Which of the following are uses for radiation and radioactivity that are of benefit to us?

When electrolyzing copper (ll) chloride, what reaction takes place at the anode? What reaction takes place at the cathode?

klemol [59]

Answer:

Copper ions are reduced into copper atoms.

Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎

Explanation:

During electrolysis, the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.

At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)

Thus the half equation is as follows:

Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎

5 0

3 years ago

Explain how the fire spreading from tree to tree provides a model for conduction.

Inessa05 [86]

the fire spreading is what represents the form of heat transferring from one atom to another within an object and direct contact

6 0

3 years ago

Calculate the hydronium ion concentration in an aqueous solution with a ph of 9.85 at 25Â°c.

UkoKoshka [18]

Answer:

1.41 × 10⁻¹⁰ M

Explanation:

We have a solution with a pH of 9.85 at 25 °C. We can calculate the concentration of H⁺ using the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH

[H⁺] = antilog -9.85

[H⁺] = 1.41 × 10⁻¹⁰ M

H⁺ is usually associated with water molecules forming hydronium ions.

H⁺ + H₂O → H₃O⁺

Then, the concentration of H₃O⁺ ions is 1.41 × 10⁻¹⁰ M.

8 0

3 years ago

Read 2 more answers

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

Iron and carbon:steel:copper and tin:___________

vesna_86 [32]

Answer:

iron and carbon:steel:copper and tin: metal

4 0

2 years ago