Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. neutralization reaction: Suppose 0.250 L of 0.450 M H2SO4 is mixed with 0.200 L of 0.240 M KOH . What concentration of sulfuric acid remains after neutralization

0.197 M

Explanation:

The reaction equation is:

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

number of moles of H2SO4 = 0.25 L * 0.45 M = 0.1125 moles

number of moles of KOH = 0.2 L * 0.24 M = 0.048 moles

since H2SO4 is the reactant in excess;

2 moles of KOH reacts with 1 mole of H2SO4

0.048 moles of KOH reacts with 0.048 * 1/2 = 0.024 moles of H2SO4

Amount of excess H2SO4 left unreacted = 0.1125 - 0.024 = 0.0885 moles

Total volume = 0.25 L+ 0.2 L = 0.45 L

concentration of H2SO4 = 0.0885/0.45 = 0.197 M

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.