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>
Answer:
13: if thats one of the options
Explanation:
B/c for pH to be that high you have to have alot of OH ions and almost non H+. pH won't be 0 or 2 cuz that implies ALOT of H ions. pH wont be 7 cuz that implies we have 1*10^-7 H ions.
<span><span>NaOH + HCl --> NaCl + H2Oso here 1 mole of NaOH reacts with 1 mole of HCl. Hence it is required to convert the amount of HCl and NaOH in to moles.Number of HCl moles Concentration of the solution = 2.00M = 2 mol/ literVolume of the solution = 22.4 mlNumber of HCl moles = (2mol/1000 ml)*22.4 ml = 0.0448 mol</span></span>
Answer:
true
Explanation: pull in relatively cool water to cool their machinery and let the relatively warm water flow back into the river or lake or sea. ... Water on hot paved surfaces gets hot, then runs off into nearby bodies of water, raising the water temperature
Answer:
520mL
Explanation:
Data obtained from the question include:
Molarity of stock solution (M1) = 12M
Volume of stock solution (V1) = 130mL
Molarity of diluted solution (M2) = 3M
Volume of diluted (V2) =..?
The volume of the diluted solution can be obtained as follow:
M1V1 = M2V2
12 x 130 = 3 x V2
Divide both side by 3
V2 = 12 x 130 / 3
V2 = 520mL.
Therefore, 520mL of the diluted solution can be prepared.