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Murljashka [212]

2 years ago

5

A 25.0 mL sample of 0.105 M HCl was titrated to the equivalence point with 31.5 mL of NaOH. What is the concentration of the NaO

H

1 answer:

Andrei [34K]2 years ago

5 0

Answer: The concentration of $NaOH$ is 0.08 M.

Explanation:-

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HCl$ solution = 0.105 M

$V_1$ = volume of $HCl$ solution = 25 ml

$M_2$ = molarity of $NaOH$ solution = ?

$V_2$ = volume of $NaOH$ solution = 31.5 ml

$n_1$ = valency of $HCl$ = 1

$n_2$ = valency of $NaOH$ = 1

$1\times 0.105M\times 25=1\times M_2\times 31.5$

$M_2=0.08M$

Therefore, the concentration of $NaOH$ is 0.08 M.

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2 years ago

The instruction booklet for your pressure cooker indicates that its highest setting is 11.8 psipsi . You know that standard atmo

zavuch27 [327]

<u>Answer:</u> The temperature at which the food will cook is 219.14°C

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=14.7psi\\T_1=273K\\P_2=(14.7+11.8)psi=26.5psi\\T_2=?$

Putting values in above equation, we get:

$\frac{14.7psi}{273K}=\frac{26.5psi}{T_2}\\\\T_2=\frac{26.5\times 273}{14.7}=492.14K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$492.14=T(^oC)+273\\\\T(^oC)=219.14^oC$

Hence, the temperature at which the food will cook is 219.14°C

7 0

2 years ago

When a skin diver is on the top of the water at 1.03 atm, her lung volume is 3.62L. How will the volume of her lungs change as s

Anon25 [30]

Answer:

–2.23 L

Explanation:

We'll begin by calculating the final volume. This can be obtained as follow:

Initial pressure (P₁) = 1.03 atm

Initial volume (V₁) = 3.62 L

Final pressure (P₂) = 2.68 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

1.03 × 3.62 = 2.68 × V₂

3.7286 = 2.68 × V₂

Divide both side by 2.68

V₂ = 3.7286 / 2.68

V₂ = 1.39 L

Finally, we shall determine the change in volume. This can be obtained as follow:

Initial volume (V₁) = 3.62 L

Final volume (V₂) = 1.39 L

Change in volume (ΔV) =?

ΔV = V₂ – V₁

ΔV = 1.39 – 3.62

ΔV = –2.23 L

Thus, the change in the volume of her lung is –2.23 L.

NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!

3 0

2 years ago

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Katarina [22]

False is the correct answer

4 0

2 years ago

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What is the ph of a solution labeled .30 trimethylamine k for trimethylamine is 7.42 x 10^4

valentina_108 [34]

Answer:the pH is 12

Explanation:

First We need to understand the structure of trimethylamine

Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline

C3H9N+ H2O --> C3H9NH + OH-

$k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}$

Then:

The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine

[OH-]= $\sqrt{ 0.3*7.42x10^{-4}}$

[OH-]=0.01

pH=14-(-log[OH-])

pH=12

5 0

2 years ago