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nasty-shy [4]
3 years ago
7

Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and

products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
CrO_4^2- (aq) + N_2O(g) rightarrow Cr^3+ (aq) + NO(g) [acidic]
The oxidizing agent is:_______.
The reducing agent is:_______.
Chemistry
1 answer:
N76 [4]3 years ago
8 0

Answer:

Oxidizing agent - CrO4^2-

Reducing agent- N2O

Explanation:

Let us look at the equation closely;

CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]

The reduction half equation is;

CrO4^2- (aq) + 3e -------->Cr^3+ (aq)

Oxidation half equation is;

3N2O(g) ------>3 NO(g) +3 e

Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.

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What is the pH when 10.0 mL of 0.20 M potassium hydroxide is added to 30.0 mL of 0.10 M cinnamic acid, HC9H7O2 (Ka = 3.6 × 10–5)
astra-53 [7]

Answer:-

Solution:- As is clear from the given Ka value, Cinnamic acid is a weak acid. let's calculate the moles of acid and KOH added to it from their given molarities and mL.

For KOH,  10.0mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})

= 0.002 mol

For Cinnamic acid,  30.0mL(\frac{1L}{1000mL})(\frac{0.10mol}{1L})

= 0.003 mol

Acid and base react as:

HC_9H_7O_2(aq)+KOH(aq)\rightleftharpoons KC_9H_7O_2(aq)+H_2O(l)

The reaction takes place in 1:1 mol ratio. Since the moles of acid are in excess, the acid is still remaining when all the kOH is used.

0.002 moles of KOH react with 0.002 moles of Cinnamic acid to form 0.002 moles of potassium cinnamate. Excess moles of Cinnamic acid = 0.003 - 0.002 = 0.001

As the solution have weak acid and it's salt(or we could say conjugate base), it is a buffer solution and the pH of the buffer solution could easily be calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa could be caluted from given Ka value using the formula:

pKa = - log Ka

pKa=-log3.6*10^-^5

pKa = 4.44

let's plug in the values in Handerson equation and calculate the pH:

pH=4.44+log(\frac{0.002}{0.001})

pH = 4.44+0.30

pH = 4.74

So, the first choice is correct, pH is 4.74.

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3 years ago
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In the first chemical reaction, CH₄ (g) → CH₄ (l)

Since, the reactants are in gaseous phase and the products are in liquid phase, the entropy is decreasing during the reaction.

In the second chemical reaction,

N₂H₄ (g) → N₂ (g) + 2H₂ (g)

Since, one mole of the reactant in gaseous phase is decomposing to produce three moles of product in gaseous phase, the entropy is increasing during the reaction.

In the third chemical reaction, H₂O  (s) → H₂O (g)

Since, the reactant is in solid phase and product are in gaseous phase. The entropy is increasing during the reaction.

Thus, the entropy is increasing in the 2nd and 3rd reactions.


3 0
3 years ago
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