Acontains 2.60g of impure h2so4 in 5oocm³ of a solution 5.30g of na2co3 in 1.00dm³ of solution
an average of 27.65cm of a neutralized completely 25.0cm of b
calculate the
a)concentration of b in moldm³
b)concentration of a in 1)moldm³ 2)g/dm³
c)%purity of h²so4

a. 0.05mol/dm3

bi. 0.045mol/dm3

bii. 4.41g/dm3

c. 84.8%

Explanation:Please see attachment for explanation

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon 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