Answer:
Different atmospheric pressure. When there is a different atmospheric pressure, air moves from the higher pressure to the lower pressure area which results in what you call <u>WIND</u> but can result in various speeds and pressure.
Hope this helped and if it did, please give my answer a brainliest.
Answer:
N2
Explanation:
Rate of effusion is defined by Graham's Law:
(Rate 1/Rate 2) = (sqrt (M2)/ sqrt (M1))
(Where M is the molar mass of each substance. )
Molar Mass of oxygen, O2, is 32 (M1).
Rate of effusion of O2 to an unknown gas is .935(Rate 1).
Rate 2 is unknown so put 1.
Solve for x (M2).
.935/1 = sqrt x/ sqrt32
.935 x sqrt 32 = sqrt x
5.29 = sq rt x
5.29^2 = 27.975 = 28
N2 has a molar mass of 28 so it is the correct gas.
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O =
= 0.13moles
Number of moles of NH₃ =
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g