Write the balanced neutralization reaction that occurs between H 2 SO 4 and KOH in aqueous solution. Phases are optional. neutralization reaction: H_{2}SO_{4}(aq) +KOH(aq)->K_{2}SO_{4}(aq) +H_{2}O(l) H 2 SO 4 ( aq ) + KOH ( aq ) ⟶ K 2 SO 4 ( aq ) + H 2 O ( l ) Suppose 0.950 L of 0.400 M H 2 SO 4 is mixed with 0.900 L of 0.260 M KOH . What concentration of sulfuric acid remains after neutralization?

Concentration H2SO4 remaining after neutralization = 0.142 M

Explanation:

Step 1: Data given

Volume of a 0.400 M H2SO4 = 0.950 L

Volume of a 0.260 M KOH = 0.900 L

Step 2: The balanced equation

H2SO4(aq) +2KOH(aq) → K2SO4(aq) + 2H2O(l)

Step 3: Calculate moles

Moles = molarity * volume

Moles H2SO4 = 0.400 M * 0.950 L

Moles H2SO4 = 0.380 moles

Moles KOH = 0.260 M * 0.900 L

Moles KOH = 0.234 moles

Step 4: Calculate limiting reactant

For 1 mol H2SO4 we need 2 moles KOH to produce 1 mol K2SO4 and 2 moles H2O

KOH is the limiting reactant. It will completely be consumed (0.234 moles). H2SO4 is in excess. There will react 0.234/2 =0.117 moles

There will remain 0.380 -0.117 = 0.263 moles H2SO4

Step 5: Calculate concentration of sulfuric acid remains after neutralization

Concentration H2SO4 = moles H2SO4 / volume

Concentration H2SO4 = 0.263 moles / (0.950L+ 0.900L)

Concentration H2SO4 = 0.142 M

the answer is either b or c

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."