Force = Mass * Acceleration = 59 kg * 9.75 m/s^2 = 575.25 N
Answer:
a) The work done is 10.0777 kJ
b) The water's change in internal energy is -122.1973 kJ
Explanation:
Given data:
1 mol of liquid water
T₁ = temperature = 100.9°C
P = pressure = 1 atm
Endothermic reaction
T₂ = temperature = 100°C
1 mol of water vapor
VL = volume of liquid water = 18.8 mL = 0.0188 L
VG = volume of water vapor = 30.62 L
3.25 moles of liquid water vaporizes
Q = heat added to the system = -40.7 kJ
Questions: a) Calculate the work done on or by the system, W = ?
b) Calculate the water's change in internal energy, ΔU = ?
Heat for 3.25 moles:
The work done:
The change in internal energy:
The answer is transition metals because they have no specific charge except for 1 or 2 of them
<h3>Answers:</h3>
1) 2 Units of Ozone
2) 3 Units of Ozone
3) 9 Units of Ozone
<h3>Solution:</h3>
1) From 6 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
6 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 2 Units of Ozone
2) From 9 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
9 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 3 Units of Ozone
3) From 27 Oxygen Particles;
As given,
3 Oxygen Particles form = 1 Unit of Ozone
So,
27 Oxygen Particles will form = X Units of Ozone
Solving for X,
X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles
X = 9 Units of Ozone
Answer:
i got you dawg just gimme one sec i'll get to you fr g
Explanation:
