Given :
2NOBr(g) - -> 2NO(g) + Br2(g)
Initial pressure of NOBr , 1 atm .
At equilibrium, the partial pressure of NOBr is 0.82 atm.
To Find :
The equilibrium constant for the reaction .
Solution :
2NOBr(g) - -> 2NO(g) + Br2(g)
t=0 s 1 atm 0 0
1( 1-2x) 2x x
So ,
At equilibrium :
Hence , this is the required solution .
The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:
Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:
Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:
And the resulting concentration of KOH and OH ions as this is a strong base:
And the resulting pH is:
Regards!