Answer:
The Equilibrium constant K is far greater than 1; K>>1
Explanation:
The equilibrium constant, K, for any given reaction at equilibrium, is defined as the ratio of the concentration of the products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their stoichiometric coefficients.
It tells us more about how how bigger or smaller the concentration of products is to that of the reactants when a reaction attains equilibrium. From the given data, as the color of the reactant mixture (Br2 is reddish-brown, and H2 is colourless) fades, more of the colorless product (HBr is colorless) is being formed as the reaction approaches equilibrium. This indicates yhat the concentration of products becomes relatively higher than that of the reactants as the reaction progresses towards equilibrium, the equilibrium constant K, must be greater than 1 therefore.
We’d have to be very careful because if we had our skeletons on the outside it’d be very easy to injure ourselves
Answer:
there are 20 oxygen atoms in 4.00 moles of Dinitrogen pentoxide
Explanation:
there are 2 atoms in an oxygen molecule , so each oxygen molecules has at least 2. Dinitrogen pentoxide is N2O5, which has 7 atoms, 2 nitrogen and 5 oxygen. 1 molecule of N2O5 has 5 oxygen atoms, so 4 of then would be 20
Answer:
a free swimming larval stage in which a parasitic fluke passes from an intermediate host to another intermediate host
Answer:
Percentage abundance of 121 Sb is = 57.2 %
Percentage abundance of 123 Sb is = 42.8 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, 121 Sb :
% = x %
Mass = 120.9038 u
For second isotope, 123 Sb:
% = 100 - x
Mass = 122.9042 u
Given, Average Mass = 121.7601 u
Thus,

Solving for x, we get that:
x = 57.2 %
<u>Thus, percentage abundance of 121 Sb is = 57.2 %
</u>
<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>