Answer:
Equilibrium constant of the given reaction is 
Explanation:
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The given reaction can be written as summation of the following reaction-


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Equilibrium constant of this reaction is given as-
![\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%5BBr_%7B2%7D%5D%7D)
![=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B%5BNOBr%5D%7D%7B%5BNO%5D%5BBr_%7B2%7D%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%29%5E%7B2%7D%28%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%29)


Answer:
The carbons of the acetyl group oxidize which generate CO2, and in turn H2O.
Explanation:
The pyruvic acid that is generated during glycolysis enters the mitochondria. Inside this organelle, the acid molecules undergo a process called oxidative decaborxylation in which an enzyme of several cofactors is involved, one of which is coenzyme A. Pyruvic acid is transformed into an acetyl molecule and these are been introduced to the begining of the Krebs Cycle where the acetyl-group (2C) from acetyl-CoA is transferred to oxaloacetate (4C) to produce citrate (6C). As the molecule cycles the two carbons of the acetyl oxidize and are released in the form of CO2. Then the energy of the Krebs cycle becomes sufficient to reduce three NAD +, which means that three NADH molecules are formed. Although a small portion of energy is used to generate ATP, most of it is used to reduce not only the NAD + but also the FAD which, if oxidized, passes to its reduced state, FADH2
Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :

Now put all the given values in this formula, we get the mass of solute KOH.


Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Question:
Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.
A. 0.272 mol
B. 0.04 mol
C. 0.997 mol
D. 0.139 mol
E. 0.0681 mol
Answer:
The correct option is;
E. 0.0681 mol
Explanation:
The equation for the reaction is
Zn + HCl = H₂ + ZnCl₂
Vapor pressure of the liquid = 18 torr = 2399.803 Pa
Total pressure of gas mixture H₂ + liquid vapor = 0.987 atm
= 100007.775 Pa
Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation
Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa
Volume of H₂ = 1.7 L = 0.0017 m³
Temperature = 20 °C = 293.15 K
Therefore,

Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.
for HClO, pKa = 7.54
for HNO_2, pKa = 3.15
for CH3 COOH, pKa = 4.74
Explanation:
The concentration of the solution given is 0.1 M has a pH closest to 7
The mixtures are weak acids and their salts except
HNO_3 and NaNO_3 pH = pH is near to '1'
for buffers( acidic) pH = pKa + log [salt] / [acid]
therefore [salt] = [acid] = 0.1 pH = pKa + log 0.1 / 0.1 = pKa
pH = pKa
for HClO, pKa = 7.54
for HNO_2, pKa = 3.15 therefore HClO and NaclO
mixture hs a pH closest to '7'
for CH3 COOH, pKa = 4.74