Answer:The molar heat of vaporization of molten lead is 178.03 kJ/mol.
Explanation:
The molar heats of fusion of lead
...(1)
The molar heats of sublimation of lead
...(2)
Adding (1) and (2)
The molar heat of vaporization of molten lead is 178.03 kJ/mol.
Answer:
0.373 moles of ammonium carbonate
Explanation:
To solve this question we must find the molar mass of ammonium carbonate. With the molar mass and the mass we can find its moles, as follows:
(NH₄)₂CO₃ contains: 2 moles N, 8 moles H, 1 mole C, 3 moles O. Molar mass:
2N = 14.0g/mol*2 = 28.0
8H = 1.0g/mol*8 = 8.0
1C = 12.0g/mol*1 = 12.0
3O = 16.0g/mol*3 = 48.0
Molar mass: 28.0 + 8.0 + 12.0 + 48.0 = 96.0g/mol
The moles of ammonium carbonate in 35.8g are:
35.8g * (1mol / 96.0g) =
<h3>0.373 moles of ammonium carbonate</h3>
Each mole of a substance contains 6.02 x 10²³ particles
Atoms of Fe = 4.5 x 6.02 x 10²³
= 2.709 x 10²⁹ atoms
Answer:
10moles of kcl
Explanation:
2
K
C
l
O3 → 2
K
C
l + 3
O
2
Notice that you have a 2
:
3 mole ratio between potassium chlorate and oxygen gas, which means that, regardless of how many moles of the former react, you'll always produce 2/3 times more moles of the latter.
15 mol of O2 * ((2mol of KCLO3)/(3mol of O2))= 15*2/3=10 Mol
