Rutherford used gold for his scattering experiment because gold is the most malleable metal and he wanted the thinnest layer as possible. The goldsheet used was around 1000 atoms thick. Therefore, Rutherford selected a Gold foil in his alpha scatttering experiment.
Answer:
14.4g
Explanation:
First, we need to write a balanced equation for the reaction between Fe and O2 to produce Fe2O3. This is illustrated below:
4Fe + 3O2 —> 2Fe2O3
From the balanced equation,
4moles of Fe produced 2moles of Fe2O3.
Therefore, 0.18mol of Fe will produce = (0.18x2) /4 = 0.09mol of Fe2O3.
Now we need to find the mass present in 0.09mol of Fe2O3. This can be achieved by doing the following:
Molar Mass of Fe2O3 = (56x2) + (16x3) = 112 + 48 = 160g/mol
Number of mole of Fe2O3 = 0.09mol
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Fe2O3 = 0.09 x 160 = 14.4g
Mn₂O
Explanation:
The oxide that will most likely form colored solutions is Mn₂O.
This is because most transition metals form colored compounds. Manganese is a transition metal belonging to the d-block on the periodic table.
- Other examples of transition metals are scandium, titanium, iron, copper, cobalt, nickel, zinc
- They belong to the d-block on the periodic table.
- They have variable oxidation states.
- Most of their solutions are always colored.
Learn more:
Periodic table brainly.com/question/8543126
#learnwithBrainly
68000 = 6.8 * 10000 = 6.8 * 10^4
hope this helps? c;
