Answer: A) 1.0 x 10-10 M
Explanation: The pH is 4.0, and pOH + pH = 14, so pOH = 10. [OH-] = 10^-pOH, so for this drink, [OH-] = 10^-10 M, or 1.0 x 10^-10 M.
Answer:
The final temperature will be close to 20°C
Explanation:
First of all, the resulting temperature of the mix can't be higher than the hot substance's (80°C) or lower than the cold one's (20°C). So options d) and e) are imposible.
Now, due to the high heat capacity of water (4,1813 J/mol*K) it can absorb a huge amount of heat without having a great increment in its temperature. On the other hand, copper have a small heat capacity (0,385 J/mol*K)in comparison.
In conclusion, the copper will release its heat decreasing importantly its temperature and the water will absorb that heat resulting in a small increment of temperature. So the final temperature will be close to 20°C
<u>This analysis can be done because we have equal masses of both substances. </u>
Answer:
Coefficients represents no of moles while subscripts represent no of atoms.
Answer:
Ca(OH)2 will not precipitate because Q<Ksp
Explanation:
Ksp for Ca(OH)2 has already been stated in the question as 8.0 x 10-8mol2dm-6
The value of the reaction quotient depends heavily on the concentration of the reactants. As the initial concentration of the calcium carbide decreases considerably, the reaction quotient decreases until Q<Ksp hence the Ca(OH)2 will not precipitate from solution.
The reaction equation is:
CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂
From
Ca(OH)2= Ca2+ + 2OH-
Concentration of solution= 0.064×1/64= 1×10-3
Since [Ca2+] = 1×10-3
[OH-]= (2×10-3)^2= 4×10^-6
Hence Q= 4×10^-9
This is less than the Ksp hence the answer.
