Answer:
The calculated density will be larger
Explanation:
The calculated density will be <u>larger</u>. Because, the volume is taken accurately, by the water displacement method. But, when we the took the mass, the water was present on the unknown solid. So, the mass of that water was added to the original mass of the solid. Hence, the mass measured was larger than the original mass. We, know from the formula of density that density is directly proportional to the mass of the object.
Density = Mass/Volume
Hence, the larger measured mass means the larger value of density.
In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.
Glyceraldehyde 3-phosphate dehydrogenase Simultaneous oxidation and phosphorylation of G3P produce 1,3-bisphosphoglycerate (1,3-BPG) and nicotine adenine dinucleotide (NADH).
The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.
This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.
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Answer:
81 molecules
Explanation:
The reaction between C5H12 and O2 is a combustion reaction and is represented by the following equation;
C5H12 + 8O2 --> 5CO2 + 6H2O
The ratio of C5H12 to O2 from the above equation is 1 : 8.
Aplying the conditins of the question; 24 molecules each of C5H12 and O2 we have;
3C5H12 + 24O2 --> 15CO2 + 18H2O
This means we have 24 - 3 = 21 molecules of C5H12 that are unreacted.
Total molecules is given as;
3(C5H12) + 24(O2) + 15(CO2) + 18(H2O) + 21(Unreacted C5H12) = 81 molecules
