The answer is c Yep Allll day
As we have the balanced reaction equation is:
N2O4 (g) ↔ 2NO2(g)
from this balanced equation, we can get the equilibrium constant expression
KC = [NO2]^2[N2O4]^1
from this expression, we can see that [NO2 ] is with 2 exponent of the stoichiometric and we can see that from the balanced equation as NO2
is 2NO2 in the balanced equation.
and [N2O4] is with 1 exponent of the stoichiometric and we can see that from the balanced equation as N2O4 is 1 N2O4 in the balanced equation.
∴ the correct exponent for N2O4 in the equilibrium constant expression is 1
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V1 = 2.0 L
T1 = 25.0 oC = 298 K V2 = V1T2 = (2.0 L)(244 K) = 1.6 L
V2 = ? t1(298 K)
T2 = –28.9 oC = 244 K
Answer:
13mL
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
HNO3 + KOH —> KNO3 + H2O
From the balanced equation above, we obtained the following data:
Mole ratio of the acid (nA) = 1
Mole ratio of the base (nB) = 1
Step 2:
Data obtained from the question.
This includes the following:
Molarity of the acid (Ma) = 6M
Volume of the acid (Va) =?
Volume of the base (Vb) = 39mL
Molarity of the base (Mb) = 2M
Step 3:
Determination of the volume of the acid.
Using the equation:
MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
6 x Va / 2 x 39 = 1/1
Cross multiply to express in linear form
6 x Va = 2 x 39
Divide both side by 6
Va = (2 x 39)/6
Va = 13mL
Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL
