1) mass composition
N: 30.45%
O: 69.55%
-----------
100.00%
2) molar composition
Divide each element by its atomic mass
N: 30.45 / 14.00 = 2.175 mol
O: 69.55 / 16.00 = 4.346875
4) Find the smallest molar proportion
Divide both by the smaller number
N: 2.175 / 2.175 = 1
O: 4.346875 / 2.175 = 1.999 = 2
5) Empirical formula: NO2
6) mass of the empirical formula
14.00 + 2 * 16.00 = 46.00 g
7) Find the number of moles of the gas using the equation pV = nRT
=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)
=> n = 0.01769 moles
8) Find molar mass
molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol
9) Find how many times the mass of the empirical formula is contained in the molar mass
92.14 / 46.00 = 2.00
10) Multiply the subscripts of the empirical formula by the number found in the previous step
=> N2O4
Answer: N2O4
Hey there!
Balance the equation:
SiCl₄ + H₂O → H₄SiO₄ + HCl
Balance H.
2 on the left, 5 on the right. Add a coefficient of 3 in front of H₂O and a coefficient of 2 in front of HCl.
SiCl₄ + 3H₂O → H₄SiO₄ + 2HCl
Balance O.
3 on the left, 4 on the right. Change the coefficient of 3 in front of H₂O to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 2HCl
This unbalanced our H, so change the coefficient of 2 in front of HCl to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Balance Cl.
4 on the left, 4 on the right. Already balanced.
Balance Si.
1 on the left, 1 on the right. Already balanced.
Our final balanced equation:
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Hope this helps!
The balanced equation is 2
AlI
3
(
a
q
)
+
3
Cl
2
(
g
)
→
2
AlCl
3
(
a
q
)
+
3
I
2
(
g
)
.
<u>Explanation:</u>
- Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- .
Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.
- Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.
Balancing the equation, we get:
2AlI
3( aq ) + 3Cl2
( g ) → 2AlCl3
( aq )
+ 3
I
2 ( g )
-
Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.
-
Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.
-
We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
Solubility data of a certain solute with a certain solvent is empirical. There are constant values for this at varying temperatures. For KCl in water at 25°C, the solubility is 35.7 g/100 mL of water. When you compare this with the solubility data of KCl with ethanol, this means that KCl is more soluble in water than in ethanol. This is true because KCl is an ionic salt which is very soluble in water.
