The oxidation state (or number) is zero. This is true for any element in any of its allotropic elemental forms.
zero- H2 is the elemental form- by definition the ON# is zero
The oxidation state of magnesium in magnesium hydroxide is +2 no matter what reaction you are looking at.
The oxidation state of magnesium in magnesium hydroxide is +2 no matter what reaction you are looking at.
Out of these possible answers A) Ba(NO3)2(aq) + K2SO4(aq) -> BaSO4(s) + 2KNO3(aq) B) HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l) C) MgO(s) + H2)(l) -> Mg(OH)2(s) D) 2SO2(g) + O2(g) -> 2 SO3(g) E) 2H2O(l) -> 2H2(g) + O2(g)
For 2HCl(g) ==> H2(g) + Cl2(g) the Keq = [H2][Cl2]/[HCl]^2
In:(H2)g oxidation state: 0 In:(O2)g oxidation state: 0 In:(H2O)l oxidation state: H: +1 and O: -2
zero- H2 is the elemental form- by definition the ON# is zero
The answer is 0
The oxidation state of magnesium in magnesium hydroxide is +2 no matter what reaction you are looking at.
Simply remember OIL RIG. Oxidation Is Loss (of electron) and Reduction Is Gain (of electron). In the case of MnO4-(aq) + 8H+(aq) + 5e- --> Mn2+(aq) + 4H2O(l) H is the oxidizing agent, because it causes Mn to be oxidized to Mn2+.
The oxidation state of magnesium in magnesium hydroxide is +2 no matter what reaction you are looking at.
The oxidation state of magnesium in magnesium hydroxide is +2 no matter what reaction you are looking at.
36g2h2 = 2*2*3*3*g*g*h*h provided g and h are prime.
2 Fe (s) + 3 H 2SO 4 (aq) → Fe 2(SO 4) 3 (s) + 3 H 2 (g). This is an oxidation-reduction (redox) reaction: 6 H I + 6 e - → 6 H 0 (reduction).
Al starts at 0 oxidation. then becomes Al 3+ O remains -2 on both sides. H starts at 1+ then becomes 0
Out of these possible answers A) Ba(NO3)2(aq) + K2SO4(aq) -> BaSO4(s) + 2KNO3(aq) B) HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l) C) MgO(s) + H2)(l) -> Mg(OH)2(s) D) 2SO2(g) + O2(g) -> 2 SO3(g) E) 2H2O(l) -> 2H2(g) + O2(g)
In the algebraic equation for a circle. (x - g)^2 + (y - h)^2 = r^2 'g' & 'h' are the centre of rotation.