THE MOLECULE HAS A C=C AND AN -OH GROUP, SO IT IS CALLED AN ENE/OL, I.E., AN ENOL. ENOLS CAN BE FORMED ONLY FROM CARBONYL COMPOUNDS WHICH HAVE ALPHA HYDROGENS. THEY CAN BE FORMED BY ACID OR BASE CATALYSIS, AND ONCE FORMED ARE HIGHLY REACTIVE TOWARD ELECTROPHILES, LIKE BROMINE.
Answer:
3 moles
Explanation:
To solve this problem we will use the Avogadro numbers.
The number 6.022×10²³ is called Avogadro number and it is the number of atoms, ions or molecules in one mole of substance. According to this,
1.008 g of hydrogen = 1 mole = 6.022×10²³ atoms.
18 g water = 1 mole = 6.022×10²³ molecules
we are given 36 g of C-12. So,
12 g of C-12 = 1 mole
24 g of C-12 = 2 mole
36 g of C-12 = 3 mole
So 3 moles of C-12 equals to the number of particles in 36 g of C-12.
Answer:
1.089%
Explanation:
From;
ν =1/2πc(k/meff)^1/2
Where;
ν = wave number
meff = reduced mass or effective mass
k = force constant
c= speed of light
Let
ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl
ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl
The between the two is obtained from;
ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2
Therefore;
ν' - ν /ν = [meff/m'eff]^1/2 - 1
Substituting values, we have;
ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1
ν' - ν /ν = -0.01089
percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;
ν' - ν /ν * 100
|(-0.01089)| × 100 = 1.089%
Moles of methanol = 9.27x10^24/6.02x10^23 = 15.398 moles.
Mass of methanol = moles of methanol x molar mass of methanol
= 15.398 x 32.042
= 493.38 grams.
Hope this helps!
