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2 years ago

Given the following equation: 2C6H6 + 1502 - 12CO2 + 6H20

Chemistry

1 answer:

Margarita [4]2 years ago

5 0

Answer:

a) 30 moles

Explanation:

2C6H6 + 1502 -------> 12CO2 + 6H20

from reaction 2 mol 15 mol

given 4.0 mol x mol

x = 4.0*15/2 = 30. mol

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2.4 × 10⁻² is the pH of a solution

What is the short definition of pH?

The pH scale determines how acidic or basic water is. The range is 0 to 14, with 7 representing neutrality. Acidity is indicated by pH values below 7, whereas baseness is shown by pH values above 7.
In reality, pH is a measurement of the proportion of free hydrogen and hydroxyl ions in water.

Equation representing the i onization of Na₃PO₄ is as fallow

Na₃PO₄ → PO₄³⁻ + 3Na⁺

Concentration of Na₃PO₄ = $\frac{26.9 * \frac{1 mol}{164 g} }{1 L} = 0.164 M$

Since 1 mol of Na₃PO₄ forms 1 mol of PO₄³⁻ [ PO₄³⁻ ] = [ Na₃PO₄ ] = 0.164 M

Equation representing the i onization of HPO²⁻₄ is as fallow

HPO²⁻₄ → PO₄³⁻ + H⁺

Acidionization constant Ka₃ of HPO²⁻₄ = 4. 17 × 10⁻¹³

Baseionization constant Kb₃ of PO₄³⁻ = Ka₃/ Kb₃

= 1 .0 × 10⁻¹⁴/ 4. 17 × 10⁻¹³ = 2.4 × 10⁻²

Learn more about pH

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1 year ago

What type of plant is thought to be the first plant?

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Angiosperms can produce flowers and fruits to aid in seed dispersal.

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2 years ago

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Two atoms of the same substance wi the different numbers of neutrons are called what

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They are called ISOTOPES.
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2 years ago

What elements does lithium resemble (in terms of electrons) when it loses 1 electron?

tigry1 [53]

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2 years ago

Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8). 2. What is the pH of a solution containing a hydrogen ion concentrat

Pepsi [2]

Answer:

$\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}$

Explanation:

1. Solubility of CaF_2

(a) Molar solubility

CaF₂ ⇌ Ca²⁺ + 2F⁻

$K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}$

(b) Mass solubility

$\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF$_{2}$ is $\large \boxed{\textbf{0.17 g/L}}$}$

2. pH

pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52

3. Oxidizing and reducing agents

Zn + Cl₂ ⟶ ZnCl₂

$\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}$