We calculate the number of moles in 90 g of glucose:
180 g--------------------1 mol
90 g----------------------- x
x=(90 g * 1 mol)/180 g=1/2 moles of glucose
A molecule of glucose have 6 atoms of C ;
Therefore
1 mol of glucose have 6 moles of C
But, we have 1/2 moles of glucose therefore:
1 mol of glucose----------------6 moles of C
1/2 mol of glucose------------- x
x=(1/2 mol of glucose * 6 moles of C) / 1 mol of glucose=3 moles of C
Answer: we have 3 moles of C in 90 g of glucose.
Other method:
1)We calculate the number of moles in 90 g of glucose
180 g-------------1 mol
90 g--------------- x
x=(90 g * 1 mol)/180 g=1/2 moles=0.5 moles
2)we calculate the number of molecules in 0.5 moles of clucose.
1 mol=6.022*10²² molecules
Then:
1 mol-------------------6.022*10²³ molecules
0.5 moles------------- x
x=(0.5 moles * 6.022.10²³ molecules)/1 mol=3.011*10²³ molecules.
3) we calculate the number of atoms in 0.5 moles of glucose.
A molecule of glucose have 6 atoms of C; therefore:
the number of atoms of C =(6 atoms/1 molecule)(3.011*10²³ molecules)=
1.8066*10²⁴
4)We calculate the number of moles of C
atomic mass (C)=12 u
molar mass=12 g/1 mol
1 mol of C=6.022*10²³ atoms of C
Therefore:
1 mol--------------------------6.022.10²³ atoms of C
x--------------------------------1.8066*10²⁴ atoms of C
x=(1 mol * 1.8066.10²⁴ atoms of C) / 6.022*10²³ atoms of C=
=3
Answer: we have 3 moles of C in 90 g of glucose.
Answer:
sunlight
Explanation: i go to school
To illustrate the Lewis structure,
P has 5 valence electrons
O has 6 valence electrons (each for 4 oxygen)
And finally, for every negative charge, there is an additional valence electron
We should add these all up = 5 + 24 + 3 = 32 valence electrons
With this, we can be guided to illustrate the lewis structure as P as central atom and the 3 oxygen each with a single bond with P and 1 oxygen with a double bond with P. We place the valence electrons until octet rule is satisfied,
we will be left with 12 lone pairs for phosphate ion.
V1/T1 =V2/T2 (using charles law)
V1=6.00
V2=?
T1=273
T2=273
Making V2 the subject of formula the equation then becomes
V2= V1xT2/T1
6.00x263/273=6.0L