Which type of rock does B represent?

Extend the aufbau sequence through an element that has not yet been identified, but whose atoms would completely fill 7p orbital

Thepotemich [5.8K]

Answer:

Number of electrons = 118

Electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Explanation:

Given: A chemical element that has completely filled 7p orbital.

According to this, the principal occupied electron shell or the valence shell of such an element is 7p.

⇒ the principal quantum number (n) for the valence shell is 7.

∴ this element belongs to the period 7 of the periodic table.

Also, an element that has completely filled p-orbital belongs to the group 18 of the p-block.

Therefore, an element that belongs to the group 7 and period 18 of the periodic table, should have a completely filled 5f, 6d, 7s and 7p orbitals.

Therefore, the electronic configuration should be: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

And, the number of electrons = atomic number of radon (Rn) + 14 + 10 + 2 + 6 = 86 + 32 = 118.

Therefore, the given element has atomic number 118 and has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶. Thus the given element can be Oganesson.

7 0

3 years ago

For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.

Rudik [331]

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

8 0

3 years ago

An 80.0g sample of an unknown metal is at an initial temperature of 55.5oC. Afer 540 J of energy is absorbed by the metal, the t

Lunna [17]

Answer:

Specific heat of metal = 0.26 j/g.°C

Explanation:

Given data:

Mass of sample = 80.0 g

Initial temperature = 55.5 °C

Final temperature = 81.75 °C

Amount of heat absorbed = 540 j

Specific heat of metal = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

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

ΔT = 81.75 °C - 55.5 °C

ΔT = 26.25 °C

540 j = 80 g × c × 26.25 °C

540 j = 2100 g.°C× c

540 j / 2100 g.°C = c

c = 0.26 j/g.°C

7 0

3 years ago

You were tasked to check if the liquid sample you have is substance or a mixture.which among these tests is the BEST way to do s

mr Goodwill [35]

D. 1 and 3 only because you never taste, and a liquid cannot melt

7 0

3 years ago

If you burned 4.10 x 10^24 molecules of butane (C4H10), what mass of ethane did you burn?

LenKa [72]

Answer:

Explanation:

Molar mass of C4H10 = 12x 4 + 1 x 10= 48+ 10=58g.

Also note that molar mass of a substance is equivalent to Avogadro's number (6.02 X 10^-23)

Hence,

6.02 X 10^-23 molecules of ethane was burned in 58g of ethane

So, 4.10 X 10^24 molecules of ethane will burn in xg of ethane.

Cross multiply:

6.02 X 10^-23x = 58 X 4.10 X 10^24

6.02 X 10^-23x = 2.378 X 10 ^26

Dividing both sides by the coefficient of 'x'

x = 2.378 X 10^26/(6.02X 10^-23)

x = 3.95 X10^48g approximately

x = 4.0 X 10^48g

Or

Mole = number of elementary

particless / Avogadro's number

Where,

No of elementary particles= 4.10 X 10^24 molecules

Avogadro's number= 6.02 x 10^-23

Therefore,

Mole = 4.10 X 10^24/ (6.02 X 10^-23)

= 6.81 X 10^ 46 moles.

But mole = mass / molar mass

Mass = mole x molar mass

where

Molar mass of C4H10 = 12x 4 + 1 x 10= 48+ 10=58g/mol

Mass= 6.81 X 10^46 X 58

= 3.95 X 10^48g approximately

= 4.0 X 10^ 48g.

Thanks

8 0

2 years ago