Answer:
Explanation:
<u>1) Calculate the mass of carbon (C)</u>
- mass of C = % of C × mass of sample / 100
- mass of C = 37.51% × 5.91 g / 100 = 2.21 g
<u>2) Calculate the number of moles of C</u>
- number of moles = mass in grams / molar mass
- number of moles of C = 2.21 g / 12.01 g/mol = 0.184 moles
<u>3) Calculate the number of moles of hydrogen atoms, H:</u>
- number of moles = number of atoms / Avogadro's number
- number of moles of H = 1.4830 × 10²³ / 6.022 × 10²³ = 0.24626 moles
<u>4) Calculate the number of moles of oxygen atoms, O:</u>
- number of moles = number of atoms / Avogadro's number
- number of moles of O = 1.2966 × 10²³ / 6.022 × 10²³ = 0.21531 moles
<u>5) Find the </u><u>mole ratios</u><u>:</u>
Summary of moles:
Divide every amount by the smallest number, which is 0.184:
- H: 0.24626 / 0.184 = 1.34
- O: 0.21531 / 0.184 = 1.17
Multiply by 3 to round to integer numbers:
Multiply by 2 to round to integer numbers:
Use the mole ratios as superscripts to write the empirical formula
Just as a reference, you can search in internet and find that one compound with that empirical formula is citric acid.
Take note of the whole number beside each substance in the reaction because these will be used in the stoichiometric calculations below:
Molar mass of <span>k[ag(cn)2]:199 g/mol
Molar mass of Zn(Cn)2: 117.44 g/mol
Avogadro's number: 6.022</span>×10²³ molecules/mol
a.)
35.27 g*(1 mol/199 g)*(1 mol Zn(Cn)₂/ 2 mol K[Ag(CN)₂])*(6.022×10²³ molecules/mol) = <em>5.34×10²² molecules of Zn(Cn)₂</em>
b.)
35.27 g*(1 mol/199 g)*(1 mol Zn(Cn)₂/ 2 mol K[Ag(CN)₂])*(117.44 g/mol) = <em>10.41 g of Zn(Cn)₂</em>
Answer:
Supervision of weights and measures promotes accurate measurements of goods and services to ensure that everybody gets a fair trade in the marketplace. Not so coincidentally it also is a deterrent to ensure that traders are being honest in their trade practises.
Explanation: