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Artist 52 [7]
2 years ago
15

How should the ph of a 0.1 m solution of nac2h3o2 compare with that of a 0.1 m solution of kc2h3o2?

Chemistry
1 answer:
Semenov [28]2 years ago
7 0

We have that for the Question it can be said that the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)

From the question we are told

how should the ph of a 0.1 m solution of <em>nac2h3o2</em> compare with that of a 0.1 m <u>solution </u>of kc2h3o2?

Generally

with  the ph of a 0.1 m solution of <em>nac2h3o2</em> compared with that of a 0.1 m <u>solution </u>of kc2h3o2 ,we see that the salt produce  is a weak acid and strong akali salt

We see that the salt produced in water gives a base from the derived weak acid

The salt produce is CH_3COONa

Therefore

the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)

For more information on this visit

brainly.com/question/17756498

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What volume will 2.5 mol of a gas at STP occupy ?
kvv77 [185]

Answer:

56

Explanation:

1 mole of gas at STP occupies 22.4 L of the gas

2.5 mole of the gas at STP occupies 22.4×2.5 L of the gas

so 2.5 mole of the gas at STP occupies 56 L of the gas .

6 0
2 years ago
An electrochemical cell is constructed with a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver c
lana [24]

Answer:

Q = 12.38

Explanation:

The Nernst equation is given as; Ecell = E°cell - (2.303RT/nF) log Q  ;where Q is the reaction quotient.

The reaction quotient, Q  in a reaction, is the product of the concentrations of the products divided by the product of the concentrations of the reactants.

In an electrochemical cell, Q is the ratio of the concentration of the electrolyte at the anode to that of the electrolyte at the cathode.

Q = [anode]/[cathode]

therefore , Q = 0.052/0.0042 = 12.38

6 0
3 years ago
He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o
Xelga [282]

Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

Solution :  

The relation between the rate constant the activation energy is,  

\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = initial rate constant = 4.55\times 10^{-5}L/mole\text{ s}

K_2 = final rate constant = 8.75\times 10^{-3}L/mole\text{ s}

T_1 = initial temperature = 195^oC=273+195=468K

T_2 = final temperature = 258^oC=273+258=531K

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

Ea=17.285\times 10^4kJ/mole

Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

8 0
3 years ago
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