Answer:
- Dipole interactions
- London dispersion forces
- Hydrogen bonds
Credit goes to: chem.libretexts.org
Answer:
Explanation:
A) False.
Glucosidase (not calnexin nor calreticulin) helps to remove glucose residue.
Both calnexin and calreticulin rather have an affinity for last glucose residue of misfolded protein (Only misfolded proteins are marked by glycosyltransferase by attaching glucose residue). They attach with misfolded protein and with the help of other proteins like ERp57 (a type of protein disulfide isomerase) and try to fold it properly. If protein is properly folded then glucosidase removes the glucose residue thereby releasing the properly folded protein from calnexin or calreticulin. and now protein is transported to the Golgi body. If folding is still not proper then the same cycle of glycosylation -binding of calnexin/calreticulin and effort to fold it properly is repeated.
B) True.
Transketolase is a key enzyme of the pentose phosphate pathway. It contains thiamine diphosphate (TPP) as a cofactor. it does transfer 2 carbon residue from a ketose to aldose. So, effectively it converts one ketose sugar to aldose with 2 carbonless and aldose to ketose with 2 carbon more.
C) True.
Theoretically, for the evolution of one molecule of oxygen, only 8 photons are required. But in practice, it is known that there are many variants like wavelength and the energy of the photon. The larger the wavelength, like the one which is used in PS1 (more than 700nM), the lesser the energy. Secondly, the energy of the photon is also wasted as heat energy. Because of these factors, more than 8 photons are needed in reality.
D) Wrong.
Fructose 2,6 bisphosphate is a key substrate and affects both the enzymes- phosphofructokinase and fructose bisphosphatase allosterically during gluconeogenesis. It strongly favors the breakdown of glucose during glycolysis by activating phosphofructokinase but it inhibits fructose bisphosphatase. Hence it activates the kinase enzyme while inhibiting the phosphatase and maintains a huge supply of glucose in the system.
E) Wrong.
The Calvin cycle shares similarity with the pentose phosphate pathway as both are involved in the synthesis of sugar (Triose and Ribose). However, it does not share similarity with enzymes of glycolysis (which is primarily focused on the breakdown of glucose) and gluconeogenesis.
In ionic bonding, an arrow is often drawn on the diagram to show the direction the electrons move to form the ions.
Answer:
5.471% As₂O₃ in the sample.
Explanation:
<em>...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).</em>
<em />
First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:
<em>Moles KSCN = Moles Ag⁺ in the filtrate:</em>
0.01127L * (0.100mol / L)= 0.001127moles Ag⁺
<em>Total moles Ag⁺:</em>
0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺
<em>Moles of Ag⁺ in the precipitate:</em>
0.0031564 - 0.001127 = 0.0020294 moles Ag⁺
<em>Moles AsO₄ = Moles As:</em>
0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄
<em>Moles As₂O₃:</em>
6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =
3.382x10⁻⁴ moles As₂O₃
<em>Mass As₂O₃:</em>
3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃
Percent is:
0.0669g As₂O₃ / 1.223g sample * 100 =
<h3>5.471% As₂O₃ in the sample</h3>
<em />
Answer:
The correct answer is 8.79 × 10⁻² M.
Explanation:
Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,
No. of moles of NaI = Weight of NaI/ Molecular mass
= 4.11 / 149.89
= 0.027420
The vol. of the solution given is 312 ml or 0.312 L
The molarity can be determined by using the formula,
Molarity = No. of moles/ Volume of the solution in L
= 0.027420/0.312
= 0.0879 M or 8.79 × 10⁻² M
