Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium
Heat gained by water = 100 × 4.186 × 30.5
= 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5
= 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
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
Hello.
The answer would be <span>A. Three feet of concrete.
Have a nice day.</span>
Answer:
Its a small dense spherical structure in the nucleus of a cell during interphase.
Answer:
2.94 x 10⁻¹⁴L
Explanation:
To solve this problem, we have to assume that the condition of this water is at standard temperature and pressure, STP.
At STP;
1 mole of gas have a volume of 22.4L
So, let us find the number of moles of this water first;
6.02 x 10²³ molecules can be found in 1 mole of a substance
7.89 x 10⁸ molecules will contain
= 1.31 x 10⁻¹⁵mole of water
So;
Volume of water = 22.4 x 1.31 x 10⁻¹⁵ = 2.94 x 10⁻¹⁴L