Answer:
for this reaction at this temperature is 0.029
Explanation:
Moles of
= 2.00 mole
Volume of solution = 4.00 L
Initial concentration of
The given balanced equilibrium reaction is,

Initial conc. 0.500 M 0 M 0 M
At eqm. conc. (0.500-2x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5Ctimes%20%5BBr_2%5D%7D%7B%5BHBr%5D%5E2%7D)
Equilibrium concentration of
= x = 0.0955 M
Now put all the given values in this expression, we get :


Thus
for this reaction at this temperature is 0.029
84.34 grams of grams of iron (III) chloride that can be produced is maximum because Fe is the limiting reagent in this reaction and chlorine gas is excess reagent.
Explanation:
Balanced chemical equation:
2 Fe + 3 Cl2 → 2 FeCl3
DATA GIVEN:
iron = atoms
mass of chlorine gas = 67.2 liters
mass of FeCl3 = ?
number of moles of iron will be calculated as
number of moles = 
number of moles = 
number of moles = 0.52 moles of iron
moles of chlorine gas
number of moles = 
Putting the values in the equation:
n =
(atomic mass of chlorine gas = 70.96 grams/mole)
= 947.01 moles
Fe is the limiting reagent so
2 moles of Fe gives 2 moles of FeCl3
0.52 moles of Fe will give
= 
0.52 moles of FeCl3 is formed.
to convert it into grams:
mass = n X atomic mass
= 0.52 x 162.2 (atomic mass of FeCl3 is 162.2grams/mole)
<h3> = 84.34 grams </h3>
Answer:
37.65mL
Explanation:
Given parameters:
density of liquid Z = 0.9237g/mL
Mass of liquidZ + mass of cup = 50.7g
Mass of cup= 15.92g
Volume of liquid Z in cup=?
Solution:
Density is the mass per unit volume of a substance. It is mathematically expressed as shown below:
Density = 
To find the volume of liquid Z, we know the density of the liquid but we dont know the mass yet.
Mass of liquidZ = 50.7g - mass of cup = 50.7g - 15.92g = 34.98g
Therefore:
Volume of liquidZ = 
=
= 37.65mL
Answer:
The answer to be filled in the respective blanks in question is
3 and 1
Explanation:
So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.
Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1