Percent composition by mass is calculated (mass of element within compound)/(mass of compound)*100. The lower the total molar mass of the compound, the greater the percent composition of sulfur. In this case, MgS would be that compound, since Mg has the lowest molar mass of the four elements bonded to S.
Given problem:
S₂O₈²⁻
Find the oxidation number of S;
Oxidation number presents the extent of oxidation of each atom of elements a molecular formular or formula unit or an ionic radical.
For radicals:
"the algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion "
S₂O₈²⁻; oxidation number of O is usually -2
2(S) + 8(-2) = -2
2S - 16 = -2
2S = -2 + 16
2S = 14
S = +7
The oxidation state of S in the radical is +7
Missing question:
<span>A. [PdZn(H2O)2(CO)2]Br4.
B. [Zn(H2O)2(CO)2]2[PdBr4].
C. [Pd(H2O)2][Zn(CO)2]Br4.
D. [Pd(H2O)2]2[Zn(CO)2]3Br4.
E. [Zn(H2O)2(CO)2][PdBr4].
</span>Answer is: E. [Zn(H2O)2(CO)2][PdBr4]..
In this complex diaqua means two waters (H₂O), <span>dicarbonyl means two carbonyl groups (CO), zinc(Zn) and palladium (Pd) are central atoms or metals, bromine has negative charge -1. Bromine, water and carbonyl are ligands.</span>
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
Answer:
1) 0.009 61 g C; 2) 0.008 00 mol C
Step-by-step explanation:
You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.
M_r: 12.01 44.01
C + ½O₂ ⟶ CO₂
m/g: 0.352
1) <em>Mass of C
</em>
Convert grams of CO₂ to grams of C
44.01 g CO₂ = 12.01 g C
Mass of C = 0.352 g CO₂ × 12.01 g C/44.01 g CO₂
Mass of C = 0.009 61 g C
2) <em>Moles of C
</em>
Convert mass of C to moles of C.
1 mol C = 12.01 g C
Moles of C = 0.00961 g C × (1 mol C/12.01 g C)
Moles of C = 0.008 00 mol C
All the carbon comes from Compound A, so there are 0.008 00 mol C in Compound A.
