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Ilya [14]
3 years ago
9

An electron na hydrogen atom moves from level to level 4 The electron then drops from level 4 to level 2. Which statement

Chemistry
1 answer:
Vsevolod [243]3 years ago
3 0

Answer:

The energy absorbed in the first move is greater than the energy released in the second move.

Explanation:

The greater the level number the greater the energy.

A movement up a level requires energy so energy id absorbed.

Dropping down a level releases energy.

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A compound's properties are different than the properties of the elements that make it up.
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The answer is True because elements in a compound combine and become an entirely different substance with its own unique properties.
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What volume and mass of steam at 100 Celsius and 760. torr (or 1 atm) would release the same amount of energy as heat during con
erastova [34]
  
<span>As we know that
1 cu cm H2O = 1 mL H2O = 1g H2O 
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Heat of fusion of water = 79.8 cal/g 
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Heat of vaporization of water = 540 cal/g 

Atomic weight of water : H=1 O=16 H2O=18 
now by calculating and putting values

65.5gH2O x 79.8cal/gH2O x 1gH2O/540cal = 9.68g H2O (steam) 

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3 0
3 years ago
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The pH of a vinegar solution is 4.15. What is the H+ concentration of the solution
marishachu [46]
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A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

5 0
3 years ago
1 point If the boy is pushing with 50N of force and the static friction resistance is 70N of force, what will happen? *​
solniwko [45]
Nothing, he shouldn’t be able to move it. Think about it like this say you try really hard to push something that is 5,000 pounds and you push as hard as you can. Well you can’t move it bc it weighs more than you can push. I’m sure their is a equation you can use to see how much you can push (body weight=force?)
6 0
3 years ago
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