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Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.

Dovator [93]

Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0

3 years ago

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The "air bags" that are currently installed in automobiles to prevent injuries in the event of a crash are equipped with sodium

stellarik [79]

Answer:

0.0177 L of nitrogen will be produced

Explanation:

The decomposition reaction of sodium azide will be:

$2NaN_{3}(s)--->2Na(s)+3N_{2}(g)$

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas

The molecular weight of sodium azide = 65 g/mol

The mass of sodium azide used = 100 g

The moles of sodium azide used = $\frac{mass}{molarmass}=\frac{100}{65}=1.54mol$

so 1.54 moles of sodium azide will give = $\frac{3X1.54}{2}=2.31$ mol

the volume will be calculated using ideal gas equation

PV=nRT

Where

P = Pressure = 1.00 atm

V = ?

n = moles = 2.31 mol

R = 0.0821 L atm / mol K

T = 25 °C = 298.15 K

Volume = $\frac{P}{nRT}=\frac{1}{2.31X0.0821X298.15}=0.0177L$

3 0

3 years ago

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:

kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g = 0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3 = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂ I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0

3 years ago

The Best Answer Will Be Rated Brainliest!

statuscvo [17]

The answer is flourine

flourine some what sounds like flow so

flourine is a flowing element

4 0

3 years ago

"An unsaturated solution is formed when 80 grams of salt is dissolved in 100 grams of water at 40° celsius. This salt could be..

olchik [2.2K]

Answer:

3. NaNO3

Explanation:

Below the line on TABLE G in the reference tables at 40C

4 0

4 years ago