Answer:
The <u>equilibrium constant</u> is:
Explanation:
The correct equation is:
Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.
The equation for the equilibrium constant is:
![k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}](https://tex.z-dn.net/?f=k_c%3D%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ccdot%20%5BH_2%5D%5E3%7D)
Substituting:
First, we will need to find the density of the object, take the mass and divide it by the dispplaced water:
128/424 = 0.302 grams/milliliters
Convert that to kg/m3
We get: 302kg/m3
Divide that to the density of water: 1000kg/m3
302/1000 = 0.302
(thats a pretty darn light weighted metal)
Answer:
The number ratio is 4:7
Explanation:
Step 1: Data given
Compound 1 has 50.48 % oxygen
Compound 2 has 36.81 % oxygen
Molar mass oxygen = 16 g/mol
Molar mass manganese = 54.94 g/mol
Step 2: Calculate % manganes
Compound 1: 100 - 50.48 = 49.52 %
Compound 2: 100 - 36.81 = 63.19 %
Step 3: Calculate mass
Suppose mass of compounds = 100 grams
Compound 1:
50.48 % O = 50.48 grams
49.52 % Mn = 49.52 grams
Compound 2:
36.81 % O = 36.81 grams
63.19 % Mn = 63.19 grams
Step 4: Calculate moles
Compound 1
Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles
Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles
Compound 2
Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles
Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles
Step 5: calculate mol ratio
We will divide by the smallest amount of moles
Compound 1
O: 3.155/0.9013 = 3.5
Mn: 0.9013 / 0.9013 = 1
Mn2O7
Compound 2
O: 2.301 / 1.150 = 2
Mn: 1.150 / 1.150 = 1
MnO2
The number ratio is 2:3.5 or 4:7
A: 12 N
B: 150 N
C: 100 N
D: 150 N
E: 220 N
