The organism would no longer grow.
Density = m/v therefore v = m/d v = 25/0.85 v = 29.4117... cm^3 ( = 2.94 * 10^-5 m^3
The solubility of carbon dioxide at 400 kPa at room temperature is ;
( B ) 0.61 CO2/L
<u>Given data </u>
pressure of CO₂ = 400 Kpa = 3.95 atm
Kh of CO₂ = 3.3 * 10⁻² mol/L.atm
<h3>Calculate the solubility of carbon dioxide </h3>
Solubility = pressure * Kh value of CO₂
= 3.95 atm * 3.3 * 10⁻² mol / L.atm
= 0.13 mol/l CO₂
= 0.61 CO₂ / L
Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l CO₂.
Learn more about solubility : brainly.com/question/23946616
From the options the closest answer is ( B ) 0.61 CO₂ / L
Answer:
About 5 times faster.
Explanation:
Hello,
In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:
By replacing the corresponding values we obtain:
Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.
Best regards.
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
