There are 424 grams of oxygen in 477 grams of water. 8 times 53 equals 424.
25.78g
if 14 grams of nitrogen is formed, then 8 grams of oxygen, add those two together and you get 22. and that's 22 of the 40 grams used, so 40 subtracted by 22 is 18. 18 grams of water would be formed.
2.55 grams = 2,550 grams
2068 grams
To determine the grams of aluminum hydroxide obtained from 17.2 grams of aluminum sulfide, we need to consider the stoichiometry of the reaction between aluminum sulfide and water to form aluminum hydroxide. Given the balanced chemical equation, we can calculate the molar mass of aluminum hydroxide and use it to convert the mass of aluminum sulfide to grams of aluminum hydroxide formed.
16 grams of oxygen how many moles is 0,5 moles.
The answer is 224,141 grams oxygen.
320 grams of oxygen is the equivalent of 10 moles.
There are 192 grams of oxygen in 6 moles of O2.
There are 424 grams of oxygen in 477 grams of water. 8 times 53 equals 424.
12.8 grams oxygen (1 mole O/16.0 grams) = 0.800 moles of oxygen
There are three oxygen atoms in 3.45 grams of water (H2O).
1,9 grams
We need 14,8 g oxygen.
There are 0.47 grams of oxygen in 0.53 grams of NaHCO3.
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.