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Ket [755]
3 years ago
5

A neutral atom in the ground state of Sulfur has its outer most valence electron in which orbital? f d

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
3 0

<u>Answer:</u> The outermost valence electron enters the p orbital.

<u>Explanation:</u>

Valence electrons are defined as the electrons which are present in outer most orbital of an atom.

Sulfur is the 16th element of the periodic table having 16 electrons.

Electronic configuration of sulfur atom is 1s^22s^22p^63s^23p^4

The number of valence electrons are 2 + 4 = 6

These 6 electrons enter s-orbital and p-orbital but the outermost valence electron will enter the p-orbital.

Hence, the outermost valence electron enters p orbital.

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Please help..... who is up for a zoom to help me with balancing equations ​
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Energy of reactants is higher than the products , I am not sure through
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How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
ahrayia [7]
<span>Answer: 100 ml
</span>

<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>

iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>

<span>2) Use the percent yield to calculate the theoretical amount:
</span>

<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
</span>

<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
</span>

<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
</span>

<span>i) Molarity formula: M = n / V
</span>

<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>

3 0
3 years ago
Read 2 more answers
You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.2750 M NaOH sol
oksano4ka [1.4K]

Answer:

a. 0.393M CH₃COOH.

b. 2.360% of acetic acid in the solution

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

<em>That means 1 mole of acid reacts per mole of NaOH.</em>

Moles of NaOH to reach the equivalence point are:

35.75mL = 0.03575L × (0.2750mol / L) = <em>9.831x10⁻³ moles of NaOH</em>

As 1 mole of acid reacts per mole of NaOH, moles of CH₃COOH in the acid solution are 9.831x10⁻³ moles.

a. As the volume of the acetic acid solution is 25.00mL = 0.02500L, the molarity of the solution is:

9.831x10⁻³ moles / 0.02500L =

<h3>0.393M CH₃COOH</h3>

b. Molar mass of acetic acid is 60g/mol. The mass of 9.831x10⁻³ moles is:

9.831x10⁻³ moles ₓ (60g / mol) = <em>0.590g of CH₃COOH</em>.

As volume of the solution is 25.00mL, the percentage of acetic acid is:

(0.590g CH₃COOH / 25.00mL) ₓ 100 =

<h3>2.360% of acetic acid in the solution</h3>

8 0
2 years ago
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