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wolverine [178]
3 years ago
13

Since acids have 1 more proton (H+ - ions) than base, and the acid gives it away, doesn't that mean that they switch roles? Acid

becomes base, and base becomes acid...?
Thanks for clarifying .. :)
Chemistry
1 answer:
andreev551 [17]3 years ago
6 0

Answer:

In an acid-base equilibrium, acid becomes a conjugate base and base becomes a conjugate acid.

Explanation:

Let's remember the Bronsted-Lowry theory to answer this specific question. According to the theory, acid is a proton donor, while a base is a proton acceptor.

Consider an acid in a form HA (aq) and base in a form of B (aq). Since acid is a proton donor, it will donate its hydrogen ion to the base, B. The resultant products would be A^{-} (aq) and BH^{+} (aq).

Remember that an acid-base reaction is an equilibrium reaction. This means we may also look at this proton transfer reaction from the product side towards the reactants. Summarizing what has been said, we may write the equilibrium as:

HA (aq) + B (aq) ⇄ BH^{+} (aq) + A^{-} (aq)

Now acid, HA, donates a proton to become a conjugate base. The conjugate base, if we look from the reverse equation side, is actually a base, since it can accept a proton to become HA. Similarly, B accepts a proton to become a conjugate acid. Looking from the reverse reaction, it can now donate a proton, so in reality we can consider it a base.

To summarize, your logic is correct.

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A gas occupies a volume of 1.00 L at 25.0°C. What volume will the gas occupy at 1.00 x10^2 °C?
Leno4ka [110]

Answer : The volume of gas occupy at 1.00\times 10^2^oC is, 1.25 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=1.00L\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=1.00\times 10^2^oC=((1.00\times 10^2)+273)K=373K

Putting values in above equation, we get:

\frac{1.00L}{298K}=\frac{V_2}{373K}\\\\V_2=1.25L

Therefore, the volume of gas occupy at 1.00\times 10^2^oC is, 1.25 L

3 0
3 years ago
To make 340mL of a 0.4 M solution of Lici, how many mL of a 8 M solution should be used?
ioda

Answer:

Volume is the quantity of three-dimensional space enclosed by a closed surface, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic metre.

Explanation:

3 0
3 years ago
Predict what would be formed
Anarel [89]

Answer:

No

Explanation:

T6tgbv. 55678 r4fyx a

8 0
3 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
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5 0
3 years ago
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
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