According to the reversible reaction equation:
2Hi(g) ↔ H2(g) + i2(g)
and when Keq is the concentration of the products / the concentration of the reactants.
Keq = [H2][i2]/[Hi]^2
when we have Keq = 1.67 x 10^-2
[H2] = 2.44 x 10^-3
[i2] = 7.18 x 10^-5
so, by substitution:
1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2
∴[Hi] = 0.0033 M
Ionic: transfer of electrons
Covalent: sharing of electrons
Metallic: sharing of free electrons in a structure of cations
<u>Given:</u>
Concentration of Ba(OH)2 = 0.348 M
<u>To determine:</u>
pOH of the above solution
<u>Explanation:</u>
Based on the stoichiometry-
1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion
Therefore, concentration of OH- ion = 2*0.348 = 0.696 M
pOH = -log[OH-] = - log[0.696] = 0.157
Ans: pOH of 0.348M Ba(OH)2 is 0.157
Answer:Answer: (B) Archaebacteria
Explanation:
Explanation: Archaebacteria, like all prokaryotes, have no membrane bound organelles. This means that the archaebacteria are without nuclei, mitochondria, endoplasmic reticula, lysosomes, Golgi complexes, or chloroplasts. Because these organisms have no nucleus, the genetic material floats freely in the cytoplasm
