Answer:
12 moles of F₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of F₂ needed to produce 8 moles of NF₃. This can be obtained as illustrated below:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, Xmol of F₂ will react to produce 8 moles of NF₃ i.e
Xmol of F₂ = (3 × 8)/2
Xmol of F₂ = 12 moles
Thus, 12 moles of F₂ is needed for the reaction.
Answer:
ICI 204448 hydrochloride | C23H27Cl3N2O4 | CID 129407 - structure, chemical names, physical and chemical properties, classification, patents, literature, etc...
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Assuming that the combustion formula is
CH4 + 2O2 --> 2H2O + CO2<span>,
That means for every 1 molecule of methane(CH4) there will be one molecule of carbon dioxide(</span>CO2) produced. Methane molecular weight 16, carbon dioxide molecular weight is 44. Then the percent yield should be:
1 * (0.374/ 16) /(0.983/44)= 0.374*44/ 0.983 * 16= 104.6%
You sure the number is correct? Percent yield should not exceed 100%
Answer:
0.50 mol
Explanation:
The half-life is <em>the time required for the amount of a radioactive isotope to decay to half that amount</em>.
Initially, there are 8.0 moles.
- After 1 half-life, there remain 1/2 × 8.0 mol = 4.0 mol.
- After 2 half-lives, there remain 1/2 × 4.0 mol = 2.0 mol.
- After 3 half-lives, there remain 1/2 × 2.0 mol = 1.0 mol.
- After 4 half-lives, there remain 1/2 × 1.0 mol = 0.50 mol.
