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juin [17]
2 years ago
14

Which of the following is a reasonable ground-state electron configuration?

Chemistry
2 answers:
Veronika [31]2 years ago
8 0

Answer is: 1s22s22p5.

1) Electron configuration 1s²1p⁶2d², this is not reasonable because 1p and 2d oritals do not exist.

2) Electron configuration 1s²2s⁴2p⁶, this is not reasonable because s orbitals only contain 2 electrons.

3) Electron configuration 1s²2s²2p⁵ is of an element fluorine. Fluorine (F) has atomic number 9, which means it has 9 protons and 9 electrons.

4) Electron configuration 1s²2s²2d⁶, this is not reasonable because 2d orbitals do not exist.

Ivan2 years ago
7 0

Answer:

C) 1s22s22p5

Explanation:

I took the test and got it right

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Express each aqueous concentration in the unit indicated.
MAXImum [283]

Answer:

a. ppb of trichloroethylene = 3 × 10⁶ ppb

b. ppm of Cl₂ = 3.8 ppm

c. Molarity = 0.0002 mol / L

d. Molarity = 0.0007 mol / L

e. For trace amount of concentrations

Explanation:

a. Given data

mass of trichloroethylene = 25 mg

Volume of water = 9.5 L

ppb of trichloroethylene = ?

Solution

As we know that

1 L = 1000 milliliters

9.5 L = 9.5 × 1000

9.5 L =  9500 millileters (ml)

we consider 25 mg = 25 millileters

<em>ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)</em>

ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000

ppb of trichloroethylene = 0.003 × 1000,000,000

ppb of trichloroethylene = 3 × 10⁶ ppb

B. Given data

Mass of Cl₂ = 38 g

volume of water = 1.00 × 10⁴ L ( 10000 L)

ppm of Cl₂ = ?

Solution

Volume of water in ml = 1 L = 1000 ml

Volume of water in ml =  10000  × 1000

Volume of water in ml = 10000000 ml

we take 38 g = 38 ml

Now we convert it to ppm

<em>ppm = (mass of solute / mass of solvent) × 1000000 (1 million)</em>

ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000

ppm of Cl₂ = 0.0000038 × 1000000

ppm of Cl₂ = 3.8 ppm

C. Given data

Concentration of F⁻ ( Fluoride ion) = 2.4 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

2.4 ppm = 2.4 × 0.001 g/L

2.4 ppm = 0.0024 g/L

Mass of flouride ions = 0.0024 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of F⁻ = 19 g/mol

moles of F⁻ = 0.0024 g / 19 g/mol

moles of F⁻ = 0.0002 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0002 mol / 1 L

Molarity = 0.0002 mol / L

D. Given data

Concentration of NO₃⁻ ( nitrate ion) = 45 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

45 ppm = 45 × 0.001 g/L

45 ppm = 0.045 g/L

Mass of nitrate ions = 0.045 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of NO₃⁻ = 62 g/mol

moles of NO₃⁻ = 0.045 g / 62 g/mol

moles of F⁻ = 0.0007 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0007 mol / 1 L

Molarity = 0.0007 mol / L

E. Reason of expressing concentration in ppm and ppb

Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.

As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.

Example

Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.

Lets change this in to molarity

mass = 10 μg

10 μg = 10 / 1000000

10 μg = 0.00001 g

now find out moles of Arcenic

moles = mass / molar mass

molar mass of arcenic = 75 g/mol

<em>moles = mass / molar mass</em>

moles of arcenic = 0.00001 g / 75 g/mol

moles of arcenic = 0.00000012 mol

<em>Molarity = moles of solute / litres of solution</em>

Molarity = 0.00000012 mol / 1 L

Molarity = 0.00000012 mol/ L

As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb

7 0
2 years ago
A piece of iron (mass = 25.0 g) at 398 k is placed in a styrofoam coffee cup containing 25.0 ml of water at 298 k. assuming that
Firlakuza [10]
The final temperature of the water is the equilibrium temperature, or the also the final temperature of the iron after a long period of time. Applying the conservation of energy:

m,iron*C,iron*ΔT = - m,water*C,water*ΔT

The density of water is 1000 g/mL.

(25 g)(0.449 J/g·°C)(T - 398 K) = - (25 mL)(1000 g/mL)(4.18 J/g·°C)(T - 298)
Solving for T,
<em>T = 298.01 K</em>
6 0
1 year ago
Read 2 more answers
COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at
mestny [16]

<u>Answer:</u> The value of K_p for the reaction at 690 K is 0.05

<u>Explanation:</u>

We are given:

Initial pressure of COCl_2 = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

                  COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial:            1                    -         -

At eqllm:       1-x                 x        x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

[(1 - x) + x+ x]=1.2\\\\x=0.2atm

The expression for K_p for above equation follows:

K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}

p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm

Putting values in above equation, we get:

K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05

Hence, the value of K_p for the reaction at 690 K is 0.05

3 0
1 year ago
Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2] [I–] [H+] rate I 0.100 M 5.00 × 10–4 M
otez555 [7]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

4 0
1 year ago
Ozone reacts completely with NO, producing NO2 and O2. A 13.0 L vessel is filled with 1.30 mol of NO and 1.30 mol of O3 at 401.0
alex41 [277]

Answer:

6.58 atm total

3.29 atm NO2

3.29 atm O2

Explanation:

Balanced equation:

O3 + NO → NO2 + O2

There are equal numbers of moles of both reactants, so neither is in excess and either could be considered the limiting reactant.

( 1.30 mol NO) x (1 mol NO2 / 1 mol NO) =  1.30 mol NO2

( 1.30 mol NO) x (1 mol O2 / 1 mol NO) =  1.30 mol O2

<em>Total pressure by using the formula;</em>

P = nRT / V

= ( 1.30 mol +  1.30 mol) x (0.08205746 L atm/K mol) x (401.0 K) / (13.0 L)  

= 6.58 atm

<em>Partial pressure for NO2;</em>

(6.58 atm) x (1.30 mol NO2) / (1.30 mol + 1.30 mol)

= 3.29 atm NO2

<em>Partial pressure for O2</em>

6.58 atm total - 3.29 atm NO2

= 3.29 atm O2

5 0
1 year ago
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