This is a double replacement reaction.
2koh + h2so4 = k2so4 + 2h20
SeO2 (s) + 2KOH(aq) > K2SeO3(aq) +H2O(l) SeO2(s)+2KOH(aq)→K2SeO3(aq)+H2O(l)
The answer is one mole.
1
No Potassium oxide is extremely basic (alkaline).
CuSO4 + 2 KOH = Cu(OH)2(s) + K2SO4Only copper hydroxide is insoluble in water; other compounds are soluble in water.
This problem is a Double Reactant. "K+ H20 ----> KOH + H2" would equal "2K + 2H20 ---> 2KOH + H2"
2koh+co2--k2co3+h2o
It is a simple AB + CD = AD + CB reaction so we would have: NaOH + KCl -> NaCl + KOH
CuSO4(aq) + 2KOH(aq) --> Cu(OH)2(s) + K2SO4(aq)
H2so4+2koh=k2so4+2h2o
2koh + h2so4 > k2so4 + h2o
2koh + h2so4 = k2so4 + 2h20
SeO2 (s) + 2KOH(aq) > K2SeO3(aq) +H2O(l) SeO2(s)+2KOH(aq)→K2SeO3(aq)+H2O(l)
K2SO4 + 2NaOH ==> Na2SO4 + 2KOH
2koh+co2--k2co3+h2o
2ki + h2o2 = 2koh +i2