Who is single lol? I'm just curious (yes ik this is app is for school karens)

Use the following information to calculate the concentration, Ka and pka for an unknown monoprotic weak acid. (8 pts.) 20.00 mL

lisov135 [29]

Answer:

Concentration: 0.185M HX

Ka = 9.836x10⁻⁶

pKa = 5.01

Explanation:

A weak acid, HX, reacts with NaOH as follows:

HX + NaOH → NaX + H2O

Where 1 mole of HX reacts with 1 mole of NaOH

To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).

18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX

In 20.0mL = 0.0200L =

0.00370 moles HX / 0.0200L = 0.185M HX

The equilibrium of HX is:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

And Ka is defined as:

Ka = [H⁺] [X⁻] / [HX]

Where [H⁺] = [X⁻] because comes from the same equilibrium

As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M

Replacing:

Ka = [H⁺] [H⁺] / [HX]

Ka = [1.349x10⁻³M]² / [0.185M]

Ka = 9.836x10⁻⁶

pKa = -log Ka

6 0

3 years ago

Protons are blank charged particles

Fittoniya [83]

Protons have a positive charge.
How I remember is Protons and Positive both start with the letter P.

7 0

3 years ago

A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the

cupoosta [38]

Answer:

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

Explanation:

Pressure in the container is P and volume is V.

Temperature of the helium gas molecules = $T_1$

Molecules helium gas = x

Moles of helium has = $n_1= \frac{x}{N_A}$

PV = nRT (Ideal gas equation)

$PV=n_1RT_1$ ...[1]

After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.

Molecules of helium left after removal = $\frac{x}{4}$

Moles of helium has left after removal = $n_2= \frac{x}{4\times N_A}$

$PV=n_2RT_2$ ...[2]

$n_1RT_1=n_2RT_2$

$\frac{x}{N_A}\times T_1=\frac{x}{4\times N_A}\times T_2$

$T_1=\frac{T_2}{4}$

$T_2=4T_1$

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

6 0

3 years ago

I’m a paragraph of no less than 125 words describe the steps you can take to effectively organize your writing

Tanya [424]

how many sentences must it be?

5 0

3 years ago

Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with

emmasim [6.3K]

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

It is a stichiometry problem.

2Na + 2H₂O → 2NaOH + H₂,

The balanced equation shows that 2.0 moles of Na metal react with 2.0 moles of water to produce 2.0 moles of NaOH and 1.0 mole of H₂,

Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: PV = nRT,

where, P is the pressure of the gas in atm (P at STP = 1.0 atm),

V is the volume of the gas in L (V = 7.80 L),

n is the number of moles in mole,

R is the general gas constant (R = 0.082 L.atm/mol),

T is the temperature of the gas in K (T at STP = 0.0 °C + 273 = 273.0 K).

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

Using cross multiplication:

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.

6 0

3 years ago

Who is single lol? I'm just curious (yes ik this is app is for school karens)​

Who is single lol? I'm just curious (yes ik this is app is for school karens)