The solubility of potassium chloride in at room temperature is approximately 34 g per 100 g of water. Therefore, the maximum amount that could be dissolved would be 34/100 ( 200) = 68 g of KCl. When more than this amount is added, excess potassium would not dissolve forming crystals in the solution.
Answer:
46.3g H2O
Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)
then use factor label method to solve
82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
This is given by Avogagro number: 1 mol = 6.02*10^23 particles
Then you can do whichever to these two relations, because they are equivalent:
- 1mol / 6.02*10^23 representative particles, and
- 6.02*10^23 representative particle /1 mol
Only the second option of the question includes one of the valid conversion factors. Then, the conversion factor of the second option is the right answer
Answer:
The products are Calcium oxide and Carbon dioxide.
Explanation:
When calcium carbonate is heated, thermal decomposition occurs.
Calcium calcium → Calcium oxide + Carbon dioxide