Answer:
<h2>
The equilibrium constant Kc for this reaction is 19.4760</h2>
Explanation:
The volume of vessel used=
ml
Initial moles of NO=
moles
Initial moles of H2=
moles
Concentration of NO at equilibrium=
M

Moles of NO at equilibrium= 
=
moles
2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)
<u>Initial</u> :1.3*10^-2 2.6*10^-2 0 0 moles
<u>Equilibrium</u>:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles
∴
⇒
![Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH2O%5D%5E2%5BN2%5D%7D%7B%5BH2%5D%5E2%5BNO%5D%5E2%7D%20%28volume%20of%20vesselin%20litre%29)
<u>Equilibrium</u>:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles
⇒
⇒
Answer:
The correct answer is 28.2 %.
Explanation:
Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,
Ptotal = PO2 + PN2 + PHe
= 55 + 90 + 50
= 195 atm
The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100
Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,
%VO2 = (55/195) * 100 = 28.2%
%VN2 = (90/195) * 100 = 46%
%VHe = (50/195) * 100 = 25.64%
Hence, the percent oxygen by volume present in the blend is 28.2 %.
The chemical equation representing the reaction between baking soda and acetic acid :

According to the Law of conservation of mass, the overall mass is conserved in a chemical reaction. That is total mass of reactants is equal to the mass of the products.
Given here Mass of Baking soda = 15 g
Mass of acetic acid = 10 g
Total mass of the reactants = 15 g +10 g = 25 g
After the reaction, total mass of products = 23 g
This difference in masses must be because one of product carbon dioxide is a gas, which is lost and hence the mass of products is lower than the mass of reactants. Therefore, the mass is conserved in this reaction.
Answer:
Each specific property of motif and domain is explained.
Explanation:
Domain;
- May retain a 3D structure when separated from rest of the protein.
- Unit of tertiary structure because alpha helix and beta sheets are units of secondary structure.
- Stable globular units like pyruvate kinase
- May be distinct functional units in a protein
Motif;
- Repetetive supersecondary structure because they contain cluster of secondary structure.
- Beta Alpha Beta unit is an example of motif
- Clusters of secondary structure
Both Motif and Domain;
- Stabilized by hydrophobic interactions like hydrogen bonding stabilize the both.
- Depends on primary structure like the arrangement of amino acid in polypeptide chain determine the secondary and tertiary structure of proteins.