precipitate is the answer
Answer:
<em> = 0.2 mL.</em>
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
=
× 0.010
<em> = 0.2 mL.</em>
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
MgqI% = the formula and mass for H20 and that should be your answer for water
<h2>
Answer: 6 moles</h2>
<h3>
Explanation:</h3>
3 H₂ + N₂ → 2 NH₃
↓ ↓
4 mol 3 mol
Since the moles of N₂ is the smaller of the two reactants, then N₂ is the limiting factor (the reactant that will decide how much ammonia is produced since it has the smaller amount of moles). ∴ we have to use it in calculating the number of moles of ammonia
The mole ratio of N₂ to NH₃ based on the balanced equation is 1 to 2.
∴ the moles of NH₃ = moles of N₂ × 2
= 3 moles × 2
= 6 moles