Mass SrBr2 = 247.42 g/mol x 1.36 mol
<span>= 336 g</span>
Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation:
The number of hydrogen atoms that are in 4.40 mol of ammonium sulfide is 2.12 x10^25 atoms
calculation
find the number of moles of Hydrogen in ammonium sulfide (NH4)2S
that is 4.40 x number of hydrogen atoms in (NH4)2S ( 4x2= 8 atoms)
moles is therefore= 4.40 x8= 35.2 moles
by use of Avogadro's law constant
that is 1mole = 6.02 x10^23 atoms
35.2 moles=?
by cross multiplication
{35.2 moles x 6.02 x10^23} /1 mole = 2.12 x10^25 atoms
It’s deceleration hope that helps!
