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3 years ago

Why is seawater a better conductor of electricity than freshwater?

Chemistry

2 answers:

Andre45 [30]3 years ago

7 0

Answer:

Because it contains Na+ and Cl- ions, Which are conductors of electricity

Explanation:

GenaCL600 [577]3 years ago

3 0

Answer:

please refer to the attachment above;)

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What is the wavelength of a light of frequency 6.42 x 1014 Hz?

Dahasolnce [82]

Answer:

467 nm is the answer

Explanation:

IT'S the answer for

8 0

3 years ago

Choose whether the statements about oil sands are true or false. The viscosity of bitumen is about 100 times greater than the vi

Alexxx [7]

The First 2 statements stated above were false whereas the third one is a true statement.

Explanation:

The viscosity of bitumen is about 100 times greater than the viscosity of water - False

Reason - The viscosity of bitumen is about not 100 times greater than the viscosity of water, it is actually 100, 000 times greater.

Oil from oil sand deposits is only obtained by first heating the sands at high temperatures is False.

Reason- Oil from oil sand deposits is not obtained by first heating the sands at high temperatures but by using steams

Oil sands contain sand, water, and light crude oil is true.

6 0

3 years ago

Which of these phenomena cause uneven heating of the earth

Ostrovityanka [42]

Answer:

The angle of incoming sunlight varies at different places

8 0

3 years ago

Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be

AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

$\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles$

$\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

$\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles$

The concentration of benzoic acid is:

$C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M$

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺

0.14 - x x x

$Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}$

$Ka = \frac{x*x}{0.14 - x}$

$6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0$ (2)

By solving equation (2) for x we have:

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

$pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52$

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻ (3)

The number of moles of C₆H₅CO₂⁻ is:

$\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles$

The volume of NaOH added is:

$V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L$

The concentration of C₆H₅CO₂⁻ is:

$C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M$

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻

0.14 - x x x

$Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}$

$(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0$

$(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0$

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

$pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33$

$pH = 14 - pOH = 14 - 5.33 = 8.67$

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

$\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:

$\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles$

The concentration of NaOH is:

$C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M$

Therefore, the pH is given by this excess of NaOH:

$pOH = -log([OH^{-}]) = -log(0.067) = 1.17$

$pH = 14 - pOH = 12.83$

I hope it helps you!

4 0

3 years ago

Valence electrons are found:

eduard

Valence electrons are found AT THE OUTERMOST SHELL OF AN ATOM.
Valence electron refers to those electrons that are located furthest away from the nucleus of an atom, that is, they are found at the outermost shell. Valence electrons determine the chemical properties of an element and they are the only ones that get involved in chemical reactions with other elements.

7 0

3 years ago