How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead
1 answer:
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Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
At constant speed, we will have;
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
Answer:
Explanation:
One mole of a substance contains the same amount of representative particles. These particles can be atoms, molecules, ions, or formula units. In this case, the particles are atoms of titanium.
Regardless of the particles, there will always be <u>6.02*10²³</u> (also known as Avogadro's Number) particles in one mole of a substance.
Therefore, the best answer for 1 mole of titanium is D. 6.02*10²³ atoms.
Answer:
1. 24375 N/C
2. 2925 V
Explanation:
d = 12 cm = 0.12 m
F = 3.9 x 10^-15 N
q = 1.6 x 10^-19 C
1. The relation between the electric field and the charge is given by
F = q E
So,
E = 24375 N/C
2. The potential difference and the electric field is related by the given relation.
V = E x d
where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.
By substituting the values, we get
V = 24375 x 0.12 = 2925 Volt