Answer: the more big it is
Explanation:
Answer:
34g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H2S + 2AgNO3 —> 2HNO3 + Ag2S
Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.
This is illustrated below:
From the balanced equation above,
We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.
Finally, we shall convert 1 mole of H2S to grams. This is shown below:
Number of mole H2S = 1 mole
Molar mass of H2S = (2x1) + 32 = 34g/mol
Mass = number of mole x molar Mass
Mass of H2S = 1 x 34
Mass of H2S = 34g
Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.
Answer:
5.004kg
Explanation:
Combustion of carbon
C+O2=CO2
from the relationship of molar ratio
mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)
mass of carbon =1000kg
atomic mass of carbon =12
volume of CO2 produced=1000×22.4/12
volume of CO2 produced =1866.6dm3
from the combustion reaction equation provided
CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)
applying the same relationship of molar ratio
no of mole of CO2=no of mole of urea
therefore
vol of CO2\22.4=mass of urea/molar mass of urea
molar mass of urea=60.06g/mol
from the first calculation
vol of CO2=1866.6dm3
mass of urea=1866.6×60.06/22.4
mass of urea=5004.82kg
This is an incomplete question, here is a complete question.
The reduction of iron(III) oxide to iron during steel-making can be summarized by this sequence of reactions:
;
;
The net reaction is:
;
Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K be sure you use their standard symbols.
Answer : The equilibrium expression will be:
Explanation :
The intermediate reaction are:
(1) ;
(2) ;
The net reaction is:
;
Now we are multiplying reaction 1 by 3 and reaction 2 by 2 and then adding both the reaction we get the net reaction.
(1) ;
(2) ;
If the equation is multiplied by a factor of '2' or '3', the equilibrium constant will be the power of 2 or 3 of the equilibrium constant of initial reaction.
If the equations are added then the equilibrium constant of the reactions will be multiplied.
Thus the equilibrium expression will be: