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Aleksandr-060686 [28]
3 years ago
10

Which element has a partially filled f orbital? sm os ba bi

Chemistry
2 answers:
Sladkaya [172]3 years ago
8 0
Sm is the answer

i hope i helped :)

asambeis [7]3 years ago
6 0

Answer:

Sm (Samarium) = [Xe] 4f6 6s2

Explanation:

Noble gas electron configuration of Sm (Samarium) = [Xe] 4f⁶ 6s²

Noble gas electron configuration of Os (Osmium) = [Xe] 4f¹⁴5d⁶6s²

Noble gas electron configuration of Ba (Barium) = [Xe] 6s²

Noble gas electron configuration of Bi (Bismuth) = [Xe] 4f¹⁴5d¹⁰6s²6p³

A completely filled s, p, d, and f orbitals have 2, 6, 10 and 14 electrons respectively. Half filled s, p, d, and f orbitals have 1, 3, 5 and 7 electrons respectively.  Bi and Os have completely filled f orbitals. Sm has partially filled f orbital.  

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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
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Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

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