To determine the fraction of carbon in morphine, we need to know the chemical formula of morphine. From my readings, the chemical formula would be <span>C17H19NO<span>3. We assume we have 1 g of this substance. Using the molar mass, we can calculate for the moles of morphine. Then, from the formula we relate the amount of carbon in every mole of morphine. Lastly, we multiply the molar mass of carbon to obtain the mass of carbon. We calculate as follows:
1 g </span></span> <span>C17H19NO<span>3 ( 1 mol / 285.34 g ) ( 17 mol C / 1 mol </span></span> <span>C17H19NO3</span>) ( 12.01 g C / 1 mol C) = 0.7155 g C
Fraction of carbon = 0.7155 g C / 1 g <span>C17H19NO<span>3 = 0.7155</span></span>
Answer:
A) 54.04%
B) 13-karat
Explanation:
A) From the problem we have
<em>1)</em> Mg + Ms = 9.40 g
<em>2)</em> Vg + Vs = 0.675 cm³
Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.
We can rewrite the first equation using the density values:
<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40
So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:
We <u>express Vg in terms of Vs</u>:
We <u>replace the value of Vg in equation 3</u>:
- Vg * 19.3 + Vs * 10.5 = 9.40
- (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
- 13.0275 - 19.3Vs + 10.5Vs = 9.40
Now we <u>calculate Vg</u>:
- Vg + 0.412 cm³ = 0.675 cm³
We <u>calculate Mg from Vg</u>:
- 0.263 cm³ * 19.3 g/cm³ = 5.08 g
We calculate the mass percentage of gold:
- 5.08 / 9.40 * 100% = 54.04%
B)
We multiply 24 by the percentage fraction:
- 24 * 54.04/100 = 12.97-karat ≅ 13-karat
The
equation for the photosynthesis reaction in which carbon dioxide and water
react to form glucose is .
The hear reaction is the difference between the bond dissociation energies in
the products and the bond dissociation energies of the reactants
The
reactant molecules have 12 C = O, 12 H - O bonds while the product molecules
have 5 C - C, 7 C – O, 5 H – O, and 6 O = O bonds. The average bond
dissociation energies for the bonds involved in the reaction are 191 for C = O,
112 for H – O, 83 C –C, 99 C – H, 86 C – O, 119 O = O.
Substitute
the average bond dissociation energies in the equation for and
calculate as follows
=
[12 (C=O) + 12 (H-O)] – [5(C-C) + 7(C-H) + 7 (C-O) + 5(H-O) + 6(O=O)]
=
[12x191 kcal/mol + 12x112 kcal//mol] – [5x83 kcal/mol + 7x99 kcal/mol + 7x86
kcal/mol + 5x112 kcal/mol + 6x119 kcal/mol]
=
3636 kcal/mol – 2984 kcal/mol = 652 kcal/mol x 4.184 Kj/1kcal = 2.73x10^3 kJ/mol
So,
enthalpy change for the reaction is 652 kcal/mol or 2.73x10^3 kJ/mol
<span> </span>
