Answer : The equilibrium concentration of NO is, 0.0092 M.
Solution :
First we have to calculate the concentration of NO.
The given equilibrium reaction is,
Initially conc. 0 0 0.1576
At eqm. (x) (x) (0.1576-2x)
The expression of 
will be,
![K_c=\frac{[NO]^2}{[N_2][O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%7D)
By solving the term, we get:
Neglecting the 0.0839 value of x because it can not be more than initial value.
Thus, the value of 'x' will be, 0.0742 M
Now we have to calculate the equilibrium concentration of NO.
Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M
Therefore, the equilibrium concentration of NO is, 0.0092 M.
Answer:
1. Theoretical yield = 2.03g
2. Actual yield 1.89g
Explanation:
Let us write a balanced equation. This is illustrated below:
Zn + 2HCI —> ZnCl2 + H2
Molar Mass of HCl = 1 +35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 2 x 36.5 = 73g
Molar Mass of H2 = 2x1 = 2g/mol
1. From the equation,
73g of HCl produced 2g of H2.
Therefore, 74g of HCl will produce = (74 x 2)/73 = 2.03g
Therefore, theoretical yield = 2.03g
2. %yield = 93%
Theoretical yield = 2.03g
Actual yield =?
%yield = Actual yield /Theoretical yield x100
Actual yield = %yield x theoretical yield
Actual yield = 93% x 2.03 = (93/100)x2.03 = 1.89g
Actual yield =1.89g
Answer:
plaster is important because of the many uses you can make out of plaster
Answer:
the number of molecules in 7g nitrogen
Ans) 7 gram of Nitrogen contains 1.5×10^23 molecules =1.5e+23
First, we need to convert the grams of gold into mole using the molar mass (molecular weight) of gold, and then into particles using avogadro's number (1 mol= 6.02 x 10^23)
molar mass of gold= 197 grams/ mole
100.0 grams (1 mole/ 197 grams) x (6.02 x 10^23 particles/ 1 mole) = 3.06 x 10^23
