All categories

1 year ago

An element with a relatively large amount of electrons in the valence ring is considered to be a good conductor. True False

Chemistry

1 answer:

galben [10]1 year ago

5 0

Answer:

False

Explanation:

In every substance, there is a valence band and there is a conduction band. The gap between the valence band and the conduction band determines Whether a substance is a conductor or an insulator.

Conductors have a narrow gap between the valence band and the conduction band while insulators have a large gap between conduction band and insulation band. However, elements with a large number of electrons in the valence band are non conductors.

You might be interested in

You are writing a safety contract for your class. List 10 things you would include in the contract.

andre [41]

1.always listen to teacher

2 no eating in class

3 always have attentive listening

4 always have proper safety material

5 wear eyeglasses when needed

6 let others be able to listen

sorry I could only think of 6

7 0

2 years ago

Read 2 more answers

A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl

Alecsey [184]

Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

$Kr+2Cl_2\rightarrow KrCl_4$

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, $n=\frac{0.500*15.0}{0.0821*623.8}$

n = 0.146 moles

for chlorine, $n=\frac{1.50*15.0}{0.0821*623.8}$

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

$0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})$

= 32.94 g of $KrCl_4$

So, 32.94 g of the product will form.

5 0

2 years ago

A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the

DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × $10^{-46}$ KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × $10^{-18}$ ( $\frac{1}{n^{2}_{final} }$ - $\frac{1}{n^{2}_{initial} }$ )

(1/434 × $10^{-9}$ ) = -2.178 × $10^{-18}$ ( $\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }$ )

⇒ 434 × $10^{-9}$ = (1/-2.178 × $10^{-18}$ ) $\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }$

But, $n_{final}$ = 2

434 × $10^{-9}$ = (1/2.178 × $10^{-18}$ ) $\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }$

434 × $10^{-9}$ × 2.178 × $10^{-18}$ = $(\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })$

⇒ $n_{initial}$ = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

= (hc) ÷ λ

= (6.626 × 10−34 × 3.0 × $10^{8}$ ) ÷ (434 × $10^{-9}$ )

= (1.9878 × $10^{-25}$ ) ÷ (434 × $10^{-9}$ )

= 4.58 × $10^{-19}$ J

= 4.58 × $10^{-22}$ KJ

But 1 mole = 6.02× $10^{23}$ , then;

energy in KJ/mole = (4.58 × $10^{-22}$ KJ) ÷ (6.02× $10^{23}$ )

= 7.61 × $10^{-46}$ KJ/mole

7 0

2 years ago

How many liters of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa)?

OLga [1]

Using ideal gas equation,

P\times V=n\times R\times T

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=100 kPa

T=293 K

R=8.314472 L kPa K⁻¹ mol⁻¹

Number of moles of gas=3.43 mole

Putting all the values in the above equation,

$V=\frac{3.43\times 8.314\times 293}{100}$

V=83.55 L

So the volume will be 83.55 L.

83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).

4 0

2 years ago

What is the percent CdSO4 by mass in a 1.00 m aqueous CdSO4 solution?

Svet_ta [14]

I don't know but I'm wasting 5 seconds of your time you can't take back sorry

3 0

2 years ago