Answer:
oof idk I would suggest looking at an example and try going off that
102 grams of ammonia is formed when 3 moles of nitrogen and 6.7 moles of hydrogen reacts.
Explanation:
The equation given is of Haeber's process in which the nitrogen is limiting factor in the ammonia formation and hydrogen if in excess gets delimited.
We know that 1 mole of Nitrogen gives 2 moles of ammonia.
We have 3 moles of nitrogen here,
So, 6 moles of ammonia will be form
so from the formula
no of moles=mass/atomic mass
mass= no. of moles*atomic mass
= 6*17
= 102 grams of ammonia will be formed.
So, 6 moles or 102 grams of ammonia is formed when 3 mole of nitrogen and 6.7 mole of hydrogen reacts.
It would be called the crest.
Happy to help! Have a great evening.
~Brooke❤️
Answer:
THERE IS NOTHING MENTION HERE HOW CAN ANYONE KNOW ABOUT IT?]
Explanation:
<span>Answer: 0.094%
</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />
<span>Only the ionization of the formic acid is the important part.
</span><span />
<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />
<span>2) Mass balance:
</span><span />
<span> HCOOH(aq) HCOO⁻(aq) H⁺(aq).
Start 0.311 0.189
Reaction - x +x +x
Final 0.311 - x 0.189 + x x
3) Acid constant equation:
</span><span />
<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />
<span>= (0.189 + x )x / (0.311 - x) = 0.000177
4) Solve the equation:
You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />
<span>With that approximation the equation to solve becomes:
</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M
5) With that number, the percent of ionization (alfa) is:
</span><span />
<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
