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Lisa [10]
3 years ago
14

The electronic configuration of phosphorous is 2.8.5. Explain, in terms of its electronic configuration, why phosphorus is in gr

oup 5 of the periodic table.
PLEASE HELP I WILL GIVE BRAINLIEST​
Chemistry
1 answer:
Setler [38]3 years ago
6 0

Answer:

Explanation:

phosphorus belongs to group 5 of the periodic table because it has 5 electron in its outermost shell the number of electron in the outermost shell of electron determine the group of the element in the periodic table

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a solution must be at a higher temperature than a pure solvent to boil. what colligative property can be employed to achieve thi
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Boiling-point elavation.

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A crucible is a severe test or trial. It is also a vessel in which materials are melted at high temperatures to produce a more r
zhannawk [14.2K]

In this drama, a crucible would represent the constant building action leading to the climax of the pot.

What is crucible?

A crucible is a ceramic as well as metal vessel used to melt metals or subject other substances to extremely high temperatures. Crucibles can be made from any material that can withstand temperatures high enough to melt and otherwise change their contents, despite historically being most frequently made of clay. The crucible's shape has changed over time, with designs that take into account both regional variation and the process for that they are used. The earliest crucible forms date to Iran and Eastern Europe in the sixth or fifth millennia B.C.

Similar to the contents of a crucible, the characters seem to be constantly enduring severe trials; in this case, the witching trials. The crucible would, in my opinion, symbolize the town itself and its residents. The symbolism of this suggests a theme of the townspeople going through the witching trials and coming out with a "refined" lesson. The unnecessary accusations and trials will teach a hard lesson to the people who live within the town and influence their future.

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4 0
1 year ago
) determine the henry's law constant for ammonia in water at 25°c if an ammonia pressure of 0.022 atm produces a solution with a
Nataly [62]

Answer:

a. 59 m/atm

Explanation:

  • To solve this problem, we must mention Henry's law.
  • <em>Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.</em>
  • It can be expressed as: C = KP,

C is the concentration of the solution (C = 1.3 M).

P is the partial pressure of the gas above the solution (P = 0.022 atm).

K is the Henry's law constant (K = ??? M/atm),

∵ C = KP.

∴ K = C/P = (1.3 M)/(0.022 atm) = 59.0 M/atm.

3 0
3 years ago
What is the relative rate of effusion for hydrogen iodide (128 g/mol) compared to gaseous hydrochloric acid (36.5 g/mol)
MArishka [77]

This question is hard but I found the answer from merit nation

8 0
3 years ago
An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu
chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

The density of a proton is 6.278\times 10^{14} g/cm^3.

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\pi r^3..[1]

Diameter of the nucleus ,d' = 9.00\times 10^{-5}\AA

Radius of the nucleus ,r' = 0.5 d'=0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA

Volume of nucleus = V'

V=\frac{4}{3}\pi r'^3..[2]

Dividing [2] by [1]

\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}

=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}

\frac{V'}{V}=4.6656\times 10^{-14}

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

Diameter of the proton ,d = 1.72\times 10^{-15} m = 1.72\times 10^{-13} cm

1 m = 100 cm

Radius of the proton,r = 0.5 d=0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3

Mass of proton, m = 1.0073 amu = 1.0073\times 1.66054\times 10^{-24} g

1 amu = 1.66054\times 10^{-24} g

Density of the proton : d

d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3

The density of a proton is 6.278\times 10^{14} g/cm^3.

5 0
3 years ago
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