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3 years ago

What is the complete ground state electron configuration for the selenium atom?

1 answer:

3 0

Se
1s² 2s²2p⁶ 3s²3p⁶ 4s²3d¹⁰3p⁴

You can do it just follow the periodic table
1)groups 1 and 2 give s-electrons , and number of the level equals number of the period
2)groups from 13 to 18 give p-electrons, and number of the level equals number of the period
3) groups from 3 to 12 give d-electrons and level one less than number of the period

You might be interested in

Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

o-na [289]

Answer:

pKa = 7.46

Explanation:

1) Data:

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

2) Strategy:

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

3) Solution:

a) pH

pH = - log [H₃O⁺]

4.64 = - log [H₃O⁺]

$[H_3O^+]= 10^{-4.64} = 2.29.10^{-5}$

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0

4 years ago

Could someone please help!

bezimeni [28]

Im pretty sure that the current will increase so the answer is A . Hope that helps

8 0

4 years ago

What is the empirical formula of a hydrocarbon if complete combustion or 2.900 mg of the hydrocarbon produced 9.803 mg of CO2 an

kow [346]

Answer:

The empirical formula of the hydrocarbon is CH.

Explanation:

The following data were obtained from the question:

Mass of hydrocarbon = 2.9 mg

Mass of CO2 = 9.803 mg

Mass of H2O = 2.006 mg

Next, we shall determine the mass of carbon (C) and hydrogen (H) in the compound since hydrocarbon contains carbon and hydrogen only.

This is illustrated below:

Molar mass of CO2 = 12 + (16x2) = 12 + 32 = 44 g/mol

Mass of CO2 = 9.803 mg

Mass of C in the compound =?

Mass of C in the compound =

12/44 x 9.803

= 2.674 mg

Molar mass of H2O = (2x1) + 16 = 2 + 16 = 18 g/mol

Mass of H2O = 2.006 mg

Mass of H in the compound =

2/18 x 2.006

= 0.223 mg

Finally, we shall determine the empirical formula of the hydrocarbon as follow:

Carbon (C) = 2.674 mg

Hydrogen (H) = 0.223 mg

Divide by their molar mass

C = 2.674 /12 = 0.223

H = 0.223 / 1 = 0.223

Divide both side by the the smallest

C = 0.223/0.223 = 1

H = 0.223/0.223 = 1

Therefore, the empirical formula of the hydrocarbon is CH.

5 0

3 years ago

When did ernest rutherford contribute to the atomic theory?

charle [14.2K]

In 1911, Rutherford performed experiment to determine the structure on atom

6 0

4 years ago

A 5.00 g piece of metal is heated to 100.0 C then placed in a beaker containing 20.0 g of water at 10.0 C. The temperature of th

Nataly [62]

Calculate the heat gained by the water first.

q = mCpΔT
m = 20.0 g
Cp = 4.186 J/g°C
ΔT = T(final) - T(initial) = 15.0°C - 10.0°C = 5.0°C

q = (20.0)(4.186)(5.0) = 419 J

This is equal to the heat lost by the metal, so calculate Cp for the metal, given:
q = -419 J (negative because heat was lost)
m = 5.00 g
ΔT = 15.0°C - 100.0°C = -85.0°C (negative because the temperature decreased)

q = mCpΔT —> Solve for Cp —> Cp = q/mΔT

Cp = -419 / (5.00 • -85.0) = 0.986 J/g°C

5 0

3 years ago