Answer: Sediment Transport by Wind
Explanation: Like flowing water, wind picks up and transports particles. Wind carries particles of different sizes in the same ways that water carries them (Figure below). Tiny particles, such as clay and silt, move by suspension. They hang in the air, sometimes for days.
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
Answer:
0.302 moles
Explanation:
Data given
Mass of Pb(NO₃)₂ = 100 g
Moles of Pb(NO₃)₂ = ?
Solution:
To find mole we have to know about molar mass of Pb(NO₃)₂
So,
Molar mass of Pb(NO₃)₂ = 207 + 2[14 + 3(16)]
= 207 + 2[14 + 48]
= 207 + 124
Molar mass of Pb(NO₃)₂ = 331 g/mol
Formula used :
no. of moles = mass in grams / molar mass
Put values in above formula
no. of moles = 100 g / 331 g/mol
no. of moles = 0.302 moles
no. of moles of Pb(NO₃)₂ = 0.302 moles
