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3 years ago

In relation to chemical solutions, which term does not express a degree of saturation?

Chemistry

2 answers:

denis-greek [22]3 years ago

7 0

D: oversaturated is correct option.

pav-90 [236]3 years ago

3 0

The best and most correct answer among the choices provided by the question is the fourth choice "oversaturated"

Degree of Saturation is the ratio of the humidity ratio of moist air - to the humidity ratio of saturated moist air at the same temperature and pressure.

I hope my answer has come to your help. God bless and have a nice day ahead!

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3 years ago

Consider the following reaction:

zzz [600]

Answer:

Question 1: Both the rate of reaction of H₂O₂ and the rate of formation of H₂O are:

$6.6\cdot 10^{-3}mol/(liter.s)$

Question 2: 0.60 moles of O₂ are formed in the first 50 s of reaction.

Explanation:

The coefficients of the balanced chemical equation tells the relation of formation of the reactants into the products, which is also the relation between their rates of reaction.

The balanced chemical equation (given) is:

$2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)$

Then, in the same time that 2 moles of H₂O₂ react, 2 moles of H₂O and 1 mol of O₂ are formed.

1. Question 1.

That means that the rate of formation of O₂ is half the rate of reaction of H₂O₂ and half the rate of formation of H₂O.

Hence, if the instantaneous rate of formation of O₂ is 3.3×10⁻³moles/(liters×seconds), then the rate of reaction of H₂O₂ and the rate of formation of H₂O is twice:

$2\times 3.3\cdot 10^{-3}mol/(liter.s)=6.6\cdot 10^{-3}mol/(liter.s)$

2. Question 2.

First you must find how much the concentration has changed.

You can read the initial concentration of H₂O₂ on the graph. It is the y-intercept. It is 1.0M.

You can read the approixmate concentration of H₂O₂ at the time 50 s. It is about 0.2M.

Then the change in concentration in the first 50 s of reaction is about 1.0M - 0.2M = 0.8M.

Now you can convert the concentration 0.8M into number of moles, using the volume (1.5 liters), assuming the liquid volume does not change, and the molarity definition (molarity is the number of moles of solute in one liter of solution)

$\Delta [H_2O_2]=\Delta moles/Volume(liters)$

$0.8M=\Delta moles/1.5liter\\\\\Delta moles=0.8M\times 1.5liter=1.2mol$

Hence, 1.2 mol of H₂O₂ have reacted in the first 50s of reaction.

From the coefficients of the chemical equation, the number of moles of O₂ formed is half the number of moles of H₂O₂ that react.

Thus, 1.2mol/2 = 0.60 moles of O₂ are formed in the first 50 s of reaction.

7 0

3 years ago

Chloroacetic acid, ClCH2COOH, has a pKa of 2.87. What are [H3O+], pH, [ClCH2COO−], and [ClCH2COOH] in 1.55 M ClCH2COOH?

Mademuasel [1]

Answer:

See explanation below

Explanation:

first to all, this is an acid base reaction where the chloroacetic acid is being dissociated in water. Therefore, is an equilibrium reaction.

ClCH₂COOH has a pKa of 2.87 so, the Ka would be:

pKa = -logKa ---> Ka = antlog(-pka)

Ka = antlog(-2.87)

Ka = 1.35x10⁻³

Now that we know the Ka, we need to write the chemical reaction and then, an ICE chart:

ClCH₂COOH + H₂O <----------> ClCH₂COO⁻ + H₃O⁺ Ka = 1.35x10⁻³

i) 1.55 0 0

c) -x +x +x

e) 1.55-x x x

Writting now the equilibrium reaction:

Ka = [H₃O⁺] [ClCH₂COO⁻] / [ClCH₂COOH]

Replacing the values of the chart:

1.35x10⁻³ = x² / 1.55-x

1.35x10⁻³(1.55-x) = x²

2.0925x10⁻³ - 1.35x10⁻³x = x²

x² + 1.35x10⁻³x - 2.0925x10⁻³ = 0 --> a = 1; b = 1.35x10⁻³x; c = 2.0925x10⁻³

From here we use the general equation for solve x in a quadratic equation which is:

x = -b±√(b² - 4ac) / 2a

Replacing the values we have:

x = -1.35x10⁻³ ±√(1.35x10⁻³)² - 4*1*(-2.0925x10⁻³) / 2

x = -1.35x10⁻³ ±√(8.37x10⁻³) / 2

x = -1.35x10⁻³ ± 0.091 / 2

x1 = -1.35x10⁻³ + 0.091 / 2 = 0.045 M

x2 = -1.35x10⁻³ - 0.091 / 2 = -0.046 M

In this case, we will take the positive value of x, in this case, x1.

With this value, the equilibrium concentrations are the following:

[H₃O⁺] = [ClCH₂COO⁻] = 0.045 M

[ClCH₂COOH] = 1.55 - 0.045 = 1.505 M

Finally the pH:

pH = -log[H₃O⁺]

pH = -log(0.045)

pH = 1.35

4 0

3 years ago