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4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3 is the reaction for preparation of silicon by the reduction of K₂SiF6 with Al.
AlF3xH2O-based inorganic compounds are referred to as aluminium fluoride. They are all solids without colour. Aluminium fluoride is a crystalline (sand-like), odourless, white, or colourless powder. In addition to being used to make aluminium, it also functions as a flux in welding processes and in ceramic glazes and enamels.
Silicon (Si) is created by reducing potassium silicofluoride with aluminium as the reducing agent (K2SIF6). While K2SiF6 is reduced to Si in this equation, aluminium is oxidised to aluminium fluoride. As a result, the balanced equation describing aluminum's reduction of K2SiF6 to silicon non-metal is as follows: 4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3
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Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :

where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X= 
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)

Therefore, the molal freezing point depression constant of X is 
Answer:
Explanation:
What would the answer be?
We will use this formlula: Mass in grams = Number of moles x Molecular mass of 1 mole.
Since, we know the avagadro number is 6.02 x 10²³, we only have two unknown values left which are the molecular mass of CH3OH and its mole.
Molecular Mass: C = 12, H= 1, O = 16, since we have C=12, H4 = 4, O = 16, we will add them up: 12 + 4 + 16 =32
We know that one mole of anything = 6.02 x 10²³.
So we will use this formula to find the mole of methanol: Number of moles = Number of molecules / Avagadro number
Number of moles of CH3OH = (9.79 x 10^24)/6.02 x 10²³) = 16.263 moles.
Now we know that the molecular mass = 32 and the mole is = 16.263.
Now we can find its mass by using this formula: <span>Mass in grams = Number of moles x Molecular mass of 1 mole.
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Mass in grams = 16.263 x 32 = 520g