Explanation:We have momemtum = mass X velocity
p = mv
OR, p/m = v
v = (125kg m/s)25kg
v = 125/25 m/s
v = 5 m/s
Answer:
it's a segment
Explanation:
it has multiple end points
Answer:
Dude im not 100% sure but I think its b and c im sorry if im wrong its just that im not really sure which ones are.
Explanation:
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
ANSWER : 108 + 10 = 118
118 + ( 5 + 3 )
118 + 8 = <u>1</u><u>2</u><u>6</u>
<u>=</u><u> </u><u>1</u><u>2</u><u>6</u><u> </u><u>g</u>