How many atoms are in 34.2 grams of carbon?

Zingerone, one of the flavor molecules in ginger,

MissTica

The molecular formula of the compound is C12H15O3 hence the molar mass of the compound is 207 g/mol.

We need to obtain the number of moles of carbon, hydrogen and oxygen in the compound;

Carbon = 24.91 g/44g/mol × 1 mole of carbon = 0.566 moles

Mass of carbon = 0.566 moles × 12 g/mol = 6.792 g

Number of moles of hydrogen = 6.522 g/18 g/mol × 2 moles = 0.725 moles

Mass of hydrogen = 0.725 moles × 1 g/mol = 0.725 g

Mass of oxygen = 10 - (6.792 g + 0.725 g) = 2.483 g

Number of moles of oxygen = 2.483 g/16 g/mol = 0.155 moles

Now we must divide through by the lowest number of moles;

C - 0.566/0.155 H - 0.725/0.155 O - 0.155/0.155

C - 4 H - 5 O - 1

The simplest formula is C4H5O Recall that the molar mass of the compound lies between 150.0 and 220.0 g/mol

4(12) + 5(1) + 16 = 69

Hence; n = 3 and the molecular formula of the compound is C12H15O3

The molar mass of the compound is; 12(12) + 15(1) + 3(16) = 207 g/mol

Learn more: brainly.com/question/15180604

3 0

2 years ago

What is the scientific method?

AysviL [449]

C. a plane for forming a scientific hypotheses

3 0

3 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

3 years ago

Which chemical class does phenobarbital belong to?

drek231 [11]

<span>Phenobarbital is derivative of Barbituric Acid and Barbituric Acid is derivative of Urea. (structures shown in Fig below)

Urea has H</span>₂N- group attached to Carbonyl Group (C=O), and such class of comounds conataining H₂N-C=O bond are called as Amides.

Result:
So, <span>Phenobarbital belongs to Amides.</span>

3 0

3 years ago

Gather data: On the CONTROLS pane, set the Reactant concentration to 2.0 mol/L, the Surface area to Maximum, and the Temperature

Vikentia [17]

Answer:

Explanation:

We can use the Arrhenius equation to relate the activation energy and the rate constant, k, of a given reaction:

k=Ae−Ea/RT

In this equation, R is the ideal gas constant, which has a value 8.314 J/mol/K, T is temperature on the Kelvin scale, Ea is the activation energy in joules per mole, e is the constant 2.7183, and A is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.

Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term,

e−Ea/RT, is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.

6 0

3 years ago