All categories

4 years ago

A plastic rod becomes negatively charged when it is rubbed with a piece of wool. how does the rod become charged?

Chemistry

2 answers:

Lera25 [3.4K]4 years ago

4 0

Answer;

c.electrons move from the wool to the rod.

Explanation;

-Atoms are generally neutral (have the same number of protons as electrons). Substances lose electrons to become positively charged, while other substances gain electrons to become negatively charged.

-When a plastic rod is rubbed with woolen cloth or wool, a transfer of electrons occurs (from the wool to the rod). The rod gains electrons; thus, it is negatively charged.

Anarel [89]4 years ago

4 0

Electrons move from the wool to the rod

You might be interested in

A fluid that vaporizes and condenses inside the tubing of a heat pump is called the.

miss Akunina [59]

I believe that is refrigerant!
Hope this helps!

4 0

3 years ago

Read 2 more answers

Consider a mixture of NaCl and NaNO3 that is 31.8% Na by mass. Calculate the percentage by mass of NaCl in the mixture.

lutik1710 [3]

Answer: The percentage by mass of NaCl in the mixture is 38.5 %

Explanation:

To calculate the mass percentage of element in compound, we use the equation:

$\text{Mass percent of element}=\frac{\text{Mass of element}}{\text{Mass of compound}}\times 100$ .......(1)

For NaCl:

Mass of sodium element = 23 g

Mass of NaCl = 58.5 g

Putting values in equation 1, we get:

$\text{Mass percent of sodium element}=\frac{23g}{58.5g}\times 100=39.32\%$

Mass fraction of sodium metal in NaCl = 0.3932

For $NaNO_3$ :

Mass of sodium element = 23 g

Mass of $NaNO_3$ = 85 g

Putting values in equation 1, we get:

$\text{Mass percent of sodium element}=\frac{23g}{85g}\times 100=27.1\%$

Mass fraction of sodium metal in sodium nitrate = 0.271

Let us assume the mass fraction of NaCl in the mixture is 'x'

So, the mass fraction of $NaNO_3$ in the mixture will be '(1-x)'

We are given:

Percent by mass of Na in the mixture = 31.8 %

Mass fraction of Na in the mixture = 0.318

Evaluating the mass fraction of NaCl in the mixture:

$[(x\times 0.3932)+((1-x)\times 0.271)]=0.318$

x = 0.385

Percent by mass of NaCl in the mixture will be = $(0.385\times 100)=38.5\%$

Hence, the percentage by mass of NaCl in the mixture is 38.5 %

4 0

3 years ago

50 Points: What is the mass of 6.12 moles of arsenic (As)?

ozzi

The mass of 6.12 moles of arsenic (As) is calculated to be approximately 459g.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles of a substance by its molar mass. That is;

Mass of Arsenic = no. of moles of As × molar mass of As.

According to this question, 6.12 moles of arsenic was given and its molar mass is 74.92g/mol.

Mass of As = 6.12 mol × 74.92g/mol

Mass of As = 459g

Therefore, the mass of 6.12 moles of arsenic (As) is calculated to be approximately 459g.

Learn more about mass calculations at: brainly.com/question/8101390

4 0

3 years ago

During a period of discharge of a lead-acid battery, 405 g of Pb from the anode is converted into PbSO4 (s).

Alexus [3.1K]

Answer:

The answers to the question are as follows

First part

The mass of PbO2 (s) reduced at the cathode during the period is = 467.55_g

Second part

The electrical charge are transferred from Pb to PbO2 is 377186.86_C or 3.909 F

Explanation:

To solve this, we write the equation for the discharge of the lead acid battery as

H₂SO₄ → H⁺ + HSO₄⁻

Pb (s) + HSO⁻₄ → PbSO₄ + H⁺ + 2e⁻

at the cathode we have

PbO₂ + 3H⁺ + HSO⁻₄ + 2e⁻ → PbSO₄ + 2H₂O

Summing the two equation or the total equation for discharge is

Pb (s) + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

From the above one mole of lead and one mole of PbO₂ are consumed simultaneously hence

Number of moles of lead contained in 405 g of Pb with molar mass = 207.2 g/mole = (405 g)/ (207.2 g/mole) = 1.95 mole of Pb

Hence number of moles of PbO₂ reduced at the cathode = 1.95 mole

mass of PbO₂ reduced at the cathode = (number of moles)×(molar mass)

= 1.95 mole × 239.2 g/mol = 467.55 g of Lead (IV) Oxide is reduced at the cathode

Part B

Each mole of Pb transfers 2e⁻ or 2 electrons, therefore 1.95 moles of Pb will transfer 2 × 1.95 = 3.909 moles of electrons transferred

Each electron carries a charge equal to -1.602 × 10⁻¹⁹ C or one mole of electrons carry a charge equal to 96,485.33 coulombs

hence 3.909 moles carries a charge = 3.909 × 96,485.33 coulombs =377186.86 Coulombs of electrical charge

or transferred electrical charge = 377186.86 C or 3.909 Faraday

6 0

3 years ago

a 45g sample of water at 12°c is heated with 15kJ of energy. What will be the final temperature of the water?

Marizza181 [45]

Answer:

Final temperature = 91.75 °C

Explanation:

Given data:

Mass of water = 45 g

Initial temperature = 12 °C

Energy required = 15 Kj (15000 j)

Final temperature = ?

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Solution:

Q = m.c. ΔT

15000 j = 45 g × 4.18 j/g.°C × (T2 -T1)

15000 j = 45 g × 4.18 j/g.°C × (T2- 12 °C)

15000 j = 188.1 J/°C × (T2- 12 °C)

15000 j / 188.1 J/°C = T2- 12 °C

79.745 °C = T2 - 12 °C

T2 = 79.745 °C + 12 °C

T2 = 91.75 °C

4 0

3 years ago