Adirect measurement of the standard free‑energy change associated with the hydrolysis of atp is technically demanding because the minute amount of atp remaining at equilibrium is difficult to measure accurately. the value of δ′∘ can be calculated indirectly, however, from the equilibrium constants of two other enzymatic reactions having less favorable equilibrium constants glucose 6‑phosphate+h2o⟶glucose+pi k′eq1=270 atp+glucose⟶adp+glucose 6‑phosphate k′eq2=890 using this information for equilibrium constants determined at 25∘c, calculate the standard free energy of hydrolysis of atp.

Someone offer some answers to this, i will give 98 coins and mark as brainliest! i will put the rest of the lab down in the comments, solutions pre-lab questions: in this lab, you will make fruit drinks with powdered drink mix. complete the pre-lab questions to get the values you need for your drink solutions. calculate the molar mass of powered fruit drink mix, made from sucrose (c12h22o11).using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 m solution of 100 ml. (hint: use molarity = to find the moles of drink mix, then convert moles to grams using a mole conversion.)what mass of powdered drink mix is needed to make a 0.5 m solution of 100 ml?

Each of the following reactions is allowed to come to equilibrium and then the volume is changed as indicated. predict the effect (shift right, shift left, or no effect) of the indicated volume change. drag the appropriate items to their respective bins. co(g) + h2o(g) < => co2(g) + h2(g) (volume is decreased) pcl3(g) + cl2(g) < => pcl5(g) (volume is increased) caco3(s)< => cao(s) + co2(g) (volume is increased)

The molecular weight of carbon dioxide, co2, is 44.00 amu, and the molecular weight of nitrous dioxide, no2, is 46.01 amu, so no2 diffuses co2