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spayn [35]
3 years ago
12

E)¿Cuántos átomos de oxigeno hay en la molécula de glucosa, al reaccionar 12moleculas de agua?

Chemistry
1 answer:
elena55 [62]3 years ago
6 0

Answer:

12 átomos de oxígeno hay presentes

Explanation:

Basados en la reacción:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

<em>6 moles de agua producen 1 mol de glucosa</em>

<em />

Si reaccionan 12 moleculas de agua, se producirán:

12 moleculas H₂O * (1 mol C₆H₁₂O₆ / 6 mol H₂O) =

2 moléculas de glucosa se producen.

Como cada molécula de glucosa tiene 6 átomos de oxígeno:

2 moléculas C₆H₁₂O₆ * (6 átomos Oxígeno / 1 molécula C₆H₁₂O₆) =

<h3>12 átomos de oxígeno hay presentes</h3>

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When 231. mg of a certain molecular compound X are dissolved in 65. g of benzene (CH), the freezing point of the solution is mea
Elan Coil [88]

Answer:

Molar mass X = 18.2 g/mol

Explanation:

Step 1: Data given

Mass of compound X = 231 mg = 0.231 grams

Mass of benzene = 65.0 grams

The freezing point of the solution is measured to be 4.5 °C.

Freezing point of pure benzene = 5.5 °C

The freezing point constant for benzene is 5.12 °C/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C

⇒i = the Van't hoff factor = 1

⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m

⇒m = the molality = moles X / mass benzene

m = 1.0 / 5.12 °C/m

m = 0.1953 molal

Step 3: Calculate moles X

Moles X = molality * mass benzene

Moles X = 0.1953 molal * 0.065 kg

Moles X = 0.0127 moles

Step 4: Calculate molar mass X

Molar mass X = mass / moles

Molar mass X = 0.231 grams / 0.0127 moles

Molar mass X = 18.2 g/mol

5 0
3 years ago
How many neutrons does the isotope lithium have<br> A.) 3 <br> B.) 5 <br> C.) 8 <br> D.) 4
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The answer is D they have 4 neutrons
6 0
2 years ago
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What is the difference between a strong acid and a weak acid?
Jobisdone [24]

Answer:

The 3rd choice

Explanation:

A strong acid ionises COMPLETELY in water to produce high concentration of hydrogen ions

A weak acid ionises PARTIALLY in water to produce low concentration of hydrogen ions

4 0
3 years ago
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Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --
frosja888 [35]

Answer :

(a) The anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

Thus, the anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}

The overall cell reaction will be,

2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}

E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V

E^o_{[Cu^{2+}/Cu]}=0.34V

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

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

E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}

E^o=1.54V-(0.34V)=1.20V

Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

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

E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}

E_{cell}=1.022V

Therefore, the emf of cell potential is 1.022 V

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