We have that the percentage of SPT (3.10 g/mL) present in a solution with a density of 2.60 g/mL is
Percentage of spt=2\%
From the question we are told
What percent of SPT (3.10 g/mL) is present in a solution with a density of 2.60 g/mL, suitable to separate quartz and feldspar? The other component of this solution is water (1.00 g/mL)
Generally the equation for the Desired SPT is mathematically given as
Where
Density of Quartz is Generally given as 2.7g/ml
Therefore
Therefore
Percentage of spt=0.02*100
Percentage of spt=2\%
For more information on this visit
brainly.com/question/1689737
Answer:
d. increases PFK activity, decreases FBPase activity
Explanation:
Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.
Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.
Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate
What is the total number of electrons needed to completely fill all of the orbitals in an atoms second principal energy level
Answer:
See Explanation
Explanation:
Number of moles of NaOH = 0.3868 L × 1.707 M = 0.660 moles
Reaction of Al^3+ and OH^-
Al^3+(aq) + 3OH^-(aq) -----> Al(OH)3(s)
If 1 mole of Al3+ reacts with 3 moles of OH^-
x moles of Al^3+ reacts with 0.660 moles of OH^-
x = 1 × 0.660/3
x= 0.22 moles
Mass of Al^3+ = 0.22 moles × 27 g/mol = 5.94 g of Al^3+
Co^2+(aq) + 2OH^-(aq) ----> Co(OH)2(s)
1 mole of Co^+ reacts with 2 moles of OH^-
x moles of Co^2+ reacts with 0.660 moles of OH^-
x = 1 × 0.660/2 = 0.33 moles
Mass of Co^2+ = 0.33 moles × 59 g/mol = 19.47 g
Answer:
C released to the environment
Explanation:
I just did it in class