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3 years ago

What are the possible values of n and ml for an electron in a 5d orbital?

Chemistry

1 answer:

liberstina [14]3 years ago

3 0

Answer:

The answer to your question is n = 5, l = 2, m can be -2, -1, 0, 1 or 2

Explanation:

Data

orbital = 5d

values of n, l, m

Process

1.- Determine the value of n

n is the coefficient of the orbital, in this problem n = 5

2.- Determine the value of l

l takes values depending in the sublevel of energy,

if the sublevel is s then l = 0

p l = 1

d l = 2

f l = 3

For this problem l = 2

3.- Determine the value of m

when l = 2, m takes values of -2, - 1, 0, 1 or 2

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Molar mass of copper=63.5g/mol
Given mass=6.93g

$\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\longmapsto No\:of\:moles=\dfrac{6.93}{63.5}$

$\\ \sf\longmapsto No\:of\:moles=0.109\approx 1.11moles$

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2 years ago

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In a laboratory experiment, a student titrates 10.0 mL of acetic acid, CH3COOH, solution with 0.5 M NaOH. The endpoint was reach

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Answer:

Concentration of acetic acid=2.3mol/L

Explanation:

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$CH_3COOH +NaOH = Na(CH_3COO) +H_2O$

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V1 and V2 volume of CH3COOH and NaOH respectively and they are taken upto end point of titration;

we have given molarity but we need normality;

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but in case of NaOH and CH3COOH n-factor is 1 for each.

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normality=molarity;

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3 years ago

According to the reaction below, how many moles of Ba3(PO4)2(s) can be produced from 115 mL of 0.218 M BaCl2(aq)? Assume that th

Anna71 [15]

Answer:

0.01125 moles of $Ba_3(PO_4)_2(s)$ can be produced from 115 mL of 0.218 M $BaCl_2(aq)$ .

Explanation:

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Volume of the solution = 115 mL = 0.115 L ( 1 mL=0.001 L)

Molarity of the $BaCl_2$ solution = 0.218 M

$0.218 M=\frac{n}{0.115 L}$

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$0.03379 mol\times \frac{1}{3}=0.01126 mol$

0.01125 moles of $Ba_3(PO_4)_2(s)$ can be produced from 115 mL of 0.218 M $BaCl_2(aq)$ .

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3 years ago