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3 years ago

Helppp meee outtttt

Chemistry

2 answers:

andreev551 [17]3 years ago

6 0

Answer:

The answer is 80 grams

Pls mark as brainliest

MArishka [77]3 years ago

3 0

The answer is B 80 grams

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Determination of the amount of patassium hydroxide in patassium hydroxide/Sodium carbonate mixture

Eva8 [605]

Answer:

do you still need this answer?

Explanation:

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3 years ago

A compound is found to contain 50. 05% sulfur and 49. 95% oxygen by mass. What is the empirical formula for this compound? SO S2

jekas [21]

The empirical formula for the given compound has been $\rm SO_2$ . Thus, option C is correct.

The empirical formula has been the whole unit ratio of the elements in the formula unit.

<h3>Computation for the Empirical formula</h3>

The given mass of Sulfur has been, 50.05 g

The given mass of oxygen has been 49.95 g.

The moles of elements in the sample has been given by:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

Moles of Sulfur:

$\rm Moles\;S=\dfrac{50.05}{32}\\
 Moles\;S=1.56\;mol$

The moles of sulfur in the unit has been 1.56 mol.

Moles of Oxygen:

$\rm Moles\;O=\dfrac{49.95}{16} \\
Moles\;O=3.12\;mol$

The moles of oxygen in the unit has been 3.12 mol.

The empirical formula unit has been given as:

$\rm S_{1.56}O_{3.12}=SO_2$

Thus, the empirical formula for the given compound has been $\rm SO_2$ . Thus, option C is correct.

Learn more about empirical formula, here:

brainly.com/question/11588623

8 0

2 years ago

On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?

enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

$HNO_2\ (aq) \leftrightharpoons H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\ K_a = 4.0\times \ 10^{-4}$

$H^{+}=?$

Formula:

$Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}$

Let

$[H^{+}] = [NO_2^{-}] = x$ at equilibrium

$x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \ M\\\\$

therefore,

$[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M$

Calculating the % ionization:

$= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\$

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2 years ago

Question 1 (1 point)

Sergio [31]

Answer:

Sonar Pulse

Explanation:

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2 years ago

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Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic

Mademuasel [1]

Answer:

1.33%

Explanation:

In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:

CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )

Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:

Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

<em>H₂O is not taken into account in the equilibrium because is a pure liquid</em>

<em />

When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.

The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:

[CH₃CO₂H] = 0.100M - X

[H₃O⁺] = X

[CH₃CO₂⁻] = X

<em>Where X is reaction coordinate.</em>

Replacing in Ka expression:

1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

1.8x10⁻⁵ = [X] [X] / [0.100M - X]

1.8x10⁻⁶ - 1.8x10⁻⁵X = X²

1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.00135 → False solution. There is no negative concentrations.

X = 0.00133 → Right solution.

That means concentration of acetate ion is:

[CH₃CO₂⁻] = 0.00133M.

Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:

% Ionization = 0.00133M / 0.100M × 100 =

<em />

4 0

2 years ago