Answer: 0.714 moles of 
will be produced from 237.1 g of potassium iodide
Explanation:
To calculate the moles :
The balanced chemical reaction is:
According to stoichiometry :
2 moles of 
produce = 1 mole of
Thus moles of will require=
of
Thus 0.714 moles of will be produced from 237.1 g of potassium iodide
Molality is defined as the number of moles of solute dissolved in 1 kg of solvent.
To calculate molality, we need to calculate the number of moles of CaCl₂.
Mass of CaCl₂ - 5.0 g
Molar mass of CaCl₂ - 111 g/mol
The number of moles of CaCl₂ - 5.0 g / 111 g/mol = 0.045 mol
we need to then calculate the number of moles in 1 kg solvent.
number of CaCl₂ moles in 500 g water - 0.045 mol
Therefore number of moles in 1 kg water - 0.045 mol / 500g x 1000 g = 0.090 mol
Molality of CaCl₂ - 0.090 mol/kg
Answer: The correct answer is (A).
Explanation:
Potassium-trisethylene-diamine chromate(III)
Potassium is coming first in name which means that potassium is present as cation.
In the complex, 3 molecules of 'ethylene-diamine' is present as an dianion and forms a complex with Chromium (III) metal, making the overall charge on the coordination sphere as (-3).
So, to make a neutral complex, 3 potassium ions are required.
Hence, the structure of the complex is ![K_3[Cr(NH_2CH_2CH_2NH_2)_3]](https://tex.z-dn.net/?f=K_3%5BCr%28NH_2CH_2CH_2NH_2%29_3%5D)
I feel like it would be O.40 mol Mgo but I could be wrong
Answer:
CHCl₃
Explanation:
We have the following data:
C = 5.03 g
H = 0.42 g
Cl= 44.5 g
First, we divide each mass by the molar mass (MM) of the chemical element to calculate the moles:
MM(C) = 12 g/mol
moles of C = mass/MM(C) = 5.03 g/(12 g/mol) = 0.42 mol C
MM(H) = 1 g/mol
moles of H = mass/MM(H) = 0.42 g/(1 g/mol) = 0.42 mol H
MM(Cl) = 35.4 g/mol
moles of Cl = mass/MM(Cl) = 44.5 g/(35.4 g/mol) = 1.26 mol Cl
Now, we divide the moles by the smallest number of moles (0.42):
0.42 mol C/0.42 = 1 C
0.42 mol H/0.42 = 1 H
1.26 mol Cl/0.42 = 3 Cl
Thus, the C:H:Cl ratio is 1:1:3.
Therefore, the empirical formula is CHCl₃
