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3 years ago

write a chemical equation for the ionization of water. how do you indicate that this is a reversible action?

Chemistry

1 answer:

LiRa [457]3 years ago

8 0

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

You use two single-barbed arrows pointing in opposite directions (technically rightwards harpoon over leftwards harpoon) to indicate a reversible reaction.

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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

train travel from hong kong to beijing it traveled at an average speed of 160 km/m in the first 4 hours. after that it traveled

Nonamiya [84]

Answer:

7 hours

Explanation: step 1. 160x4=640

step 2. 1180-640=540

step 3. 540÷180=3

step 4. 3+4=7

5 0

3 years ago

Read 2 more answers

PLEASE HELP!!!! WILL GIVE BRAINLIEST!!!!

Marina CMI [18]

Answer:

Explanation:

It afffects changes in pressure and temperature not melting and boiling points

5 0

2 years ago

What happens to the density of group 1 when you go down the group

OLga [1]

the density increases down the group.

4 0

2 years ago

A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardm

VARVARA [1.3K]

Answer: The nuclear equations for the given process is written below.

Explanation:

The chemical equation for the bombardment of neutron to U-238 isotope follows:

$_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}$

Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

The chemical equation for the first beta decay process of $_{92}^{239}\textrm{U}$ follows:

$_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta$

The chemical equation for the second beta decay process of $_{93}^{239}\textrm{Np}$ follows:

$_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta$

Hence, the nuclear equations for the given process is written above.

6 0

3 years ago