If you clear volume in the density equation:
![\rho = \frac{m}{V}\ \to\ V = \frac{m}{\rho}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7Bm%7D%7BV%7D%5C%20%5Cto%5C%20V%20%3D%20%5Cfrac%7Bm%7D%7B%5Crho%7D)
The greater the density the lower the volume. This means, the volume of gold nugget will be smaller than the volume of iron pyrite nugget.
![V_{gold} = \frac{m}{\rho} = \frac{50\ g}{19.3\ g/cm^3} = \bf 2.59\ cm^3](https://tex.z-dn.net/?f=V_%7Bgold%7D%20%3D%20%5Cfrac%7Bm%7D%7B%5Crho%7D%20%3D%20%5Cfrac%7B50%5C%20g%7D%7B19.3%5C%20g%2Fcm%5E3%7D%20%3D%20%5Cbf%202.59%5C%20cm%5E3)
Answer:
Explanation:according to question:
. Nacl (aq) + AgNO3 (aq) --> AgCl
(s) + NaNO3 (aq).balanced.
Answer:
what is the baby flowers color?
Explanation:
if the baby flower is purple the parents were purple
Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g