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yKpoI14uk [10]
3 years ago
5

Suppose you throw a football upward. Is it true or wrong to say that you have not increased the internal energy of the air withi

n the football
Physics
1 answer:
Ahat [919]3 years ago
5 0
Wrong because the energy would decrese
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I would say by putting two fingers under your chin or putting two fingers on the back of your wrist, hope i helped ! :)
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What are close-toed shoes least likely to provide protection against?
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Hydrogen gas is harmless to your feet so since you don’t need protection against it that seems the best answer.
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2 years ago
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for the same type of atom with 8 protons, 9 neutrons, and 10 electrons, what type of atom or ion is it​
Ilia_Sergeevich [38]

Answer:

O²⁻

Explanation:

Number of protons  = 8

Number of neutrons  = 9

Number of electrons  = 10

What type of atom or ion is it = ?

Solution:

Protons are the positively charged particle in an atom

Neutrons do not carry any charges

Electrons are negatively charged particles

 For this atom, the number of protons helps to identify what specie it is; so this is an oxygen atom.

Now,

     Charge  = Number of protons  - Number of electrons

      Charge  =  8  - 10  = -2

The charge on the atom is -2 and so it is an oxygen ion with -2 charge

 The ion is O²⁻

8 0
2 years ago
If chris throws a baseball 60 meters in 4 seconds, what is the average speed of the football?
IceJOKER [234]
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6 0
3 years ago
The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization
ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65  (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

6 0
3 years ago
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