true gas molicules bounce off one another much faster liquid just slides
The correct question is
Give the coordination number, the charge of the central metal ion, and select the correct name in each coordination compound:
Na3[CoCl6]
[Ni(CO)4]
[Ni(NH3)3(H2O)2] (NO3)2
Answer:
See explanation for details
Explanation:
Na3[CoCl6]
Name: Sodium hexacholorocobalt III
Charge on the complex: -3
Central metal ion name : cobalt III
coordination number: 6
[Ni(CO)4]
Name: tetracarbonyl nickel (0)
Charge on the complex: 0
Central metal ion name : nickel (0)
coordination number: 4
[Ni(NH3)3(H2O)2] (NO3)2
Name: Diaquatriaminenickel II nitrate
Charge on the complex: +2
Central metal ion name : nickel (+2)
coordination number: 5
Producer. Hope this helps!
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:
The concentration of the solution will be much lower than 6M
Explanation:
To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.
From
n= CV
n = number of moles m/M( m= mass of solid, M= molar mass of compound)
C= concentration of substance
V= volume of solution
m=120g
M= 40gmol-1
V=500ml
120/40= C×500/1000
C= 120/40× 1000/500
C=6M
This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.
This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.
