Answer:
See the attached file for the structure.
Explanation:
Find attached for the explanation
Answer:
The answer to your question is: SiCl₄
Explanation:
Data
amount of Si 1.71 g
amount of Cl 8.63 g
MW Si = 28 g
MW Cl = 35.5
Process (rule of three)
For Si For Cl
28 g of Si ------------------ 1 mol 35.5 g of Cl --------------- 1 mol
1.71g of Si --------------- x 8.63 g of Cl -------------- x
x = 1.71 x 1 / 28 = 0.06 mol x = 8.63 x 1 / 35.5 = 0.24 mol
Now, divide both results by the lowest of them.
Si = 0.06 mol / 0.06 = 1 molecule of Si Cl = 0.24 / 0.06 = 4 molecules of Cl
Finally
Si₁ Cl₄ or SiCl₄
There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
C = 4 mol/l
v = 0.5 l
n(NaCl)=cv
n(NaCl) = 4 mol/l · 0.5 l = 2 mol
2 moles of NaCl must be dissolved
