In the given above, we have two densities which are 0.89 g/mL and 0.72 g/mL. We are also given that the liquids are immiscible. After the settlement of the liquids, they will form two layers.
The heavier substance, the one which has a higher density will be at the bottom and the lighter substance, the one which has a lower density will be at the top layer.
Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate
Answer:
it gets rust and could not run properly
