Answer:
Product
Explanation:
It is written at the right side of an arrow
Answer:
Explanation:
Hello,
In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:
The concentrations of the A and B ions in the solution are:
![[A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M](https://tex.z-dn.net/?f=%5BA%5D%3D0.099%20%5Cfrac%7BmolAB_3%7D%7BL%7D%2A%5Cfrac%7B1molA%7D%7B1molAB_3%7D%20%20%3D0.0099M)
![[B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M](https://tex.z-dn.net/?f=%5BB%5D%3D0.099%20%5Cfrac%7BmolAB_3%7D%7BL%7D%2A%5Cfrac%7B3molB%7D%7B1molAB_3%7D%20%20%3D0.000.297M)
Then, as the solubility product is defined as:
![Ksp=[A][B]^3](https://tex.z-dn.net/?f=Ksp%3D%5BA%5D%5BB%5D%5E3)
Due to the given dissociation, it turns out:
![Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}](https://tex.z-dn.net/?f=Ksp%3D%5B0.099M%5D%5B0.297M%5D%5E3%5C%5C%5C%5CKsp%3D2.59x10%5E%7B-3%7D)
Regards.
The reaction is:
NH4 (NO3) (s) ⇄ N2O (g) + 2 H2O (g)
This means that 1 mol of NH4 (NO3)s produces 3 moles of gases.
Now find the number of moles in 1.71 kg of NH4 (NO3)
Molar mass = 2*14g/mol + 4 * 1g/mol + 3*16g/mol = 80 g/mol
# moles = mass / molar mass = 1710 g / 80 g/mol = 21.375 mol of NH4(NO3)
We already said that every mol of NH4(NO3) produces 3 moles of gases, then the number of moles of gases produced is 3 * 21.375 = 64.125 mol
Now use the equation for ideal gases to fin the volume
pV = nRT => V = nRT / p = (64.125 mol)(0.082atm*liter / K*mol) * (119 +273)K / (731mmHg *1 atm/760mmHg) =
V = 2143.01 liters
Answer:
Chlorine can be found in Period 3.
Explanation:
Periods on the periodic table are the rows (horizontal sections) that group elements together. By looking across Period 3, you will find Cl (chlorine).
