The number of atoms present in 0.58 mole of magnesium, Mg is 3.49×10²³ atoms
<h3>Avogadro's hypothesis </h3>
1 mole of Mg = 6.02×10²³ atoms
<h3>How to determine the atoms in 0.58 mole of Mg </h3>
1 mole of Mg = 6.02×10²³ atoms
Therefore,
0.58 mole of Mg = 0.58 × 6.02×10²³
0.58 mole of Mg = 3.49×10²³ atoms
Thus, 3.49×10²³ atoms are present in 0.58 mole of Mg
Learn more about Avogadro's number:
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Answer:
12 moles Pb(NO₄)₂ needed.
Explanation:
3Pb(NO₃)₂ +2AlCl₃ => 3PbCl₂ + 2Al(NO₃)₃
Given => ? moles 8 moles
from reaction stoichiometry, 2 moles AlCl₃ requires 3 moles Pb(NO₄)₂ then 8 moles AlCl₃ requires 3/2(8) moles of the Pb(NO₄)₂ => 12 moles Pb(NO₄)₂ needed.
Hey there!:
The reaction is as follows:
N2(g)+ 3 H2(g) ⇌ 2 NH3(g)
At equilibrium, the concentrations of the different species are as follows.:
[NH3] = 0.105 M
[N2] = 1.1 M
[H2] = 1.50 M
The equilibrium constant for the reaction is given as follows:
Keq = [NH3]² / [N2] [H2]³
Keq = (0.105)² / [1.1] [1.50]³
Keq = 0.00296 or 0.0030
The equilibrium constant for the reaction at this temperature is 0.0030.
Hope that helps!
