Answer:
A)Boiling point constant of benzene = 2.63°C/m
B) 242.77 g/mol is the molar mass of the solute.
Explanation:
where,
=Elevation in boiling point
= boiling point constant od solvent= 3.63 °C/m
1 - van't Hoff factor
m = molality
A) Mas of solvent = 500 g = 0.500 kg
T = 80.1°C , =82.73°C
= 82.73°C - 80.1°C = 2.63°C
Boiling point constant of benzene = 2.63°C/m
B) Mass of solute = 1.2 g
Molar mas of solute = M
Mass of solvent = 50 g= 0.050 kg
i = 1
T = 80.1°C , =80.36°C
=80.36°C - 80.1°C = 0.26°C
M = 242.77 g/mol
242.77 g/mol is the molar mass of the solute.
Answer:
1) Conversion of glucose to glucose 6-phosphate by hexokinase
2) Conversion of fructose 6-phosphate to fructose 1,6-biphosphate by phosphofructokinase
3) Conversion of phosphoenolpyruvate to pyruvate by pyruvate kinase
Explanation:
There are 10 steps in the glycolysis pathway, three of which are irreversible. The enzymes controlling these reactions have not only catalytic properties but the irreversibility of the reaction gives them regulatory properties as well. These reactions serve as control points in the pathway.
<span>In a nuclear power plant control rods are used to:
</span><span>b. slow down the reaction.</span>
Atoms show us the basic proverb about the strength and the bond when they are Unity and Diversified.
<u>Explanation:</u>
- Every basic matter in the earth is composed of atoms. It is the smallest unit of the matter which is taken to observe the properties of the whole element.
- The atom consists of different energy levels and consist of protons electrons and neutrons.
- The atoms when are compactly arranged it result in the great strength required to bring the deformation in shape which shows that unity is always great.
- But in the liquid and gas, the atoms are arranged in a randomly dispersed pattern which shows that they can be separated and involved in any process easier to get the heterogeneous product easily which is an example for Diversity.
Answer:
976.8 g
Explanation:
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂0 ------------(1)
number of moles of reaction 1 can be shown as;
1 : 8 → 5 : 6
thus, 5 moles of CO₂ will form from the combustion of 1 mole of C₅H₁₂
molecular weight of CO₂ = 44 g/mol
molecular weight of C₅H₁₂ = 72 g/mol
mass of C₅H₁₂ = 320 g
number of moles of C₅H₁₂ = mass ÷ molecular weight = 320 / 72 = 4.44 moles
thus, number of moles of CO₂ formed = 5 * 4.44 = 22.2 moles
mass of CO₂ formed = 22.2* 44 = 976.8 g