Organize the following solvents by increasing polarity A. Dichloromethane, ethanol, ethyl acetate, diethyl ether
B. Diethyl ether, dichloromethane, ethyl acetate, ethanol
C. Ethyl acetate, ethanol, dichloromethane, diethyl ether
D. Ethanol, ethyl acetate, diethyl ether, dichloromethane

Explanation :

According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.

Or we can say that, conjugate acid is proton donor and conjugate base is proton acceptor.

The equilibrium reaction will be,

In this reaction, and are base and acid and and are conjugate acid and conjugate base respectively.

Ammonia is a weak base because it accept proton from the water and gives fewer hydroxide ions.

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

Final answer:

To determine the moles of KIO_3 titrated, use the balanced equation 2 KIO_3 + 5 Na_2S_2O_3 + 6 HCl → 3 I_2 + 6 NaCl + 6 NaClO + 3 H_2O. Therefore, 0.001551 mol of KIO_3 were titrated.

Explanation:

To determine the moles of KIO3 titrated, we need to use the balanced equation for the reaction:

2 KIO3 + 5 Na2S2O3 + 6 HCl → 3 I2 + 6 NaCl + 6 NaClO + 3 H2O

From the equation, we can see that 2 moles of KIO3 react with 5 moles of Na2S2O3. Therefore, the moles of KIO3 titrated can be calculated using the following proportion:

(0.0100 M KIO3 / 1 L) * (15.51 mL / 1000 mL) * (2 mol KIO3 / 5 mol Na2S2O3) = 0.001551 mol KIO3

Learn more about Moles of KIO_3 titrated here:

brainly.com/question/35371620

#SPJ12

The quantity of acetic acid that is needed to prepare the 500 mL buffer is 9.0075 grams.

Given the following data:

• Volume of acetate buffer = 500 mL to L = 0.5 L.

• Molarity of acetate buffer = 0.300 M.

• pH = 4.90.

• MW = 60.05 g/mol.

• pKa = 4.76.

How to calculate the mass of acetic acid.

First of all, we would write the equilibrium chemical reaction for acetate-acetic acid as follows:

Next, we would calculate HA by applying Henderson-Hasselbalch equation:

Where:

• HA is acetic acid.

•  is acetate ion.

Substituting the given parameters into the formula, we have;

For the concentration of both acids, we have:

For acetate ion:

At a volume of 0.5 liters, we have:

By stoichiometry:

Total moles = = 0.15 moles.

Mass = 9.0075 grams.

Read more on moles here: brainly.com/question/3173452

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀

1,38 = (1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid  ≡ 0,15 moles of acetic acid × = 9,0 g of acetic acid

I hope it helps!

i too used it

i thought it will help

nice time. .....

Explanation:

...................

Answer:

Mass = 42.6 g

Explanation:

Given data:

Mass of CF₂Cl₂ = 31.2 g

Mass of Cl₂ = ?

Solution:

Number of moles of CF₂Cl₂ = mass/molar mass

Number of moles =  31.2 g/121 gmol

Number of moles = 0.3  mol

1 mole of CF₂Cl₂ contain 2 moles of Cl atom.

0.3 mol × 2 = 0.6 mol

Mass of Cl₂:

Mass = number of moles × molar mass

Mass = 0.6 mol × 71 g/mol

Mass = 42.6 g

Answer:

a) Approximate boiling point of HBr = -60.15 °C

b) Approximate boiling point of AsH₃ = -52.25 °C

Explanation:

Döbereiner stated that some elements could be arranged in groups of 3 similar elements ( known as "triads) , and the element of the middle ( elements are ordered with respect to their atomic mass) would have properties between the other 2 ( the average value)

a) In the first case the triad would be the halogen triad ( Cl , Br and I ) . And according to Döbereiner , the boiling point of HBr should be the average of HCl and HI . Therefore

Approximate boiling point of HBr = [(- 84.9°C) + (-35.4°C)]/2 = -60.15 °C

b) Simmilarly for  AsH₃ , PH₃ and SbH₃ , the boiling point of AsH₃ would be

Approximate boiling point of AsH₃ = [(- 87.4°C) + (-17.1°C)]/2 = -52.25 °C