Calculate the molarity of each aqueous solution: a. 78.0 ml of 0.240 m naoh diluted to 0.250 l with water b. 38.5 ml of 1.2 m hno3 diluted to 0.130 l with water

We know that when we dilute a solution the number of moles remains the same just the concentration changes [moles per unit volume]

So we can equate the moles before dilution and after dilution

moles = molarity X volume

a) Initial moles = final moles [NaOH]

M1= initial molarity = 0.240 M

V1= initial volume = 78 mL

M2= final molarity = ?

V2 = final volume = 0.250 L = 250mL

therefore

final molarity = 0.075 M

b)

Initial moles = final moles [Nitric acid]

M1= initial molarity = 1.20 M

V1= initial volume = 38.5 mL

M2= final molarity = ?

V2 = final volume = 0.130 L = 130mL

therefore

final molarity = 0.355 M

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.

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