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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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At 850K, 65L of gas has a pressure of 450kPa. What is the volume (in liters) if the gas is cooled to 430K and the pressure decre
Novosadov [1.4K]

Answer:

V₂ =  45.53 L

Explanation:

Given data:

Initial temperature = 850 K

Initial volume = 65 L

Initial pressure = 450 KPa

Final temperature = 430 K

Final pressure = 325 KPa

Final volume = ?

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 450 KPa× 65 L × 430 K / 850 K × 325KPa  

V₂ = 12577500 KPa .L. K / 276250 K. KPa

V₂ =  45.53 L

8 0
3 years ago
In what two ways can an object possess energy?
-BARSIC- [3]

Explanation:

An object can possess energy in tow ways by it's motion or position

5 0
3 years ago
Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric
kaheart [24]

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas  is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

pressure\propto \frac{1}{volume} (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure  is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

7 0
3 years ago
P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:
mash [69]

Answer:

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4(-1279) - [6(-286) - 3110]

= -5116 -(-1716-3110)

= -5116-(-4826)

= -5116 + 4826 = -290KJ/mol

6 0
3 years ago
Question 8 (10 points)
I am Lyosha [343]

Answer:

positively charged elctrons

4 0
3 years ago
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