Answer:
17.76g
Explanation:
We need to write a balanced chemical equation for the reaction:
2Al(OH)3 + 3Ca(NO3)2 ——> 3Ca(OH)2 + 2Al(NO3)3
In the reaction above, it can be seen that 2 moles of aluminum hydroxide yielded 3 moles of calcium hydroxide. This is the theoretical viewpoint.
Now we need to know what actually happened. We need to calculate the actual number of moles of aluminum hydroxide reacted l. We can get this by dividing the mass by the molar mass.
The molar mass of aluminum hydroxide is 27+ 3( 16+1)
= 27 + 51 = 78g/mol
The number of moles is thus: 12.55/78 = 0.16 moles
Now if 2 moles of aluminum hydroxide gave 3 moles of calcium hydroxide, 0.16moles will give : (0.16*3)/2 = 0.24moles
Now we can calculate the mass of calcium hydroxide formed. The mass of calcium hydroxide formed is the number of moles multiplied by the molar mass.
The molar mass of calcium hydroxide is; 40 + 2(17) = 74g/mol
The mass is thus =74 * 0.24 = 17.76g
Answer: Try mass that’s the best guess I can give off the top of my head
Explanation: An atom is the smallest unit of matter that retains all of the chemical properties of an element. Atoms come together which forms molecules. These molecules interact to form solids, gases, or liquids.
I’m sorry I just really need points rn I love you
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
Answer:
9.6 moles O2
Explanation:
I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.
The molar mass of water is 18.0 grams/mole.
Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).
The balanced equation states that: 2H20 ⇒ 2H2 +02
It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.
get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2
