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11Alexandr11 [23.1K]
3 years ago
9

Use the periodic table to write the electron configuration for barium (Ba) in noble-gas notation.

Chemistry
2 answers:
natali 33 [55]3 years ago
8 0

Answer is: [Xe] 6s².

Barium is metal from second group of periodic table of elements.

Barium has atomic number 56, it has 56 protons and 56 electrons.

Electron configuration of barium atom:  

₅₆Ba 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s².

Xenon (symbol: Xe) is an element (noble gas) with atomic number 54, which means it has 54 protons and 54 electrons.

Electron configuration of xenon atom:  

₅₄Xe 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶.

torisob [31]3 years ago
4 0

The answer is [Xe] 6s2

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Please help this is very important​
rosijanka [135]

Answer:

B

Explanation:

7 0
3 years ago
Ethylene glycol (antifreeze) has a density of 1.11 g/cm3. what is the volume in liters of 3.46 kg of ethylene glycol?
Dovator [93]

Density is a physical property which describes the mass of a substance per unit of volume of the substance. It is expressed as Density = m / V and it has units like g/cm^3. We use the density given to solve the problem.

3.46x10³ g / 1.11 g/cm³ = 3117.12 cm³  (1 L / 1000 cm³) =3.12 L

7 0
3 years ago
A gas has a solubility of 1.46 g L at 8.00 atm of pressure. What is the pressure of a sample of the same gas that
uranmaximum [27]

Answer:

14.8\ \text{atm}

Explanation:

C_1 = Initial concentration = 1.46 g/L

C_2 = Final concentration = 2.7 g/L

P_1 = Initial pressure = 8 atm

P_2 = Final pressure

From Henry's law we have the relation

\dfrac{C_2}{C_1}=\dfrac{P_2}{P_1}\\\Rightarrow P_2=\dfrac{C_2}{C_1}P_1\\\Rightarrow P_2=\dfrac{2.7}{1.46}\times 8\\\Rightarrow P_2=14.8\ \text{atm}

The pressure of a sample of the same gas at the required concentration is 14.8\ \text{atm}.

8 0
3 years ago
0.07 mol sample of octane, C^6H^18 absorbed 3.5 x 10^3 J of energy. Calculate the temp increase of octane if the molar heat capa
Dovator [93]

n = number of moles of sample of octane = 0.07 mol

Q = energy absorbed by a sample of octane = 3.5 x 10³ J

c = molar heat capacity of octane = 254.0 J/K* mol

ΔT = increase in temperature of octane = ?

Heat absorbed is given as

Q = n c ΔT

inserting the values

3.5 x 10³ J = (0.07 mol) (254.0 J/K* mol) ΔT

ΔT = (3.5 x 10³ )/((0.07) (254.0))

ΔT = 196.85 K

hence increase in temperature comes out to be 196.85 K


8 0
3 years ago
What is the answer, with the correct number of decimal places, for this problem? 4.392 g+ 102.40 g + 2.51 g=
blondinia [14]
The correct answer is C
4 0
4 years ago
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