Answer: 1 mole ➡️ 6.022×10²³ atoms of si.
X mole ➡️ 2.8×10²⁴ atoms of si.
X = 2.8×10×10²³/6.022×10²³
= 28/6.022
= 4.65 moles.
Explanation:
Answer:
5.7*10^4 is equal to 57,000.
Explanation:
First, we must multiply 10 by its power, 4. That would be 10 4 times.
10*10*10*10 = 10,000.
Then, we multiply it by 5.7.
5.7*10,000 = 57,000.
Regards!
1. CaCr2O7
2. Sodium sulfite
3. (NH4)2SO3
4. Copper (II) nitrite
Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
Answer:
See explanation
Explanation:
A titration involves the addition of a titrant to an analyte solution. It is a method of volumetric analysis.
When a particular volume of titrant is added, the colour changes to signal the end point of the reaction.
The point at which the colour changes is called the equivalence point. This is the point at which the amount of titrant added is just enough to completely neutralize the analyte solution.
Hence the volume NaOH that needs to be added to the beaker containing HCl to cause a colour change is the volume of NaOH that is just enough to completely neutralize the HCl solution.
