1. Coefficient
2. Subscript
3. Products
4. Reactants
5. Balance
6. Combustion
7. Decomposition
8. Single replacement
9. Double replacement
10. Synthesis
I’m sorry if I’m wrong
Answer:
Explanation:
It is volume-volume problems that does not require the use of molar mass.
The value of ΔG° at this temperature is -18034.18 J/mol
Calculation,
Given information
formation constant (Kf)= 1.7 × 
Universal gas constant (R) = 8.314 J/K• mol
Temperature = 25° C = 25 °C + 273 = 300 K
Formula used:
ΔG° = -RT㏑Kf
By putting the valur of R,T, Kf we get the value of ΔG°
ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 ×
ΔG° = -2494.2㏑ 1.7 × = -18034.18 J/mol
So, change in standard Gibbs's free energy is -18034.18 J/mol
Learn about formation constant
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Explanation:
12 hours ago
El ácido sulfúrico H2SO4 es uno de los compuestos que se utiliza para la producción de fertilizantes como el nitrosulfato amónico. Si disponemos de 8 mL de H2SO4 al 37 %P/P (d=1,26 g /mL), los cuales se disolvieron hasta alcanzar un volumen de solución de 400 mL, con una densidad de 1,08 g/mL. (La densidad del soluto es corresponde a 1,83 g/cm³)
Answer:
Pp O2 = 82.944 KPa
Explanation:
heliox tank:
∴ %wt He = 32%
∴ %wt O2 = 68%
∴ Pt = 395 KPa
⇒ Pp O2 = ?
assuming a mix of ideal gases at the temperature and volumen of the mix:
∴ Pi = RTni/V
∴ Pt = RTnt/V
⇒ Pi/Pt = ni/nt = Xi
⇒ Pi = (Xi)*(Pt)
∴ Xi: molar fraction (ni/nt)
⇒ 0.68 = mass O2/mass mix
assuming mass mix = 100 g
⇒ mass O2 = 68 g
∴ molar mass O2 = 32 g/mol
⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2
⇒ mass He = 32 g
∴ molar mass He = 4.0026 g/mol
⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He
⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles
molar fraction O2:
⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099
⇒ Pp O2 = (X O2)(Pt)
⇒ Pp O2 = (0.2099)(395 KPa)
⇒ Pp O2 = 82.944 KPa
