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kondor19780726 [428]
3 years ago
15

UV spectroscopy measures the energy required to promote an electron from the ________ molecular orbital to the ________ molecula

r orbital.
Chemistry
1 answer:
balu736 [363]3 years ago
6 0

Answer:

UV spectroscopy measures the energy required to promote an electron from the highest occupied molecular orbital to the lowest unoccupied molecular orbital.

Explanation:

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Given that cao(s) + h2o(l) → ca(oh)2(s), δh°rxn = –64.8 kj/mol, how many grams of cao must react in order to liberate 525 kj of
USPshnik [31]

Answer:

454.3 g.

Explanation:

  • From the given data:

1.0 mol of CaO liberates → – 64.8 kJ.

??? mol of CaO liberates → - 525  kJ.

∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.

<em>∴ The no. of grams of CaO needed = no. of moles x molar mass</em> = (8.1 mol)(56.077 g/mol) = <em>454.3 g.</em>

8 0
3 years ago
How many 3-liter balloons could the 12-L helium tank pressurized to 160 atm fill? Keep in mind that an "exhausted" helium tank i
vaieri [72.5K]

Answer:

The 12L helium tank pressurized to 160 atm will fill <em>636 </em>3-liter balloons

Explanation:

It is possible to answer this question using Boyle's law:

P_1V_1=P_2V_2

Where P₁ is the pressure of the tank (160atm), V₁ is the volume of the tank (12L), P₂ is the pressure of the balloons (1atm, atmospheric pressure) And V₂ is the volume this gas will occupy at 1 atm, thus:

160atm×12L = 1atm×V₂

V₂ = 1920L

As the tank will never be empty, the volume of the gas able to fill balloons is the total volume minus 12L, thus the volume of helium able to fill balloons is:

1920L - 12L = 1908L

1908L will fill:

1908L×\frac{1balloon}{3L} = <em>636 balloons</em>

<em></em>

I hope it helps!

7 0
4 years ago
What process helps to purify water in nature?
ryzh [129]
Rocks charcoal and sand could help
4 0
3 years ago
Read 2 more answers
If the freezing point of an aqueous 0.10 m glucose solution is −x°c, what is the approximate freezing point of a 0.10 m nacl sol
maks197457 [2]
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m -  molality, moles of solute per kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
8 0
3 years ago
Refer to the electron distribution to answer 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹. What would be the period or series number of
Salsk061 [2.6K]

Answer:

C.5

Explanation:

A number of electrons present in valence shell of penultimate Shell represents the group of elements.

For s block elements: no.of group=number of valence shell electron.

p block elements: no. of group= 2 + 10 + number of valence electrons.

d block elements: no. of group= number(n-1)d electrons + number of electrons in nth shell.

Here, the differential electron is in p orbital hence, it belongs to p block

No. of group= 2+ 10 + 3=15 i.e 15th group or VA group.

5 0
3 years ago
Read 2 more answers
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