<span>1. What is the molar mass of gold?
Molar mass is a unit that expresses the mass of a molecule per one mol. The molar mass can be obtained by adding the neutron with the proton of the atoms. Gold has atomic number 79 so the proton is 79. The number of the neutron is 118. Then the molar mass would be: 79 + 118 = </span>197 g/mol<span>
</span><span>2. Calculate the number of moles of gold (Au) in the sample. Show your work.
</span>In this question, you are given the mass of the gold and asked for how many moles the sample has. To find the number of moles you just need to divide the weight by the molar mass.
For 45.39 grams of gold, the number of moles would be:
45.39 / (197g/mol)= 0.23 moles
3. Calculate the number of atoms of gold (Au) in the sample. Show your work.Moles is unit of a number of molecules but 1 mol doesn't represent 1 molecule. The number of atoms can be obtained by multiplying the number of moles with Avogadro number. The calculation would be:
0.23 moles * (6.023 * 10^23 molecules/mol)= 1.387 * 10^23 molecules
Answer:
Electrolysis (of hydrochloric acid) is a way of splitting up (decomposition) of the compound (hydrogen chloride in water) using electrical energy.
Explanation:
The electrical energy comes from a d.c. (direct current) battery or power pack supply. A conducting liquid, containing ions, called the electrolyte (hydrochloric acid), must contain the compound (hydrogen chloride) that is being broken down.
Metals are insulators and malleable. Hope this helps even though I'm late :)
The correct number of significant figures and digits is 3
First a balanced reaction equation must be established:
![4Al _{(s)} + 3 O_{2} _{(g)}](https://tex.z-dn.net/?f=4Al%20_%7B%28s%29%7D%20%20%20%20%2B%20%20%203%20O_%7B2%7D%20%20_%7B%28g%29%7D)
→
![2 Al_{2} O_{3}](https://tex.z-dn.net/?f=2%20Al_%7B2%7D%20O_%7B3%7D%20)
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[
4Al +
3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
=
3.63 mol (to 2 sig. fig.)
Thus it can be concluded that
3.63 moles of oxygen is needed to react completely with 145 g of aluminum.