The question correctly reads:
The reaction between nitrogen and hydrogen to produce ammonia is described as an equilibrium reaction. What substances are present in the reaction mixture when equilibrium has been obtained?
3H2 + N2 = 2NH3
N2
H2
NH4+
NH3
Answer:
N2,H2,NH3
Explanation:
Equilibrium is said to be attained when the rate of forward reaction equals the rate of reverse reaction. That is forward and reverse reactions occur at the same rate. The rate at which reactants are reformed, is the same rate at which products are formed. Hence at equilibrium both reactants and products are present in the reaction mixture. This implies that nitrogen, hydrogen and ammonia are all present in the reaction mixture at equilibrium.
Answer:
The rf value is 0.38
Explanation:
You do 5/13 which gives the answer 0.38
D) melting ice because u can freeze the “ice” back into that state of matter this is called a physical change. HOPE THIS HELPED
Question
<em>what cells in living organisms are diploid?</em>
Answer:
<em>Most mammals are diploid organisms, which means they have two homologous copies of each chromosome in the cells. In humans, there are 46 chromosomes. In most diploid organisms, every cell except for gametes will be diploid and contain both sets of chromosomes.</em>
Hope this helps!
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:
![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
