How does the electron transport chain use the high energy electrons from the krebs cycle?

A current of 5. 68 a is passed through a Fe(NO3)2 solution. How long, in hours, would this current have to be applied to plate o

daser333 [38]

There are 1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

Calculation ,

Given ; Current ( I ) = 5. 68 A

In $Fe(NO_{3} )_{2}$ , the valancy of Fe is +2 .

2 moles of $e^{-}$ are required for the decomposition of 1 mole of Fe .

7. 20 g of Fe in moles = 7. 20 g /55.845 g/mol =0.12 mole

x moles of $e^{-}$ are required for the decomposition of 0.128 mole of Fe .

moles of $e^{-}$ are required = 0.256 moles

Charge on 1 mole of $e^{-}$ = 96500 C

Charge on 0.256 mole of $e^{-}$ = 24704 C

Current ( I )= Q/t

t =Q / I = 24704 C/5. 68 A = 4349 sec = 1.2 hr

Therefore , there are 1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

To learn more about iron

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8 0

1 year ago

Which one of the following items does NOT characterize a reducing agent? a. A reducing agent loses electrons. b. A reducing agen

Ad libitum [116K]

Answer: D

Explanation:

A reducing agent is a species that reduces other compounds, and is thereby oxidized. The whole compound becomes the reducing agent. In other words, of a compound is oxidized, then they are the reducing agent. On the other hand, if the compound is reduced, it is an ozidizing agent.

Since we have established that a reducing agent is the compound being oxidized, we know that A is not our answer. An oxidized compound is losing electrons. Choice A states exactly this.

For B, this is true as we have established this already.

C is also correct. Since a reducing agent loses electrons, it becomes more positive. This makes the oxidation number increase.

D would be our correct answer. It is actually a good oxidizing agent is a metal in a high oxidation state, such as Mn⁷⁺.

4 0

2 years ago

What is the mass of 5.2 moles of ca(no3)2

zhannawk [14.2K]

Mass to moles
5.2 mol/Ca(no3)2 to mol
5.2 mol/Ca(no3)2 / molar mass
5.2 mol/Ca(no3)2 / 164.1= 0.032 g/Ca(no3)2

5 0

2 years ago

How do colloids that exist in nature affect people

mel-nik [20]

The exposure of a person to colloids majorly present in industrial area may results in the accumulation in the body tissue which causes dark deposit in the skin. They are not safe and is dangerous for health which may cause turning of skin blue permanently and it is life long.

Whereas, the colloid of silver nitrate is used to treat skin burns and ulcers but its overuse may result in adverse effect.

4 0

2 years ago

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide. Ca) + H2O --> Ca(OH)2 In a particular

mars1129 [50]

Answer:

Percent yield = 92.5%

Explanation:

The question asks for the percent yield which can be defined as:

$\frac{actual yield}{theoretical yield} .100$

Where the actual yield is <em>how much product was obtained</em>, in this case 6.11 g of Ca(OH)₂, and the theoretical yield is <em>how much product could be obtained with the given reactants theoretically</em>, that is if the reaction would work perfectly. So we need to calculate first the theoretical yield.

1. First lets write the chemical equation reaction correctly and check that it is balanced:

CaO + H₂O → Ca(OH)₂

2. Calculate the amount of product Ca(OH)₂ that can be obtained with the given reactants (theoretical yield), which are 5.00g of CaO and excess of water. So the amount of CaO will determined how much Ca(OH)₂ we can obtained.

For this we'll use the molar ratio between CaO and Ca(OH)₂ which we see it is 1:1. For every mol of CaO we'll obtain a mol of Ca(OH)₂. So lets convert the 5.00 g of CaO to moles:

Molar Mass of CaO: 40.078 + 15.999 = 56.077 g/mol

moles of CaO = 5.00 g / 56.077 g/mol = 0.08916 moles

As we said before from the molar ratio moles of Ca(OH)₂ = moles of CaO

So the moles of Ca(OH)₂ that can be obtained are 56.077 g/mol

We need to convert this value to grams:

Molar Mass of Ca(OH)₂ = 40.078 + (15.999 + 1.008)*2 = 74.092 g/mol

Theoretical yield of Ca(OH)₂ = 0.08916 moles x 74.092 g/mol = 6.606 g

3. Calculate the percent yield:

$\frac{actual yield}{theoretical yield} .100$

Percent yield = (6.11 g / 6.606g) x 100 = 92.5 %

5 0

3 years ago

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