Two steps by inspection 1 qt = 0.25 gallons, 13.6 g/mL = 13.6 kg/L
0.25 gallon x 3.785411784 L/gallon x 13.6 kg/L x 1 lb/0.45359237 kg = 28.374 lb.
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Answer: Concentration of 
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of 
= 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture = 
Thus concentration of in the equilibrium mixture is 0.31 M
Answer:
0.032 L or 32 mL
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 9.0 M
V1 = This is what we're looking for.
M2 = 0.145 M
V2 = 2 L
Solve for V1 --> V1 = M2V2/M1
V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L
All alkail metals react with halogens vigorously to produce salt. The alkali metal would loose an electron to form and ion with the halogen
