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3 years ago

In an aqueous solution at 25°C, if [H30+] = 3.3 * 10^4 M, then [OH-] is:

Chemistry

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:

[OH-] = 3.0 x 10^-19 M

Explanation:

[H3O+][OH-] = Kw

Kw = 1.0 x 10^-14

[H3O+][OH-] = 1.0 x 10^-14

[OH-] = 1.0 x 10^-14 / 3.3 x 10^4 = 3.0 x 10^-19

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Elabore el balance de la ecuación por oxido reducción enumerando cada paso

Rasek [7]

Answer:

${ \rm{5CaC _{2}O _{4(aq)}+ 2KMnO _{4(aq)} + 8H _{2} SO _{4(aq)} → 5CaSO _{4(s)}+ 2Mn SO _{4(aq)}+ K _{2} SO _{4(s)} + 10CO _{2(g)} + 8H _{2} O _{l} }}$

Explanation:

This is a redox reaction formed by combining two half equations.

Check the image above, sorry if my handwriting is poor :(

8 0

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! =)

3 0

3 years ago

Which of the following is an example of a chemical change?

ad-work [718]

Answer:

burning a paper is chemical change

8 0

3 years ago

What is the freezing point of a solution that contains 36.0 g of glucose ( ) in 500.0 g of water? ( for water is 1.86°c/m. the m

Marizza181 [45]

Answer: - 1.86°C

Explanation:

The depression of freezing points of solutions is a colligative property.

That means that the depression of freezing points of solutions depends on the number of molecules or particles dissolved and not the nature of the solute.

To solve the problem follow these steps:

Data:

Tf = ?
solute = glucosa (this implies i factor is 1)
mass of solue = 36.0 g
mass of water = 500 g
Kf = 1.86 °/m
mm glucose = 180.0 g / mol

2) Formulas

Tf = Normal Tf - ΔTf

ΔTf = i * kf * m

m = number of moles of solute / kg of solvent

number of moles of solute = mass in grams / molar mass

3) Solution

number of moles of solute = 36.0 g / 180.0 g/mol = 0.2 mol

m = 0.2 mol / 0.5 kg = 1.0 m

ΔTf = i * Kb * m = 1 * 1.86 °C/m * 1 m = 1.86°C

Tf = 0°C - 1.86°C = - 1.86°C

Answer: - 1.86 °C

7 0

3 years ago

A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?

prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

3 0

3 years ago