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:
Since with LiBr no precipitation takes place. So, Ag+ is absent
When we add Li2SO4 to it, precipitation takes place.
Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate
Thus, Ca2+ is present.
When Li3PO4 is added, again precipitation takes place.Reaction is:
Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate
A. Ca2+ and Co2+ are present in solution
B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)
C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)
Answer: What are you german
Explanation:
