Find the volume of 0.240 m sulfuric acid necessary to react completely with 72.8 g sodium hydroxide. express your answer with the appropriate units.

The volume neccesary to complete reaction is 3.79L

Explanation:

First step: Think the equation

H2SO4 + 2NaOH →  Na2SO4  + 2H2O

Afterwards, ballance it.

Relation in reactants is 1:2, so 2 mol of NaOH react to 1 mol of sulfuric.

Gotta find my mol of NaOH

Mass / Molar weight = Mol

72.8 g / 40 g/m = 1.82 mol NaOH

As the relation is 1:2, I have the half of sulfuric: 0.91 mol.

By a rule of three, will be

2 mol NaOH __ react __ 1 mol H2SO4

1.82 mol NaOH __react ___ (1.82 .1)/2

Finally we have to work with molarity, which means mol/L. So, the rule of three is:

0.240 mol H2SO4 __are in __ 1 L

0.91 mol H2SO4___ are in ___ (0.91  .1)/0.240 = 3.79L

n has four pairs of electrons in the valence shell, electrons repel one another ,n has 7 protons in its nucleus, and since n has 7 protons, it must fill the second shell, giving it 4 pairs of electrons. the electrons form 3 bonds and 1 lone pair of electrons. each pair of electrons repels the other pairs, so they are equally far apart.

so all of the above

the child will have heterozygous dominant chromosomes as characters of his father are dominant over the characters of his mother.

the child is heterozygous