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deff fn [24]
2 years ago
5

What is the electron configuration for sulfur? A. 1s2 2s2 2p6 3s2 3p4 B. 1s2 1p6 2s2 2p6 3s2 C. 1s2 2s2 2p6 3s2 3p3 4s1 D. 1s2 2

s2 2p6 3p1 3d5

Chemistry
2 answers:
Mnenie [13.5K]2 years ago
4 0
A.
{1s}^{2}  {2s}^{2}  {2p}^{6}  {3s}^{2}  {3p}^{4}

3241004551 [841]2 years ago
3 0
Sulfur has two filled energy levels and six electrons on the third energy level. The corresponding electron configuration is A.
B is incorrect because there are no p orbitals at the first energy level, ie, no 1p orbitals. C is incorrect because the 4s1 electron would spontaneously drop into the 3p orbitals. D is incorrect because the 3d electrons would spontaneously drop into the 3p orbitals.
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Lens coating are made of A.teflon B.bakelite C.melamine D.PVC
suter [353]

Answer:

C. melamine

Explanation:

a thin coating that eliminates reflections and glare from the front and back surfaces of your lenses.

5 0
2 years ago
What is the pH of a solution that has a [H+] = 0.010 mol/L?
Serjik [45]

Answer:

pH = 2

A 0.010 M solution of hydrochloric acid, HCl, has a molarity of 0.010 M. This means that [H+] = 1 x 10-2 M. The pH of this aqueous solution of H+ ions is pH = 2

hope this helps :3

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5 0
2 years ago
Which cannot be chemically broken down into simpler substances?
Anna71 [15]

Answer:

B.

Explanation:

elements

6 0
2 years ago
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How many grams of carbon are in 0.24 moles of carbon?
Ronch [10]

Answer:

I'm converting this if I could remember how

2.882568

2 110321/ 125000

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so I recommend not using my answer at all,

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4 0
2 years ago
Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th
Tpy6a [65]

pH = 13.5

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}

The mixture would contain

  • 0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol} of \text{OH}^{-} and
  • 0.1 \times 0.5 = 0.05 \; \text{mol} of \text{Ac}^{-}

if \text{Ac}^{-} undergoes no hydrolysis; the solution is of volume 0.1 + 0.4 = 0.5 \; \text{L} after the mixing. The two species would thus be of concentration 0.30 \; \text{mol} \cdot \text{L}^{-1} and 0.10 \; \text{mol} \cdot \text{L}^{-1}, respectively.

Construct a RICE table for the hydrolysis of \text{Ac}^{-} under a basic aqueous environment (with a negligible hydronium concentration.)

\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}

The question supplied the <em>acid</em> dissociation constant pK_afor acetic acid \text{HAc}; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant pK_b for its conjugate base, \text{Ac}^{-}. The following relationship relates the two quantities:

pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})

... where the water self-ionization constant pK_w \approx 14 under standard conditions. Thus pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3. By the definition of pK_b:

[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b =  10^{-pK_{b}}

x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}

x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}

[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}

pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5

6 0
3 years ago
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