1) since we are given percentages, we can assume we have 100 grams of the molecule.
55.6 % Cu ----> 55.6 grams Cu
16.4 % Fe------> 16.4 grams Fe
28.0% S--------> 28.0 grams S
2) convert each gram to moles using the molar masses given



3) we divide the smallest value of moles (0.293) to each one.
Cu --> 0.876 / 0.293= 3
Fe---> 0.293 / 0.293= 1
S-----> 0.875 / 0.293= 3
4) let's write the empirical formula
Cu₃FeS₃
Answer:
2H2(g)+O2(g)--->2H2O(l)
Explanation:
4 hydrogen atoms. add 2 before hydrogen to balance and give the states as shown.
It is basic because of the fact that it has a strong base which is KOH and a weak acid HCN and the strong base causes it to remain more basic.
Answer:
0.004548 M is the concentration of B at equilibrium at 500 K.
Explanation:
A(aq) ⇆ 2 B(aq)
Initially 3.00 M
At equilibrium 3.00 -x 2x
Equilibrium constant of the reaction at 500 K =
Concentration of A at 500 K at equilibrium , [A] = (3.00 -x )M
Concentration of B at 500 K at equilibrium,[B]= 2x
An expression of equilibrium constant is given as:
![K_c=\frac{[B]^2}{[A]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BB%5D%5E2%7D%7B%5BA%5D%7D)

On solving for x:
x = 0.002274 M
[B] = 2 x = 2 × 0.002274 M = 0.004548 M
[A] = (3-x) = 3 M - 0.002274 M =2.997726 M
0.004548 M is the concentration of B at equilibrium.