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eduard
3 years ago
8

Nitrogen increases volume during what physical change

Chemistry
1 answer:
atroni [7]3 years ago
4 0
Effects of changes in volume in a reversible reaction in a chemical equilibrium can be predicted using Le Chatelier's Principle. I think this might be the answer, I hope it helps.
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Plz help soon? I’m confused
nata0808 [166]
Hi,
I think the answer is metric.
6 0
3 years ago
A mixture of H2 and water vapor is present in a closed vessel at 20°C. The total pressure of the system is 755.0 mmHg.
Masja [62]

THE ANSWER IS: <u>737.5</u>

I JUST TOOK THE QUIZ!!!!

7 0
3 years ago
Read 2 more answers
.A gas occupies 25,3 mL at a pressure of 152 kPa. Find the volume if the pressure is
Strike441 [17]

Answer:

47.36mL

Explanation:

Using Boyles law equation, which states that:

P1V1 = P2V2

Where;

V1 = initial volume (mL)

V2 = final volume (mL)

P1 = initial pressure (atm)

P2 = final pressure (atm)

Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm

First, we need to convert 152kPa to atm by dividing by 101

1kPa = 0.0099atm

152kPa = 1.505atm

P1V1 = P2V2

1.505 × 25.3 = 0.804 × V2

38.08 = 0.804V2

V2 = 38.08/0.804

V2 = 47.36mL

5 0
2 years ago
He isotope 62(over28ni has the largest binding energy per nucleon of any isotope. calculate this value from the atomic mass of n
irina [24]
The atomic mass of the isotope Ni ( 62 over 28 ) = 61.928345 amu.
Mass of the electrons: 28 · 5.4584 · 10^(-4 ) amu = 0.0152838 amu ( g/mol )
Mass of the nuclei:
61.928345 amu - 0.0152838 amu = 61.913062 amu (g/mol)
The mass difference between a nucleus and its constituent nucleons is called the mass defect.
For Ni ( 62 over 28 ): Mass of the protons: 28 · 1.00728 amu = 28.20384 amu
Mass of the neutrons: 34 · 1.00866 amu = 34.299444 amu
In total : 62.49828 amu
The mass defect = 62.49828 - 61.913062 = 0.585218 amu
Nucleus binding energy:
E = Δm · c² ( the Einstein relationship )
E = 0.585218 · ( 2.9979 · 10^8 m/s )² · 1 / (6.022 · 10^23) · 1 kg / 1000 g =
= 0.585218 · 8.9874044 · 10 ^16 : (6.022 · 10^23) · 0.001 =
= ( 5.2595908 : 6.022 ) · 0.001 · 10^(-7 ) =
= 0.0008733 · 10^(-7) J = 8.733 · 10^(-11) J
The nucleus binding energy per nucleon:
8.733 · 10^(-11) J : 62 =  0.14085 · 10 ^(-11) =
= 1.4085 · 10^(-12) J per nucleon.
4 0
3 years ago
2 1.1.4 Quiz: Amplitude, Frequency, and Speed
sammy [17]
The answer is C
Sbbsshhshsgssh
8 0
3 years ago
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