Your answer is yeast, since it is a single celled organism that can be classified into a subkingdom related to sac fungi.
That looks like cells of a multicellular organism, so B.
The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
Read more on equilibrium constant;
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Hello:
In this case, we will use the Clapeyron equation:
P = ?
n = 8 moles
T = 250 K
R = 0.082 atm.L/mol.K
V = 6 L
Therefore:
P * V = n * R * T
P * 6 = 8 * 0.082* 250
P* 6 = 164
P = 164 / 6
P = 27.33 atm
Hope that helps!
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
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