Formation of ammonia by nitrogen and hydrogen is habers process wher 28g N2 results in formation of 34g NH3
so 35g N2 will form 34*35/28=42.5g NH3 where it given that reaction takes place in excess of H2
N2+3H2 gives 2NH3
Transferred to the lipoamide by an earlier intermediate in the process.
The pyruvate dehydrogenase complex (PDC) is a mitochondrial multienzyme complex composed of three different enzymes
<h3>What reaction is catalyzed by enzyme 2 of the pyruvate dehydrogenase complex ?</h3>
the pyruvate dehydrogenase complex is the bridge between glycolysis and the citric acid cycle
- Five coenzymes are used in the pyruvate dehydrogenase complex reactions: thiamine pyrophosphate or TPP, flavin adenine dinucleotide or FAD, coenzyme A or CoA, nicotinamide adenine dinucleotide or NAD, and lipoic acid.
- during the reactions catalyzed by the pyruvate dehydrogenase complex, it is first reduced to dihydrolipoamide, a dithiol or the reduced form of the prosthetic group, and then, reoxidized to the cyclic form.
Learn more about Pyruvate dehydrogenase complex here:
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This is seen in the first law of Thermodynamics stating that matter and energy cannot be destroyed nor created.
Answer:
[H+] = 1.74 x 10⁻⁵
Explanation:
By definition pH = -log [H+]
Therefore, given the pH, all we have to do is solve algebraically for [H+] :
[H+] = antilog ( -pH ) = 10^-4.76 = 1.74 x 10⁻⁵
Answer:
![\large \boxed{\text{E) 721 K; B) 86.7 g}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7BE%29%20721%20K%3B%20B%29%2086.7%20g%7D%7D)
Explanation:
Question 7.
We can use the Combined Gas Laws to solve this question.
a) Data
p₁ = 1.88 atm; p₂ = 2.50 atm
V₁ = 285 mL; V₂ = 435 mL
T₁ = 355 K; T₂ = ?
b) Calculation
![\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7Bp_%7B1%7DV_%7B1%7D%7D%7BT_%7B1%7D%7D%26%20%3D%26%5Cdfrac%7Bp_%7B2%7DV_%7B2%7D%7D%7BT_%7B2%7D%7D%5C%5C%5C%5C%5Cdfrac%7B1.88%5Ctimes285%7D%7B355%7D%20%26%3D%20%26%5Cdfrac%7B2.50%5Ctimes%20435%7D%7BT_%7B2%7D%7D%5C%5C%5C%5C1.509%26%20%3D%20%26%5Cdfrac%7B1088%7D%7BT_%7B2%7D%7D%5C%5C%5C%5C1.509T_%7B2%7D%20%26%20%3D%20%26%201088%5C%5CT_%7B2%7D%20%26%20%3D%20%26%20%5Cdfrac%7B1088%7D%7B1.509%7D%5C%5C%5C%5C%20%26%20%3D%20%26%20%5Ctextbf%7B721K%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20gas%20must%20be%20heated%20to%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B721%20K%7D%7D%24%7D)
Question 8. I
We can use the Ideal Gas Law to solve this question.
pV = nRT
n = m/M
pV = (m/M)RT = mRT/M
a) Data:
p = 4.58 atm
V = 13.0 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 385 K
M = 46.01 g/mol
(b) Calculation
![\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7DpV%20%26%20%3D%20%26%20%5Cdfrac%7BmRT%7D%7BM%7D%5C%5C%5C%5C4.58%20%5Ctimes%2013.0%20%26%20%3D%20%26%20%5Cdfrac%7Bm%5Ctimes%200.08206%5Ctimes%20385%7D%7B46.01%7D%5C%5C%5C%5C59.54%20%26%20%3D%20%26%200.6867m%5C%5Cm%20%26%20%3D%20%26%20%5Cdfrac%7B59.54%7D%7B0.6867%20%7D%5C%5C%5C%5C%20%26%20%3D%20%26%20%5Ctextbf%7B86.7%20g%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20mass%20of%20NO%24_%7B2%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B86.7%20g%7D%7D%24%7D)