Answer:
Post it on another page I can try to help.
Answer: The molar solubility of barium fluoride is 0.0183 moles/liter.
Explanation:
The equation for the reaction will be as follows:
By Stoichiometry,
1 mole of 
gives 2 moles of 
and 1 mole of 
Thus if solubility of is s moles/liter, solubility of is s moles/liter and solubility of is 2s moles/liter
Therefore,
![K_sp=[Ba^{2+}][F^{-}]^2](https://tex.z-dn.net/?f=K_sp%3D%5BBa%5E%7B2%2B%7D%5D%5BF%5E%7B-%7D%5D%5E2)
![2.45\times 10^{-5}=[s][2s]^2](https://tex.z-dn.net/?f=2.45%5Ctimes%2010%5E%7B-5%7D%3D%5Bs%5D%5B2s%5D%5E2)
Thus the molar solubility of barium fluoride is 0.0183 moles/liter.
A- Eat more beans
yea i need like 20 characters so this is so random
The answer is (CH₃)₂CH-CH(CH₃)₂.
That is the free radical generated in bromination of propane is (CH₃)₂CH° and CH₃CH₂CH₂° , but due to more stability of secondary free radical , (CH₃)₂CH° is more stable than that of CH₃CH₂CH₂°, so reaction proceeds via (CH₃)₂CH° radical. and than at termination the the product formed will be (CH₃)₂CH-CH(CH₃)₂.
Answer:
ΔH°f C₂H₅O₂N(s) = -537.2kJ
Explanation:
Based on the reaction:
4 C₂H₅O₂N(s) + 9O₂(g) → 8CO₂(g) + 10H₂O(l) + 2N₂(g)
ΔHrxn = ΔH°f products - ΔH°f reactants.
As:
ΔH°fO₂(g) = 0
ΔH°fCO₂(g) = -393.5kJ/mol
ΔH°fH₂O(l) = -285.8kJ/mol
ΔH°fN₂(g) = 0
The ΔHrxn is:
ΔHrxn = (8×-393.5kJ/mol + 10×-285.8kJ/mol) - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-6006kJ/mol - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-4×ΔH°fC₂H₅O₂N(s) = 2149kJ/mol
ΔH°fC₂H₅O₂N(s) = 2149kJ/mol / -4
<h3>ΔH°f C₂H₅O₂N(s) = -537.2kJ</h3>
