Answer:
Some formulas for calculating mole are
Mole = Mass/ Molar mass
Mole = no of particles / avogadros constant
NB : no of particles can be no of atoms , no of ions , or no of molecules 2. Avogadros number or constant = 6.02 times 10 ^23
so we will be using the second formula
Mole = no of particles / avogadros constant
Mole = 5.03 x 10 ^23/6.02 x10^23
Mole = 8.355x10^45
hope it helps :)
Explanation:
I think it's B. Molecules collide more frequently
Answer:
D. 0.75 grams
Explanation:
The data given on the iridium 182 are;
The half life of the iridium 182, 
= 15 years
The mass of the sample of iridium, N₀ = 3 grams
The amount left, N(t) after two half lives is given as follows;
For two half lives, t = 2 ×
∴ t = 2 × 15 = 30
∴ The amount left, N(t) = 0.75 grams
Answer: hello some part of your question is missing below is the missing part
when H₂O and H₂O₂ is added to Mn(OH)₂(s) and put in water bath to dissolve
answer : attached below
Explanation:
When Mn²⁺ ions are separated from the mixture, attached below are the requires reaction equations that shows the process of separation.
Mn²⁺ ions are separated to the right of the reaction equations
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
