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.
Answer is: mass number of element is 234.
<span>Alpha particle is nucleus of a </span>helium-4<span> atom, which is made of two </span>protons<span> and two </span>neutrons<span>.
</span>²³⁸U → ²³⁴Th + α (alpha particle).
Alpha decay is radioactive decay<span> in which an </span>atomic nucleus<span> emits an </span>alpha particle<span> (helium nucleus) and transforms into an atom with </span>an atomic number that is reduced by two <span>and </span>mass number<span> that is reduced by four</span><span>.</span>
Answer:
In fact, it turns out that the mass of an electron is so small relative to the masses of protons and neutrons that electrons are considered to have a negligible effect on the overall mass of an atom.
Explanation:
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Answer: Thus the cell potential of an electrochemical cell is +0.28 V
Explanation:
The calculation of cell potential is done by :
Where both 
are standard reduction potentials.
![E^0_{[Fe^{2+}/Fe]}= -0.41V](https://tex.z-dn.net/?f=E%5E0_%7B%5BFe%5E%7B2%2B%7D%2FFe%5D%7D%3D%20-0.41V)
![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.
![E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-%20E%5E0_%7B%5BFe%5E%7B2%2B%7D%2FFe%5D%7D)
Thus the cell potential of an electrochemical cell is +0.28 V
