What did the Curies conclude about radioactivity based on their studies?

How many seconds does it take to deposit 0.94 g of Ni on a decorative drawer handle when 14.9 A is passed through a Ni(NO3)2 sol

Goshia [24]

Answer:

Time take to deposit Ni is 259.02 sec.

Explanation:

Given:

Current $I = 14.9$ A

Faraday constant $= 96485$ $\frac{C}{mole}$

Molar mass of Ni $= 58.69$ $\frac{g}{mole}$

Mass of Ni $= 0.94$ g

First find the no. moles in Ni solution,

Moles of Ni $= \frac{0.94}{58.69}$

$= 0.02$ mol

From the below reaction,

$Ni^{2+} + 2e$ ⇆ $Ni_{(s)}$

Above reaction shows "1 mol of $Ni^{2+}$ requires 2 mol of electron to form 1 mol of $Ni_{(s)}$ "

So for finding charge flow in this reaction we write,

$=$ $0.02 \times \frac{2 }{1 } \times 96485 \frac{C}{mol}$

Charge flow $= 3859.4$ C

For finding time of reaction,

$I = \frac{q}{t}$

Where $q =$ charge flow

$t = \frac{q}{I}$

$t = \frac{3859.4}{14.9}$

$t = 259.02$ sec

Therefore, time take to deposit Ni is 259.02 sec.

3 0

3 years ago

Por ser la materia electrica neutra ¿como debe de ser el numero de electrones y el de protones?

kvasek [131]

Answer:

Deben ser de igual número.

Explanation:

8 0

2 years ago

Why is a thermos considered a (mostly) closed system? Question 3 options: different types of liquids stay hot thermoses come in

cestrela7 [59]

Answer:

Different types of hot or cold items can be stored in a thermos and power cannot enter or exit the system when the thermos lid is tightly closed

Explanation:

Closed systems are those that do not interact or do not exchange energy with the environment that surrounds them, that is why internal temperatures and conditions are maintained.

The human body is an open system, that is, it would be the opposite of the thermos since we constantly exchange energy with the environment through sweating, emission of gases, urine, feces, and the ingestion of food.

Thermoses are systems specially created to maintain a medium, it will be maintained if its lid is hermetically closed to prevent heat leakage or entry in situations of cold fluids.

4 0

3 years ago

Calculate the pH of a buffer solution prepared by mixing 60.0 mL of 1.00 M lactic acid and 25.0 mL of 1.00 M sodium lactate.

marshall27 [118]

This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:

pH = pKa + log[(salt/acid]

Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).

Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L

Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M

Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48

4 0

3 years ago

If a piece of aluminum with a mass of 3.99 g and a temperature of 100.0 °C is dropped

Vinil7 [7]

Answer:

The final temperature of the system is 27.3°C.

Explanation:

Heat lost by aluminum = 3.99 × 0.91 × (100-T)

= 3.631 (100-T)

Heat gained by water = 10 × 4.184 × (T-21)

= 41.84 (T-21)

As,

Heat gained = Heat loss

or, 3.631(100-T) = 41.84(T-21)

or,363.1 - 3.631 T = 41.84 T - 878.64)

or, (41.84+ 3.631) T = 878.64 +363.1

or T= $\frac{1241.74}{45.47}$

or, T = 27.3°C

Hence the final temperature is 27.3°C.

8 0

3 years ago