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4 years ago

13

What is the hydronium ion concentration of a 0.500 m acetic acid solution with ka = 1.8 × 10-5? the equation for the dissociatio

n of acetic acid is:?

1 answer:

Harlamova29_29 [7]4 years ago

5 0

<span>The equation for the dissociation of acetic acid is:
CH3CO2H(aq) + H2O(l) <---> H3O+(aq) + CH3CO2-(aq)
pH = 0.5(4.74 - lg(0.5)) = 2.52
so
[H^+] = 3.02*10^-3 M = 3.02 mM</span>

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Balance the following reaction. 2Ca3(PO4)2 + C + 6SiO2 → CaSiO3 + P4 + CO

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Answer:

2Ca₃(PO₄)₂ + 10C + 6SiO₂ → 6CaSiO₃ + P₄ + 10CO.

Explanation:

To balance a chemical reaction, we should apply the law of conservation of mass.

Law of conversation of mass states that the no. of atoms is equal in both sides of the chemical reaction.

So, the balanced chemical reaction is:

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3 years ago

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Explain the nuclear model of the atom

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Answer:

Explanation:

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3 years ago

Determine the poh of 0.01 molar solution of carbonic acid

Veronika [31]

Hello!

Data:

Molar Mass of H2CO3 (carbonic acid)

H = 2*1 = 2 amu

C = 1*12 = 12 amu

O = 3*16 = 48 amu

------------------------

Molar Mass of H2CO3 = 2 + 12 + 48 = 62 g/mol

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:

M (molarity) = 0.01 M (Mol/L) → $1*10^{-2}\:M$

Use: Ka (ionization constant) = $4.4*10^{-7}$

$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$

$4.4*10^{-7} = 1*10^{-2}*\alpha^2$

$1*10^{-2}*\alpha^2 = 4.4*10^{-7}$

$\alpha^2 = \dfrac{4.4*10^{-7}}{1*10^{-2}}$

$\alpha^2 = 4.4*10^{-7-(-2)}$

$\alpha^2 = 4.4*10^{-7+2}$

$\alpha^2 = 4.4*10^{-5}$

$\alpha = \sqrt{4.4*10^{-5}}$

$\boxed{\alpha \approx 2.09*10^{-5}}$

Now, we will calculate the amount of Hydronium [H3O+] in carbonic acid (H2CO3), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$

$[ H_{3} O^+] = 1*10^{-2}* 2.09*10^{-5}$

$[ H_{3} O^+] = 2.09*10^{-2-5}$

$\boxed{[ H_{3} O^+] = 2.09*10^{-7}}$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:

$pH = \:?$

$[ H_{3} O^+] = 2.09*10^{-7}$

apply the data to formula

$pH = - log[H_{3} O^+]$

$pH = - log[2.09*10^{-7}]$

$pH = 7 - log\:2.09$

$pH = 7 - 0.32$

$\boxed{pH = 6.68}$

Note:. The pH <7, then we have an acidic solution (weak acid).

Now, let's find pOH by the following formula:

$pH + pOH = 14$

$6.68 + pOH = 14$

$pOH = 14 - 6.68$

$\boxed{\boxed{pOH = 7.32}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

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3 years ago

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Electrochemical cells convert chemical energy into electrical energy or vice versa. Electrolytic cell is a type of electrochemical cell in which electrical energy is converted into chemical energy. ... Electrolytic cells consist of positively charged anode and a negatively charged cathode.

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3 years ago

Using this equation 2H2+O2–>2H2O when 47g of water are produced, how many grams of hydrogen must react

hoa [83]

You must react 5.3 g H_2 to produce 47 g H_2O.

<em>Step 1</em>. Calculate the <em>moles of H_2O</em>

Moles of H_2O = 47 g H_2O × (1 mol H_2O/18.02 g H_2O) =2.61 mol H_2O

<em>Step 2</em>. Calculate the <em>moles of H_2 </em>

Moles of H_2 = 2.61 mol H_2O × (2mol H_2/2 mol H_2O) = 2.61 mol H_2

<em>Step 3</em>. Calculate the <em>mass of H_2</em>

Mass of H_2 = 2.61 mol H_2 × (2.016 g H_2/1 mol H_2) = 5.3 g H_2

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3 years ago