All categories

2 years ago

What is a wet celled battery composed of? Give an example of a wet celled battery.

Chemistry

2 answers:

kvv77 [185]2 years ago

8 0

It is a rechargeable battery. It contain a liquid electrolyte such as sulfuric acid, and a dangerous liquid

svet-max [94.6K]2 years ago

3 0

Answer:

A wet-cell battery is the original type of rechargeable battery.

An example of a wet cell battery is a lead-acid battery.

Explanation:

You might be interested in

Solutions

nordsb [41]

D. Yep, D is the answer, alright.

3 0

3 years ago

1. The solubility of AgNO3 at 20°C is 222.0g AgNO3/100g H2O. What mass of AgNO3 can be dissolved in 250 g of water at 20°C? Reca

neonofarm [45]

Answer :

(1) The mass of silver nitrate is, 555 g

(2) The solubility of the gas will be, 0.433 g/L

<u>Solution for Part 1 :</u>

From the given data we conclude that

In 100 gram of water, the amount of silver nitrate = 222 g

In 250 gram of water, the amount of silver nitrate = $\frac{222}{100}\times 250=555g$

Therefore, the mass of silver nitrate is, 555 g

<u>Solution for Part 2 :</u>

Formula used : $S_1P_1=S_2P_2$ (at constant temperature)

where,

$S_1$ = initial solubility of methane gas = 0.026 g/L

$S_2$ = final solubility of methane gas

$P_1$ = initial pressure of methane gas = 1 atm

$P_2$ = final pressure of methane gas = 0.06 atm

Now put all the given values in the above formula, we get the solubility of methane gas.

$(0.026g/L)\times (1atm)=S_2\times (0.06atm)$

$S_2=0.433g/L$

Therefore, the solubility of the gas will be, 0.433 g/L

5 0

3 years ago

How many grams of solute are needed to prepare a 3.50% mass/mass solution that has a solution mass of 2.50x102 grams.

Elis [28]

One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.

Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute

5 0

3 years ago

How many particles are in a 151 g sample of Li2O?

neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

4 0

3 years ago

Which of the following questions is outside the scope of science?

Airida [17]

A. How religion and philosophy different
All the other choices (b-d) can be explained by science while a can be explain by history I think

4 0

3 years ago