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2 years ago

What is the empirical formula of a compound that is 66.6% c, 11.2% h, and 22.2% o by mass?

1 answer:

7 0

The empirical formula of a compound is the simplest ratio of components making up the compound.
In 100 g of compound,there's 66.6 g of C, 11.2 g of H and 22.2 g of O
lets calculate for 100 g of compound
C H O
mass 66.6 g 11.2 g 22.2 g
number of moles 66.6/12 g/mol 11.2/1 g/mol 22.2/ 16 g/mol
= 5.55 mol =11.2 mol =1.3875 mol
divide by the least number of moles
5.55/1.3875 11.2/1.3875 1.3875/1.3875
= 4 = 8.08 = 1
round them off to the nearest whole number
C - 4
H - 8
O - 1
Therefore empirical formula of compound is C₄H₈O

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lina2011 [118]

Answer: The above reaction is non-spontaneous.

Explanation:

For the given chemical reaction:

$Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)$

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

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2 years ago

A 360mg sample of aspirin, C9H8O4, (molar mass 180g), is dissolved in enough water to produce 200mL of solution. What is the mol

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Answer:

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Explanation:

First of all, let's determinate the moles of aspirin in that sample

Mass / Molar mass = Moles

360 mg = 0.360 g

0.360 g / 180 g/m = 0.002 moles

This moles that are included in 200 mL of solution, are also in 50 mL.

So molarity is mol/L

50 mL = 0.05 L

0.002 m / 0.05 L = 0.04M

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Ka = [H⁺] · [HCOO⁻] / [HCOOH].
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