Answer : The value of equilibrium constant for this reaction at 328.0 K is 
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = 151.2 kJ = 151200 J
= standard entropy = 169.4 J/K
T = temperature of reaction = 328.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = 95636.8 J
R = gas constant = 8.314 J/K.mol
T = temperature = 328.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 328.0 K is
That is correct c
Explanation
<span>The
kingdom, protista’s characteristics are that the organism (not a plant,
animal or fungus) are:
unicellular however some are multicellular like algae, are heterotrophic or
autotrophic, others lives in water while some live in moist areas or human body,
have a nucleus, cellular respiration is primarily aerobic, some are pathogenic
(e.g. causing Malaria) and reproduction is mitosis or meiosis. This kingdom
includes: Sacordinians – pseudopods (e.g. Amoeba, Foraminiferans<span>.)</span>, Zooflagellates – flagellates
(e.g. Trypanosoma gambiense),
Ciliaphorans – ciliates (e.g. paramecium) and Sporozoans (e.g. Plasmodium).</span>
Answer:
B. 0.2.
Explanation:
<em>n = mass/molar mass</em>
mass of CaCO₃ = 20 g, molar mass of CaCO₃ = 100.0869 g/mol.
<em>∴ n = mass/molar mass = </em>(20 g)/(100.0869 g/mol) <em>= 0.1998 ≅ 0.2 mol.</em>
<em></em>
<em>So, the right choice is: B. 0.2.</em>
