NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Answer:
i would or i have or i did
Explanation:
Answer:
This question appears incomplete
Explanation:
This question appears incomplete because the data provided only makes it possible to calculate the certainty of the acetic acid content per total volume of the vinegar. Thus, the 4% means for every 100 mL of the vinegar, there is 4 mL of acetic acid present. To calculate the volume of acetic acid in any other volume of vinegar, the formula will be
volume of acetic acid = 4/100 × total volume of vinegar
That is called condensation
Hello there,
The answer to your question is:
C. Zinc
Hope this answer helps you.
