Suppose an epa chemist tests a 200.ml sample of groundwater known to be contaminated with iron(ii) chloride, which would react with silver nitrate solution like this: fecl2 (aq) + 2agno3 (aq) → 2agcl (s) + feno32 (aq) the chemist adds 14.0m m silver nitrate solution to the sample until silver chloride stops forming. she then washes, dries, and weighs the precipitate. she finds she has collected 6.9mg of silver chloride. calculate the concentration of iron(ii) chloride contaminant in the original groundwater sample. be sure your answer has the correct number of significant digits.

The concentration of iron(II) chloride contaminant in the original groundwater sample is .

Explanation:

Amount of silver chloride recovered = 6.9 mg = 0.0069 g[/tex]

Moles of silver chloride =

According to reaction. 2 moles of silver chloride are obtained from 1 mole of iron(II) chloride.

Then moles of silver chloride will be obtained from:

Moles of iron(II) chloride =

Volume of a sample of ground water = 200 mL = 0.2 L

The concentration of iron(II) chloride contaminant in the original groundwater sample is .

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

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.