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Yuliya22 [10]
2 years ago
10

A 0. 22 m aqueous solution of the weak acid ha at 25. 0 °C has a ph of 4. 15. The value of ka for Ha is?

Chemistry
1 answer:
-Dominant- [34]2 years ago
3 0

If 0. 22 m aqueous solution of the weak acid hA at 25. 0 °C has a ph of 4. 15, the value of ka for HA is 2.27 × 10^(-8).

<h3>What is an aqueous solution?</h3>

The aqueous solution is simply defined as something which has been dissolved in water. The aqueous symbol is denoted as (aq).

Given,

pH = 4.15

Concentration of HA = 0.22 M

Firstly, we will calculate the concentration of H+ ion

pH=-log [H+]

4.15 =-log [H+]

= 7.08x 10^{-5}

Secondly we will calculate the value of Ka

for HA is

The equilibrium reaction will be:

HA----- (H+) + (A-)

Given,

Concentration of H+

= Concentration of A-

= 7.08 x 10^{-5} M

The expression of Ka for HA will be,

Ka = [A-] [H+]/[ HA]

= 2.27 × 10^(-8)

Thus we concluded that the value of ka for HA for weak acid HA is 2.27 × 10^(-8).

learn more about aqueous solution or Ka:

brainly.com/question/12481995

#SPJ4

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Answer:

\huge\blue{answer}

Polar molecules occur when there is an electronegativity difference between the bonded atoms. Nonpolar molecules occur when electrons are shared equal between atoms of a diatomic molecule or when polar bonds in a larger molecule cancel each other out.

Explanation:

hope it helps u mate

5 0
3 years ago
If you are given an ideal gas with pressure (p)259,392.00 pa and temperature (T)=200°c of 1 mole Argon gas in a volume 8.8dm3,ca
GuDViN [60]

Answer: R=4.82436 \frac{Pa. m^{3}}{mol. K}

Explanation:

The Ideal Gas equation is:  

P.V=n.R.T  (1)

Where:  

P is the pressure of the gas  

n the number of moles of gas  

R=8.3144598 \frac{Pa. m^{3}}{mol. K} is the gas constant  

T is the absolute temperature of the gas in Kelvin.

V is the volume

It is important to note that the behavior of a real gas is far from that of an ideal gas, taking into account that <u>an ideal gas is a single hypothetical gas</u>. However, under specific conditions of standard temperature and pressure (T=0\°C=273.15 K and P=1 atm=101,3 kPa) one mole of real gas (especially in noble gases such as Argon) will behave like an ideal gas and the constant R will be 8.3144598 \frac{Pa. m^{3}}{mol. K}.

However, in this case we are not working with standard temperature and pressure, therefore, even if we are working with Argon, the value of R will be far from the constant of the ideal gases.

Having this clarified, let's isolate R from (1):

R=\frac{PV}{nT}  (2)

Where:

P=259392 Pa

n=1 mole

T=200\°C=473.15 K is the absolute temperature of the gas in Kelvin.

V=8.8 dm^{3}=0.0088 m^{3}

R=\frac{(259392 Pa)(0.0088 m^{3})}{(1 mole)(473.15 K)}  (3)

Finally:

R=4.82436 \frac{Pa. m^{3}}{mol. K}  

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3 years ago
Write a balanced equation for the complete oxidation reaction that occurs when glucose (C6H12O6) reacts with oxygen. Use the sma
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Answer:

C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O

Explanation:

Glucose (C₆H₁₂O₆) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).

The equation can be written as follow:

C₆H₁₂O₆ + O₂ —> CO₂ + H₂O

The above equation can be balance as illustrated below:

C₆H₁₂O₆ + O₂ —> CO₂ + H₂O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO₂ as shown below:

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There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H₂O as shown below:

C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O

There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by 6 in front of O₂ as shown below:

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Now, the equation is balanced.

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Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

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