Answer:
The question isn't worded properly, but if 1 or 2 are DECREASED, the frequency of collisions of specified molecules will decrease.
Explanation:
Catalysts only facilitate reaction once molecules collide. Increased temperature makes molecules move more, and thus collide more. For concentration, if there are more molecules in the same amount of room/liquid, there will be more collisions because there are more of the molecules to collide.
You need to do something like that your self so sorry can help.
Answer:
18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
Explanation:
Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.
The decomposition of mercury oxide is given by the chemical equation below:
2HgO ----> 2Hg + O₂
2 moles of HgO decomposes to produce 1 mole of Hg
2 moles of HgO has a mass of 433.2 g
433.2 g of HgO produces 216.6 g of Hg
18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg
Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
Answer:
0.825 M
Explanation:
The osmotic pressure is a colligative property, that can be calculated using the following expression.
π = M × R × T
where,
π is the osmotic pressure
M is the molarity
R is the ideal gas constant
T is the absolute temperature (24°C + 273 = 297 K)
M = π / R × T = 20.1 atm / (0.08206 atm.L/mol.K) × 297 K = 0.825 M
