Answer:
Explanation:
In this question, we have to start with the calculation of the <u>amount of drug in each powder</u>:
Powder A: Total mass of 0.5 Kg percentage of 0.038%
Powder B: Total mass of 10 Kg percentage of 0.043%
<u>The total mass of powder would be</u>:
<u>The total mass of drug would be</u>:
Now we can calculate the <u>percentage</u>:
I hope it helps!
Answer:
Urea
Explanation:
Amino acids are the monomeric units of proteins. They are bi-functional because they possess both an amino group (-NH₂) and a carboxyl group(-COOH). First class protein are essential ingredients for healthy human diet e.g milk, cheese, fish etc. Such proteins are broken down in the body by enzymatic hydrolysis to their constituent amino acids from which they are degraded into energy which is needed for growth and from which other life function is being synthesized. The result of this effect leads to separation of amine groups which are converted into urea by the liver.
This process is achieved by the deamination of excess amino acids that leads to the formation of urea which are excreted via the kidneys.
Answer:
The answer is true. Hope this helps
Answer:
see explanation
Explanation:
a. Molar Mass = ∑Atomic Masses = 12C + 22H + 11O = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 g/mole
b. For 100ml of 0.10M solution => Molarity x Volume (liters) = moles needed x mole weight = grams solute needed for 100ml solution. This can be given by the expression ...
mass solute needed (g) = Molarity needed x Volume needed in liters x mole wt of solute = (0.10M)(0.100L)(342g/mole) = 3.42 grams solute.
mixing => Transfer 3.42 grams of solute (C₁₂H₂₂O₁₁) into mixing container and add solvent (water) up to but not to exceed 100 ml total volume.
c. mass needed for 100ml of 0.50M solution = M·V·f.wt. = (0.50M)(0.100L)(342g/mole) = 1.71 grams solute. => mix as in 'b'.
Answer:
A)glyceraldehyde 3-phosphate dehydrogenase reaction,
B)Thehexose bisphosphate that accumulates is fructose 1,6-bisphosphate
C)glyceraldehyde 3-phosphate dehydrogenase reaction to yield an acyl arsenate
Explanation:
The fermentation of ethanol in yeast has the following overall equation Glucose 2ADP 2Pi88n2 ethanol 2CO22ATP 2H2O which makes it clear that phosphate is required for the continued operation of glycolysis and formation of ethanol . In extracts to which glucose is added, fermentation proceeds until ADP and Pi(present in the extracts) are exhausted.(a)Phosphate is required in the glyceraldehyde 3-phosphate dehydrogenase reaction, and glycolysis will stop at this step when Piis exhausted. Because glucose remains, it will be phosphorylated by ATP, but Piwill not be released.(b)Fermentation in yeast cells produces ethanol and CO2rather than lactate . Without these reactions (in the absence of oxygen), NADH would accumu-late and no new NADwould be available for further glycolysis ). Thehexose bisphosphate that accumulates is fructose 1,6-bisphosphate; in terms of energet-ics, this intermediate lies at a “low point” or valley in the pathway, between the energy-input reactions that precede it and the energy-payoff reactions that follow.(c)Arsenate replaces Piin the glyceraldehyde 3-phosphate dehydrogenase reaction to yieldan acyl arsenate, which spontaneously hydrolyzes. This prevents formation of fructose1,6-bisphosphate and ATP but allows formation of 3-phosphoglycerate, which continuesthrough the pathway.