Answer:
3.1 x 10⁻²¹ Nm
Explanation:
When placed in an external electric filed, an electric dipole experiences a torque. and this torque is represented mathematically with the equation:
torque (τ) = dipole moment vector (P) x electric field vector (E)
τ = P. E . sin θ
where θ is the angle between the water molecule and the electric field, which in this case is 90° (because this is where the torque is maximum)
τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . sin90
τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . 1
solve for τ
τ = 3.1 x 10⁻²¹ Nm
the maximum possible torque on the water molecule is therefore 3.1 x 10⁻²¹ Nm
<h3>
Answer:</h3>
8CO₂
<h3>
Explanation:</h3>
We are given;
- Butane is a hydrocarbon in the homologous series known as alkane.
We are required to determine the other product produced in the combustion of butane apart from water.
- We know that the complete combustion of alkane yields carbon dioxide and water.
- Therefore, combustion of butane will yield carbon dioxide and water.
- The balanced equation for the complete combustion of butane will be;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Answer:
tri-
Explanation:
Examples could be Tri-angle, Tri-cycle, Tri-ceratops
Answer:
Moles NH₃: 0.0593
0.104 moles of N₂ remain
Final pressure: 0.163atm
Explanation:
The reaction of nitrogen with hydrogen to produce ammonia is:
N₂ + 3 H₂ → 2 NH₃
Using PV = nRT, moles of N₂ and H₂ are:
N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂
H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂
The complete reaction of N₂ requires:
0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>
That means limiting reactant is H₂. And moles of NH₃ produced are:
0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>
Moles of N₂ remain are:
0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>
And final pressure is:
P = nRT / V
P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L
<em>P = 0.163atm</em>