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3 years ago

15

Earnshaw’s theorem (Prob. 3.2) says that you cannot trap a charged particle in an electrostatic field. Question: Could you trap

a neutral (but polarizable) atom in an electrostatic field?

1 answer:

Ostrovityanka [42]3 years ago

7 0

Answer: NO

You cannot not trap a polarizable atom in a purely electrostatic field.

You might be interested in

A pot of boiling water on the stove is slowly releasing steam. Which process is this an example of?

Viktor [21]

evaporation. When a substance undergoes a change of matter from a solid to gas, it is the process of evaporation.

Hope this helps

8 0

3 years ago

1. BaBr2(aq) + H2SO4(aq) →?<br> please balance the equation and predict the products

Andrew [12]

Answer:

BaBr2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HBr (aq)

Explanation:

This is a precipitation reaction: BaSO4 is the formed precipitate.

4 0

3 years ago

What else is produced during the combustion of butane, C4H10?<br><br> 2C4H10 + 13O2 → <br> + 10H2O

Neko [114]

<h3>Answer:</h3>

8CO₂

<h3>Explanation:</h3>

We are given;

Butane, C₄H₁₀

Butane is a hydrocarbon in the homologous series known as alkane.

We are required to determine the other product produced in the combustion of butane apart from water.

We know that the complete combustion of alkane yields carbon dioxide and water.
Therefore, combustion of butane will yield carbon dioxide and water.

The balanced equation for the complete combustion of butane will be;

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

8 0

3 years ago

Read 2 more answers

4th one<br> Mark the brainliest

ale4655 [162]

Answer:

1. The electronic configuration of X is: 1s2 2s2 sp6 3s2

2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

3. The formula of the compound form by X and Y is given as: XY

Explanation:

For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:

1s2 2s2 sp6 3s2

To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:

Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below

1s2 2s2 2p4

The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

The formula of the compound form by X and Y is given below :

X^2+ + Y^2- —> XY

Their valency will cancel out thus forming XY

4 0

4 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

3 years ago