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3 years ago

C6H12O6 + 6O2 ---> 6H2O + 6CO2

Chemistry

1 answer:

Ivahew [28]3 years ago

3 0

Answer:

24e⁻ are transferred by the reaction of respiration.

Explanation:

C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂

This is the reaction for the respiration process.

In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION

Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION

6C → 6C⁴⁺ + 24e⁻

In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻

6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²

In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.

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Digiron [165]

Answer:

your finger becomes negatively charged

Explanation:

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3 years ago

Read 2 more answers

How many moles are in 20g of potassium (K)?

otez555 [7]

Explanation:

Hi for this one u just need to remember and use the equation.

$moles = \frac{mass}{mr}$

then u find mr of potassium which is 39.1.

then u do

$\frac{20}{39.1}$

you get the answer as 0.5115 write ur answer to 3 significant figures which will be 0.512 moles .

hope this helps :)

7 0

3 years ago

someone pls help me with my chemistry test plsss my teacher changes the questions so I can't search them up. its 21 questions so

shusha [124]

Answer:

Correct option is

5 liters of CH

(g)NO

at STP

No. of molecules=

22.4

mol=

22.4

×N

molecules

A) 5ℊ of H

(g)

No. of moles=

mol=

×N

molecules

B) 5l of CH

(g)

No. of moles of CH

22.4

mol=

22.4

molecules

C) 5 mol of O

=5N

molecules

D) 5×10

molecules of CO

(g)

Molecules of 5l NO

(g) at STP=5l of CH

(g) molecules at STP

Therefore, option B is correct.

7 0

2 years ago

Reactivity and flammability are are examples of properties.

nika2105 [10]

Answer:

they are eamples of chemical properties

Explanation:

flammibility is the chemical makeup of the element that makes it flamable and reactivitie is the chemical makeup of an element that is able to reacte to another element

3 0

2 years ago

If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of

Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.1\ g}{186.04\ g/mol}$

$Moles\ of\ ferrocene= 0.0005375\ mol$

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

$\rho=\frac{Mass}{Volume}$

or,

$Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g$

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.044\ g}{78.49\ g/mol}$

$Moles\ of\ acetyl\ chloride= 0.0005606\ mol$

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms 0.0005375 mole of monoacetylferrocene

Moles of product formed = 0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

5 0

3 years ago