 years ago
Iodine-131 decays with a half-life of 8.02
dlinn [17]
Radioactive material undergoes 1st order decay kinetics.

For 1st order decay, half life = 0.693/k

where k = rate constant

k = 0.693/half life = 0.693/8.02 = 0.0864 day-1

Now, for 1st order reaction,
k = \frac{2.303}{t} X log \frac{initial.conc}{final.conc}

Given: t = 6.01d, initial conc. = 5mg

∴0.0864 = \frac{2.303}{6.01} X log \frac{5}{final.conc}
∴ final conc. = 2.975 mg
3 0
3 years ago
Water vapor is heated and rising into the air is called what?
Lady_Fox [76]

Answer:

Evaporation

Explanation:

3 0
3 years ago
What are the gaps to this
zheka24 [161]
In a liquid, particles are close together but can move in any direction. They won't keep a definite shape like solids do.
7 0
3 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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