1.
a 1.020 g sample of sodium fluoride is dissolved in water and then precipitated with calcium
nitrate solution. if the calcium fluoride precipitate weighs 0.905 g, what is the percent
yield?
2 naf(39) + ca(no3)2(aq) > caf2 (s) + 2 nano3(aq)

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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

answer; natural resources;

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.