The total kinetic energy of the gas sample is 3.3 KJ
<h3>What is kinetic energy? </h3>
This is the energy possessed by an object in motion. Mathematically, it can be expressed as:
KE = ½mv²
Where
- KE is the kinetic energy
- m is the mass
- v is the velocity
<h3>How to determine the mass of the fluorine gas</h3>
- Molar mass of fluorine gas = 38 g/mol
- Mole of fluorine gas = 1 mole
- Mass of fluorine gas = ?
Mass = mole × molar mass
Mass of fluorine gas = 1 × 38
Mass of fluorine gas = 38 g
<h3>How to determine the KE of the gas sample</h3>
- Mass (m) = 38 g = 38 / 1000 = 0.038 Kg
- Velocity (v) = 415 m/s
- Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.038 × 415²
KE = 3272.275 J
Divide by 1000 to express in kilojoule
KE = 3272.275 / 1000
KE = 3.3 KJ
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The balanced reaction
is:
4NH3 + 3O2 --> 2N2 + 6H2O
<span>We
are given the amount of reactants to be used for the reaction. This
will be the starting point of our calculation.</span>
83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2
2.81 moles of NH3
From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.
<span>2.62 mol O2</span><span> (6 mol H2O / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>
Answer:
The lithosphere is made up of pieces of tectonic plates. These plates are constantly changing and move towards the mantle. Non-stop movement of tectonic plates causes stress on the earth's outermost layer i.e,the crust. When these stresses extends it leads to cause cracks called faults.
Explanation:
Answer:
<h2>number of moles of solute dissolved in one liter of solution. </h2>
1. 8.28
Explanation:
I'm not sure but I hope it's help
Answer:
The volume on the tank is 6, 20 L
Explanation:
We use the formula PV=nRT. We convert the units of pressure in kPa into atm and temperature in Celsius into Kelvin:
0°C=273K
101,325kPa---1 atm
275kPa --------x=(275kPax 1 atm)/101,325kPa= 2,71 atm
PV=nRT --> V=nRT/P
V= 0,750 mol x 0,082 l atm /K mol x 273 K/ 2, 71 atm= <em>6, 20 L</em>
