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3 years ago

What is the pH of a 3.5 x 10-^6 M solution of HCl?

Chemistry

1 answer:

Svetach [21]3 years ago

8 0

Answer:

5.46

Explanation:

From the question,

pH this can be defined as the acidity or alkalinity of a solution.

The expression for pH is given as

pH = -log(H⁺)...................... Equation 1

Where H⁺ = Hydrogen ion of HCL

Given: H⁺ = 3.5×10⁻⁶ M

Substitite this value into equation 1

pH = -log(3.5×10⁻⁶)

pH = 5.46

Hence the pH of HCl is 5.46

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Answer with Explanation:

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4 years ago

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BabaBlast [244]

Answer:

6. O₂ + Cu —> CuO

7. H₂ + Fe₂O₃ —> H₂O + Fe

8. O₂ + H₂ — > H₂O

9. H₂S + NaOH —> Na₂S + H₂O

10. Al + HCl —> H₂ + AlCl₃

Explanation:

6. Oxygen gas react with solid copper metal to form copper(II) oxide

Oxygen gas => O₂

Copper => Cu

copper(II) oxide => CuO

The equation is:

O₂ + Cu —> CuO

7. hydrogen gas and iron(III) oxide powder react to form liquid water and solid iron power

hydrogen gas => H₂

Iron(III) oxide => Fe₂O₃

Water => H₂O

Iron => Fe

The equation is:

H₂ + Fe₂O₃ —> H₂O + Fe

8. Oxygen gas react with hydrogen gas to form liquid water

Oxygen gas => O₂

hydrogen gas => H₂

Water => H₂O

The equation is:

O₂ + H₂ — > H₂O

9. Hydrogen sulphide gas is bubbled through a sodium hydroxide solution to produce sodium sulphide and liquid water

hydrogen sulphide => H₂S

sodium hydroxide => NaOH

Sodium sulphide => Na₂S

Water => H₂O

The equation is:

H₂S + NaOH —> Na₂S + H₂O

10. Hydrogen gas and aluminum chloride solutions are produced when solid aluminum react with hydrochloric acid

Aluminum => Al

Hydrochloric acid => HCl

hydrogen gas => H₂

Aluminum chloride => AlCl₃

The equation is:

Al + HCl —> H₂ + AlCl₃

5 0

3 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

The activation energy for the decomposition = 33813.28 J/mol

8 0

3 years ago