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1 year ago

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An atom has the following electron configuration. 1s22s22p63s23p64s? How many valence electrons does this atom have? A: 2B: 3C:

5D: 11

1 answer:

Katena32 [7]1 year ago

3 0

ANSWER

2 valence electrons

STEP-BY-STEP EXPLANATION:

The electronic configuration is given below as

$1s^22s^22p^62s^23p^64s^2$

Looking at the configuration, you will see that the last energy carries 2 electrons

Hence, the valence electrons are 2

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The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

The red line of the hydrogen emission spectrum has a wavelength of 656.1 nm. Calculate the energy of one photon

Effectus [21]

The formula we can use here is the Plancks equation:

E = h c / ʎ

where h is Plancks constant = 6.626 × 10-34 m2 kg / s, c is speed of light = 3 x 10^8 m/s and ʎ is wavelength = 656.1 x 10^-9 m

Therefore E is:

E = (6.626 × 10-34 m2 kg / s) * (3 x 10^8 m/s) / 656.1 x 10^-9 m

<span>E = 3.03 x 10^-19 J</span>

4 0

3 years ago

Calculate the theoretical yield for the bromination of both stilbenes

podryga [215]

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

4 0

2 years ago

I need help plez! Thank you!

vladimir2022 [97]

The answer should be E,C,F,B,A,D in that order.

6 0

3 years ago

What is the speed of a wave with a wavelength of 3 m and a frequency of .1 Hz?

ratelena [41]

Answer:

wave velocity= frequency × wave length

=1×3

=3m/s

Explanation:

The distance covered by the wave in one second is equal to its wavelength, therefore,

wave velocity=wavelength/time period

OR wave velocity= frequency× wavelength

You can assume velocity as speed here.

3 0

3 years ago