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mr Goodwill [35]
2 years ago
6

10.Give the possible values for the magnetic quantum number for each of the following orbitals (a) 4p: (b) 3d: (c) 2s (d) 5P

Chemistry
1 answer:
andre [41]2 years ago
5 0

Answer:

(a) 4p: ml = -3, -2, -1, 0, +1, +2, +3

(b) 3d: ml = -2, -1, 0, +1, +2

(c) 2s: ml = -1, 0, +1

(d) 5p: ml = -4, -3, -2, -1, 0, +1, +2, +3, +4

Explanation:

Manetic quantum numbers determine the spatial orientation and number of orbitals in the subshell. ml is from -l to +l.

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DNA testing which I think is the largest breakthrough.
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3 years ago
The fluid in a barometer (shown above) is mercury, Hg. Why point
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Answer:

Mercury responds to temperatures differently than water. When mercury is cold, the molecules bunch together, making it seem to be less liquid, and the opposite happens during heat.

Hope this helps!

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3 years ago
A student uses a dye to change the colour of his tee-shirt. he fills a metal bucket with water. When the bucket is heated, what
Tanya [424]

Answer:

The particles begin to vibrate faster and more.

Explanation:

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8 0
3 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
How many kj of heat are needed to completely vaporize 1.30 moles of h2o? the heat of vaporization for water at the boiling point
Cerrena [4.2K]
The Kj of heat that  are needed to completely vaporize 1.30  moles of H2O  if  the heat of vaporization  for water is 40.6 Kj/mole  is calculated as  below

Q(heat) = moles x heat  of vaporization)

=1.30 mol  x40.6 kj/mol= 52.78 Kj  is needed
6 0
3 years ago
Read 2 more answers
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