Answer:
A reduction-oxidation (redox) reaction will occur.
Chlorine gas (Cl2) will accept electrons to form Cl- ions.
The Iron(II) ions (Fe2+) will lose electrons to form Fe3+ ions. (Iron (III) ions)
Fe2+ ions are green, while Fe3+ ions are yellow,
so the observation will be the solution turning from green to yellow.
Answer:
The answer is Lyase
Explanation:
Any enzyme that catalyzes the addition or removal of the elements water (hydrogen, oxygen), ammonia (nitrogen, hydrogen), or carbon dioxide (carbon, oxygen) at double bonds, as defined in physiology. Decarboxylases, for example, remove carbon dioxide from amino acids, while dehydrases eliminate water.
Based on the assumption that the reaction involves N and O to produce NO, if 25.0 g of NO are produced, the amount of N gas used would be 11.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
N + O ---------> NO
Mole ratio of N to NO is 1:1
Mole of 25.0 g of NO = 25/30.01 = 0.833 moles
Equivalent mole of N = 0.833 moles
Mass of 0.833 moles N = 0.833 x 14 = 11.66 grams
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
The reaction can produce 287 grams of iron(II) carbonate
Explanation:
To solve this question we must find the moles of iron(II) chloride that react. Using the chemical equation we can find the moles of iron(II) carbonate and its mass -Molar mass FeCO3: 115.854g/mol-
<em>Moles FeCl2:</em>
1.24L * (2.00mol / L) = 2.48 moles FeCl2
As 1 mol FeCl2 produce 1 mol FeCO3, the moles of FeCO3 = 2.48 moles
<em>Mass FeCO3:</em>
2.48mol * (115.854g / mol) =
<h3>The reaction can produce 287 grams of iron(II) carbonate</h3>
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
