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e-lub [12.9K]
2 years ago
9

Which graphic below shows the correct orientation of each water molecule when it is near the cl- ion in the aqueous solution?

Chemistry
1 answer:
andrew-mc [135]2 years ago
8 0
The question is missing the graphics required to answer which I have attached as an image.

There are four different representations of the orientation of water molecules around chloride anion. Let's first analyze the water molecule.

We have H-O-H as the structure of water. The oxygen atom is more electronegative than the hydrogen atoms, which results in a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom.

The chloride anion is a negative charge. Therefore, the water molecules should orient themselves with the hydrogen atoms facing the chlorine atom as the partial positive charge on the hydrogen atoms will be attracted to the negative charge of the chlorine atom.

The correct representation is shown in graph 3 which shows all hydrogen atoms facing the chlorine anion.

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Which statement describes the transfer of heat energy that occurs when an ice cube is added to an insulated container with 100 m
nadya68 [22]
Number 3 is the most likely answer
4 0
2 years ago
Gamma rays are often used to kill microorganisms in food, in an attempt to make the food safer. Some people contend that this ir
nikdorinn [45]

Answer:

b . Irradiated food is shown to not be radioactive.

Explanation:

If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.

  • An irradiated food is one in which ionizing radiations have been employed to improve food quality.
  • Thus, bacteria and other food spoilers can be exterminated from the food.
  • Most irradiated food do not contain radiation and are fit for consumption.

If it can be proven, that this is true, then it will challenge the idea that irradiated foods are not safe.

4 0
1 year ago
Calculate the pH of a polyprotic acid given and sketch the titration curves for the following reaction:
victus00 [196]

Answer: (1). pH = 1.70

(2). pH = 2.3

(3). pH = 3.3

(4). pH = 4.3

(5). pH = 8.41

(6). pH = 10.22

Explanation:

we assume that the formula representation of acid is H₂A

the titration curve has reasonably sharp breaks at both equivalence points, corresponding to the reactions;

H₂A + OH⁻ → HA⁻ + H₂O

HA⁻ + OH⁻ → A²⁻ + H₂O

the volume of NaOH (V₀) at the first equivalent point is,

V₀ = (20.0 mL)(0.100M) / 0.100M = 20.0mL

where volume of NaOH at 1/2 equivalent point is,

V₀/2 = 10.0mL

also Volume of NaOH at the second equivalence (2V₀) point is 40.0mL

the volume of NaOH at 1/2 second equivalent point is,

V₀ + V₀/2 = 30.0mL

Volume of NaOH after second equivalence exceeds 40mL

therefore, at 0 mL NaOH addition;

where the extent of ionization is assumed to be x, we have

                        H₂A   ⇆     HA⁻   +   H⁺

where initial:   0.1 M       -            -

          change:   -x         +x           +x

          Equili:      0.1-x      x             x

Kаl = [HA⁻][H⁺] / [H₂A]

10⁻²³ = (x)(x) / (0.1-x)

x = 0.020

[H⁺] = 0.020 M

pH = -log [H⁺]

pH = -log(0.020)

pH = 1.70

(2). at 10 mL NaOH addition

[H₂A]ini = 0.10 M * 20.0 mL = 2 mmol

[OH⁻] = 0.1 M * 10 mL = 1 mmol

after reaction:

[H₂A] = 1 mmol

[H⁻] = 1 mmol

pH = pKa₁ + log [HA⁻] / [[HA⁻]

pH = 2.3 + log 1mmol / 1mmol

pH = 2.3

(3). pH at the first equivalence point is,

pH = 1/2 (pKa₁ + pKa₂)

pH = 1/2(2.3 + 4.3) = 3.3

pH = 3.3

(4). pH at the second 1/2 equivalence point is

pH = pKa₂ = 4.3

pH = 4.3

(5). pH at the second equivalence point;

all H₂A is converted into A²⁻

[A²⁻] = initial moles of H₂A / total volume = (20.0 mL)(0.10 M) / (20.0 + 40.0) mL = 0.033 M

at equilibrium:

                   A²⁻ + H²O    ⇆   HA⁺ OH⁻

          0.033 - x

from the Kb₁ expression,

Kb₁ = [OH⁻][HA⁻] / [A²]

Kw/Ka₂ = x²/(0.0333 - x)

10⁻¹⁴/10⁻⁴³ = x²/(0.0333 - x)

x = 2.57 * 10⁻⁶

[OH⁻] = 2.57 * 10⁻⁶M

pH = -log Kw/[OH⁻] = 8.41

pH = 8.41

(6). pH after second equivalence point;

assuming the volume of NaOH is 40.10 mL

after second equivalence point OH⁻ in excess

[OH⁻] = 0.10 M * 0.10 mL / (20 + 40.10) mL = 1.66 * 10⁻⁴ M

pH = 0=-log Kw/[OH⁻] = 10.22

pH = 10.22

8 0
2 years ago
Find the volume of a gas at STP, if its volume is 80.0 mL at 109 kPa and -12.5°C.​
exis [7]

Answer:

= 913.84 mL

Explanation:

Using the combined gas laws

P1V1/T1 = P2V2/T2

At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.

V1 = 80.0 mL

P1 = 109 kPa

T1 = -12.5 + 273 = 260.5 K

P2 = 10 kPa

V2 = ?

T2 = 273 K

Therefore;

V2 = P1V1T2/P2T1

     = (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)

     <u>= 913.84 mL</u>

5 0
2 years ago
Read 2 more answers
Titanium has five common isotopes: 46Ti (8.0%), 47Ti (7.8%), 48Ti (73.4%), 49Ti (5.5%), 50 Ti (5.3%). What is the average atomic
zhannawk [14.2K]
(46x8.0)+(47x7.8)+(48x73.4)+(49x5.5)+(50x5.3) = 4792.3

4792.3/100 = 47.923 this is the average atomic mass of Titanium
6 0
2 years ago
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