answer: A homogeneous mixture has the same uniform appearance and composition throughout. Many homogeneous mixtures are commonly referred to as solutions. A heterogeneous mixture consists of visibly different substances or phases.
My answer is....A
sorry if im wrong
Answer:
D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.
Explanation:
Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.
<u>(1) Preparatory phase</u>
During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.
When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.
<u>(2) Pay-off phase</u>
During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.
It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.
Ain't is the correct answer
Answer:
![\large \boxed{\text{-41.2 kJ/mol}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B-41.2%20kJ%2Fmol%7D%7D)
Explanation:
Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)
We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products
![\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)](https://tex.z-dn.net/?f=%5CDelta_%7B%5Ctext%7Brxn%7D%7DH%5E%7B%5Ccirc%7D%20%3D%20%5Csum%20%5Cleft%28%20%5CDelta_%7B%5Ctext%7Bf%7D%7D%20H%5E%7B%5Ccirc%7D%20%5Ctext%7Bproducts%7D%5Cright%29%20-%20%5Csum%20%5Cleft%20%28%5CDelta_%7B%5Ctext%7Bf%7D%7DH%5E%7B%5Ccirc%7D%20%5Ctext%7Breactants%7D%20%5Cright%29)
(a) Enthalpies of formation of reactants and products
![\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bcc%7D%5Ctextbf%7BSubstance%7D%20%26%20%5Ctextbf%7B%24%5CDelta_%7B%5Ctext%7Bf%7D%7D%24H%2F%28kJ%2Fmol%7D%29%20%5C%5C%5Ctext%7BCO%28g%29%7D%20%26%20-110.5%20%5C%5C%5Ctext%7BH%24_%7B2%7D%24O%7D%20%26%20-241.8%5C%5C%5Ctext%7BCO%24_%7B2%7D%24%28g%29%7D%20%26%20-393.5%20%5C%5C%5Ctext%7BH%24_%7B2%7D%24%28g%29%7D%20%26%200%20%5C%5C%5Cend%7Barray%7D)
(b) Total enthalpies of reactants and products
![\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bccr%7D%5Ctextbf%7BSubstance%7D%20%26%20%5Ctextbf%7BContribution%29%2F%28kJ%2Fmol%7D%29%26%5Ctextbf%7BSum%7D%20%5C%5C%5Ctext%7BCO%28g%29%7D%20%26%20-110.5%26%20-110.5%20%5C%5C%5Ctext%7BH%24_%7B2%7D%24O%28g%29%7D%20%26-241.8%26%20-241.8%5C%5C%5Ctextbf%7BTotal%7D%26%5Ctextbf%7Bfor%20reactants%7D%20%26%5Cmathbf%7B%20-352.3%7D%5C%5C%26%26%5C%5C%5Ctext%7BCO%7D_%7B2%7D%28g%29%20%26%20-393.5%26-393.5%20%5C%5C%5Ctext%7BH%7D_%7B2%7D%20%26%200%20%26%200%5C%5C%5Ctextbf%7BTotal%7D%26%5Ctextbf%7Bfor%20products%7D%20%26%20%5Cmathbf%7B-393.5%7D%5Cend%7Barray%7D)
(c) Enthalpy of reaction