Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.
Amount of heat required to vaporize 1 mole of lead = 177.7 kJ
Molar mass of lead = 207.2 g
Mass of lead given = 1.31 kg = 1310 g (1kg=1000g)
Heat required to vaporize 207.2 of lead = 177.7 kJ
Thus Heat required to vaporize 1310 g of lead =
Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
Answer:
The molecular formula of estradiol is: .
Explanation:
Molar mass of of estradiol = M= 272.37 g/mol
Let the molecular formula of estradiol be
Percentage of an element in a compound:
Percentage of carbon in estradiol :
x = 18.0
Percentage of hydrogen in estradiol :
y = 24.2 ≈ 24
Percentage of oxygen in estradiol :
z = 2
The molecular formula of estradiol is:
Answer:
Explanation:
A sigma bond is formed when a hybrid orbital sp overlaps with another hybrid orbital or with s- or p- orbital
Ethylene has a structural formula of H - C≡ C- H
At the ground state; we have :
C | ⇅ | |↑ | ↑
2s 2p
At the excited state, we have:
C | ↑ | |↑ | ↑ | ↑
the hybrid orbital
C = | ↑ | ↑ |
2sp
Gay Lussac's Law
Convert:
200°C = 200 + 273 = 473 K
50°C = 50 + 273 = 323 K
Input the value: