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1 year ago

9

Q3.4. complex i transfers electrons to q (coenzyme q) in one of the reactions in the electron transport chain. which molecule is

reduced and which is oxidized in this reaction?

1 answer:

xenn [34]1 year ago

5 0

A small molecular in reaction to electron

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In a 1 M solution of NH3(aq), identify the relative molar amounts of these species. Arrange them from most to least.

Lera25 [3.4K]

Answer:

928933

Explanation:

carbon hydrated

8 0

2 years ago

N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole fraction of N2 at that t

son4ous [18]

Answer : The mole fraction of nitrogen will be 0.4615.

Explanation : When nitrogen ( $N_{2}$ )and hydrogen ( $H_{2}$ )are mixed, the mole ratio becomes 1 : 1.5,

Now we know that ( $H_{2}$ ) is acting as a limiting agent.

So at the time of when 0.4 moles of ( $NH_{3}$ ) is been formed it requires 0.4 moles of ( $N_{2}$ ) and 3.4 moles of ( $H_{2}$ )

So, we find the the remaining ( $N_{2}$ ) will be 0.6 and

( $H_{2}$ ) will be 0.3 mole present in mixture.

So, the mole fraction of ( $N_{2}$ ) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615

8 0

2 years ago

Read 2 more answers

Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(c) $Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

$3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}$

Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

6 0

2 years ago

When a known quantity of compound, at a known concentration, is added to a known volume of another compound to determine the con

Vladimir [108]

Answer:

A titration

Explanation:

A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.

By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, <u>there are quantitative analyzes which are not titrations</u>. This is why the most appropiate answer is<em> a titration</em>.

5 0

1 year ago

Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base

Vesna [10]

Answer : The pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

Explanation :

First we have to calculate the value of $K_b$ .

As we know that,

$K_a\times K_b=K_w$

where,

$K_a$ = dissociation constant of an acid = $1.8\times 10^{-5}$

$K_b$ = dissociation constant of a base = ?

$K_w$ = dissociation constant of water = $1\times 10^{-14}$

Now put all the given values in the above expression, we get the dissociation constant of a base.

$1.8\times 10^{-5}\times K_b=1\times 10^{-14}$

$K_b=5.5\times 10^{-10}$

Now we have to calculate the concentration of hydroxide ion.

Formula used :

$[OH^-]=(K_b\times C)^{\frac{1}{2}}$

where,

C is the concentration of solution.

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

$[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}$

$[OH^-]=1.3\times 10^{-5}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.3\times 10^{-5})$

$pOH=4.9$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1$

Therefore, the pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

4 0

2 years ago