The stronger the pull of gravity the greater the mass
All elements in their standard states (oxygen<span> gas, solid carbon in the form of graphite, etc.) have a standard </span>enthalpy of formation<span> of </span>zero<span>, as there is no change involved in their </span>formation<span>.</span>
<span>The molecular formula that describes the problem is
2CH3COOH (aq) + Ca(OH)2 (s) ---> Ca(CH3COO)2 (aq) + 2H2O (l)
The net equation is written as follows:
2CH3COOH- (aq) + 2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 CH3COO- (aq) + 2H2O (l)
canceling out spectator ions
2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 H2O (l)</span>
Answer:
2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O
Explanation:
2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O
Oxidizing agent: -----------------------------> CrO42-
Reducing agent: ----------------------------> N2O
explanation:
in CrO4-2 oxdiation state of Cr = +6
in Cr+3 oxidation state = +3
+6 oxidation state changed from +3 it is reduction .
so CrO4-2 is oxidizing agent
atomatically
N2O should be reducing agent