Chemical formula of the glucose: C₆H₁₂O₆
We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u
atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol
2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g---------------------- x
x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles
we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)
3)We calculate the number of molecules:
Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles-------- x
x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.
4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²
number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .
number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²
Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.
3ZnCl2 + 2Al → 2AlCl3 + 3Zn
When palmitate (C16) completely oxidized, net yield is 129 ATP.
There are 7 β-oxidation cycles for palmitate to completely oxidazed.
One β-oxidation (beta-oxidation) cycle produces 1 NADH, 1 FADH2, and 1 acetyl CoA (sea the picture below).
For 7 cycles: 7 NADH, 7 FADH2, and 8 acetyl CoA.
NADH and FADH2 enter Electron Transport System (ETS) cycle to generate 3ATP and 2ATP respectively.
Acetyl CoA enter into tricarboxylic acid cycle (TCA) and each CoA gives 12 ATP.
7 NADH = 7 x 3 = 21 ATP
7 FADH2 = 7 x 2 = 14 ATP
8 acetyl-CoA = 8 x 12 = 96 ATP
Net yield of ATP = 21 ATP + 14 ATP + 96 ATP - 2 ATP (used during conversion of palmitic acid into palmitoyl CoA)
Net yield of ATP = 129 ATP
More about beta-oxidation: brainly.com/question/14993930
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The items are answered below and are numbered separately for each compound.
The freezing point of impure solution is calculated through the equation,
Tf = Tfw - (Kf)(m)
where Tf is the freezing point, Tfw is the freezing point of water, Kf is the freezing point constant and m is the molality. For water, Kf is equal to 1.86°C/m. In this regard, it is assumed that m as the unit of 0.25 is molarity.
1. NH4NO3
Tf = 0°C - (1.86°C/m)(0.25 M)(2) = -0.93°C
2. NiCl3
Tf = 0°C - (1.86°C/m)(0.25 M)(4) = -1.86°C
3. Al2(SO4)3
Tf = 0°C - (1.86 °C/m)(0.25 M)(5) = -2.325°C
For boiling points,
Tb = Tbw + (Kb)(m)
For water, Kb is equal to 0.51°C/m.
1. NH4NO3
Tb = 100°C + (0.51°C/m)(0.25 M)(2) = 100.255°C
2. NiCl3
Tb = 100°C + (0.51°C/m)(0.25 M)(4) = 100.51°C
3. Al2(SO4)3
Tb = 100°C + (0.51°C/m)(0.25 M)(5) = 100.6375°C
