Answer:
Volume of stock solution needed = 6.0299 mL
Explanation:
<u>
</u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.
This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.
<u>Data:</u>
M1 = 6.01 M stock solution concentration
M2 = 0.3624 M diluted solution concentration
V2 =100 mL diluted solution volume
V1 = ? stock solution volume
M1 * V1 = M2 * V2
Answer:
64.17 Moles of Au
Explanation:
(atoms and particles are the same)
3.85 x 10 ^25 x (1 mol
/6.02 x 10^23)
3.85 / 6 = .64166
.6416 x 10^2 = 64.166
If you round up the answer you will get 64.17
64.17 moles of Au
Answer:
falso
Explanation:
tiempo no determinado El circulation de elctrones
T A R G E T or T J M A X X
Answer:
83.64%.
Explanation:
∵ The percent yield = (actual yield/theoretical yield)*100.
actual yield of CO₂ = 2300 g.
- We need to find the theoretical yield of CO₂:
For the reaction:
<em>CH₄ + 2O₂ → 2H₂O + CO₂,</em>
1.0 mol of CH₄ react with 2 mol of O₂ to produce 2 mol of H₂O and 1.0 mol of CO₂.
- Firstly, we need to calculate the no. of moles of 1000 g of CH₄ using the relation:
<em>no. of moles of CH₄ = mass/molar mass</em> = (1000 g)/(16.0 g/mol) = <em>62.5 mol.</em>
<u><em>Using cross-multiplication:</em></u>
1.0 mol of CH₄ produces → 1.0 mol of CO₂, from stichiometry.
∴ 62.5 mol of CH₄ produces → 62.5 mol of CO₂.
- We can calculate the theoretical yield of carbon dioxide gas using the relation:
∴ The theoretical yield of CO₂ gas = n*molar mass = (62.5 mol)(44.0 g/mol) = 2750 g.
<em>∵ The percent yield = (actual yield/theoretical yield)*100.</em>
actual yield = 2300 g, theoretical yield = 2750 g.
<em>∴ the percent yield</em> = (2300 g/2750 g)*100 = <em>83.64%.</em>
